Mahesh

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B The Straight Line Solutions Exercise 3(c) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1B The Straight Line Solutions Exercise 3(c)

I.

Question 1.
Find the ratio in which the following straight lines divide the line segment joining the given points. State whether the points lie on the same side or on either side of the straight line, i) 3x – 4y = 7, (2, -7) and (-1, 3)
Solution:
3x – 4y – 7= 0

L₁₁, L₂₂ are of opposite signs. The given points lie on opposite sides of the line.

ii) 3x + 4y = 6, (2, -1) and (1, 1)
Solution:
Equation of the line is 3x + 4y – 6 = 0

The points lie on the opposite side of the line.

iii) 2x + 3y = 5, (0, 0) and (-2, 1)
Solution:
\(\frac{I}{m}=\frac{-(0+0-5)}{-4+3-5}\)
\(\frac{-5}{6}\)
The points lie on same side of the line.

Question 2.
Find,the point of intersection of the follow- ing lines.
i) 4x + 8y – 1 = 0, 2x – y + 1 = 0
Solution:
4x + 8y – 1 = 0, 2x – y + 1 = 0
point of intersection is

ii) 7x + y + 3 = 0, x + y = 0
Solution:
7x + y+ 3 = 0, x + y = 0 x y 1

Question 3.
Show that the straight lines (a – b) x + (b – c) y = c – a, (b – c)x + (c – a) y = (a – b) and (c – a)x + (a – b) y = b – c are concurrent.
Solution:
Equations of the given lines are
(a – b) x + (b – c) y = c – a …………… (1)
(b – c) x +(c – a) y = a – b …………… (2)
(c – a) x + (a – b) y = b – c …………… (3)
From (1) and (2)


Point of intersection of (1) and (2) is P (-1, -1)
Substituting in (3)
(c – a) (-1) + (a – b) (-1) = -c + a – a + b = b – c
∴ P (-1, -1) is a point on (3)
The given lines are concurrent.

Question 4.
Transform the following equation into the form L₁ + λL₂ = 0 and find the point of concurrency of the family of straight lines represented by the equation.
i) (2 + 5k) x – 3(1 + 2k)y + (2 – k) = 0
Solution:
(2 + 5k)x – 3(1 + 2k)y + (2 – k) = 0
(2x – 3y + 2) + k (5x – 6y – 1) = 0
This is of the form L₁ + λL₂ = 0
L₁ = 2x – 3y + 2 = 0
L₂ = 5x – 6y – 1 = 0

The point of concurrency is P (5, 4).

ii) (k + 1) x + (k + 2) y + 5 = 0
Solution:
(k + 1) x + (k + 2) y + 5 = 0
k (x + y) + (x + 2y + 5) = 0
i.e., (x + 2y + 5) + k (x + y) = 0
This is of the form L₁ + λL₂ = 0
∴ L₁ = x + 2y + 5 = 0
L₂ = x + y = 0

Point of concurrency is P (5, – 5).

Question 5.
Find the value of P, If the straight lines. x + p = 0, y + 2 = 0 and 3x + 2y + 5 = 0 are concurrent.
Solution:
Equations of the given lines are
x + p = 0 …………. (1)
y + 2 = 0 …………. (2)
3x + 2y + 5 = 0 …………. (3)
From (2), y = -2
Substituting in (3)
3x – 4 + 5 = 0
3x = 4 – 5 = -1
x = –\(\frac{1}{3}\)
Point of intersection of (2) and (3) is P (-\(\frac{1}{3}\), – 2)
The given lines are concurrent
P is a point on x + p = 0
–\(\frac{1}{3}\) + p = o ⇒ p = \(\frac{1}{3}\)

Question 6.
Find the area of the triangle formed by the following straight lines and the coordinate axes,
i) x – 4y + 2 = 0
Solution:
Equation of AB is
x – 4y + 2 = 0
– x + 4y = 2
\(\frac{x}{-2}+\frac{y}{\left(\frac{1}{2}\right)}=1\)
a = -2 b = \(\frac{1}{2}\)
Area of ∆OAB = \(\frac{1}{2}\)|ab|
= \(\frac{1}{2}\)|-2 × \(\frac{1}{2}\)| = \(\frac{1}{2}\) sq.units

ii) 3x – 4y + 12 = 0
Solution:
Equation of AB is
3x – 4y + 12 = 0
– 3x + 4y = 12
\(\frac{x}{-4}+\frac{y}{3}\) = 1
Area of ∆OAB = \(\frac{1}{2}\)|ab|
= \(\frac{1}{2}\)|(-4) (3)| = \(\frac{1}{2}\)(12)
= 6 sq. units

II.

Question 1.
A straight line meets the co-ordinate axes at A and B. Find the equation of the straight line, when
i) \(\overline{\mathrm{AB}}\) is divided in the ratio 2:3 at (-5, 2).

Solution:
Let OA = a and OB = b
Co-ordinates of A are (a, 0) and B are (0, b)
P divides AB in the ratio 2:3

-3x + 5y = 25
3x – 5y + 25 = 0

ii) \(\overline{\mathrm{AB}}\) is divided in the ratio 1:2 at (-5,4)
Solution:
Let OA = a and OB = b
Co-ordinates of A are (a, 0) and B are (0, b)
P divides AB in the ratio 1 : 2

– 8x + 5y = 60
8x – 5y + 60 = 0

iii) (p, q) bisects \(\overline{\mathrm{AB}}\)
Solution:
Let OA a and OB = b
Co-ordinates of A are (a, 0) and B are (0, b)

Question 2.
Find the equation of the straight line passing through the points (-1, 2) and (5, -1) and also find the area of the triangle formed by it with the axes of coordinates.
Solution:
P (-1, 2) and Q (5, -1) are the given points. Equation of
PQ is (y – y₁) (x₁ – x₂) = (x – x₁) (y₁ – y₂)
(y – 2) (-1 -5) = (x + 1) (2 + 1)
-6 (y – 2) = 3 (x+ 1)
-2y + 4 = x+ 1
x + 2y – 3 = 0

Question 3.
A triangle of area 24 sq. units is formed by a straight line and the co-ordinate axes is the first quadrant. Find the equation of the straight line, if it passes through (3, 4).

Solution:
Equation of AB in the
intercept form is \(\frac{x}{a}+\frac{y}{b}\) = 1
This line passes through P (3, 4)

Area of ∆ OAB = 24 ⇒ \(\frac{1}{2}\) |ab|= 24
\(\frac{1}{2} \frac{4 a^{2}}{a-3}=24\)
a² = 12 (a – 3)
= 12a – 36
a² – 12a + 36 = 0
(a – 6)² = 0 ⇒ a = 6
b = \(\frac{4a}{a-3}=\frac{24}{3}\) = 8
Equation of AB is \(\frac{x}{6}+\frac{y}{8}\) = 1
4x + 3y = 24
4x + 3y – 24 = 0

Question 4.
A straight line with slope 1 passes through Q(-3, 5) and meets the straight line x + y – 6 = 0 at P. Find the distance PQ.

Solution:
Given slope = 1
tan α = 1 = tan 45°
α = 45°
The line passes through Q (-3, 5)
Co-ordinates of P are (x₁ + r cos α, y₁ + r sin α)
= (-3 + r cos 45°, 5 + r sin 45°)

PQ = 2√2

Question 5.
Find the set of values of ‘a’ if the points (1, 2) and (3, 4) lie to the same side of the straight line 3x – 5y + a = 0
Solution:
P (1, 2), Q (3, 4) are the given points.
Equation of the given line is 3x – 5y + a = 0
L₁₁ = 3.1 – 5.2 + a = a – 7
L₂₂ = 3.3 – 5.4 + a = a – 11
a – 7 and a – 11 must both be positive or both negative.

Case (i) :
a – 7 > 0, a – 11 > 0
a > 7 and a > 11
∴ a > 7.11 ⇒ a∈ (11, α)

Case (ii) :
a – 7 < 0, a – 17 < 0
a < 7, a < 17
⇒ a < 17, ⇒ a∈ (-α, 27)
∴ a ∈ (-α, 7) U (11, α)

Question 6.
Show that the lines 2x + y – 3 = 0,
3x + 2y – 2 = 0 and 2x – 3y – 23 = 0 are concurrent and find the point of concurrency.
Solution:
Equations of the given lines are
2x + y – 3 = 0 ……………. (1)
3x + 2y – 2 = 0 ……………. (2)
2x – 3y – 23 = 0 ……………. (3)

From (1) and (2)

x = 4, y = – 5
Co-ordinates of P are (4, -5)
2x – 3y – 23 = 2(4) – 3(-5) – 23
= 8 + 15 – 23 = 0
P lies on (3)
The given lines are concurrent Point of concurrency is P (4, -5)

Question 7.
Find the value of p, if the following lines are concurrent.
(i) 3x + 4y = 5, 2x + 3y = 4, px + 4y = 6
(ii) 4x – 3y – 7 = 0, 2x + py + 2 = 0, 6x + 5y – 1 = 0
Solution:
(i) Equations of the lines are
3x + 4y – 5 = 0
2x + 3y – 4 = 0

x = -1, y = 2

Point of intersection of (1) and (2) is P (-1, 2)
The given lines are concurrent.
P lies on px + 4y = 6
-p + 8 = 6 ⇒ p = 8- 6 = 2

(ii) Equations of the lines
4x – 3y – 7 = 0
6x + 5y- 1 = 0

x = 1, y = -1
Co-ordinates of P are (1, -1)
The given lines are concurrent.
P is a point on
2x + py + 2 = 0
2 – p + 2 = 0
p = 4

Question 8.
Determine whether or not the four straight lines with equations
x + 2y – 3 = 0, 3x + 4y – 7 = 0,
2x + 3y – 4 = 0 and 4x + 5y – 6 = 0 are concurrent.
Solution:
Equations of the given lines are
x + 2y – 3 = 0 ………. (1)
3x + 4y – 7 = 0 ………. (2)
2x + 3y – 4 = 0 ………. (3)
4x + 5y – 6 = 0 ………. (4)
Solving (1) and (2)

x = 1, y = 1
Point of intersection of (1) and (2) is P (1, 1)
2x + 3y – 4 = 2.1 + 3.1 – 4 = 5 – 4 = 1 ≠ 0
4x + 5y- 6 = 4.1 + 5.1 – 6 = 9 – 6 = 3 ≠ 0
∴ P (1,1) is not a point on (3) and (4)
∴ The given lines are not concurrent.

Question 9.
If 3a + 2b + 4c = 0, then show that the equation ax + by + c = 0 represents a family of concurrent straight lines and find the point of concurrency.
Solution:
Given condition is 3a + 2b + 4c = 0
\(\frac{3}{4}\)a + \(\frac{1}{2}\)b + c = 0
For all values of a, b the line ax + by + c = 0
passes through the point (\(\frac{3}{4}\), \(\frac{1}{2}\))
The equation ax + by + c = 0 represents a family of concurrent lines.
Point of concurrency is (\(\frac{3}{4}\), \(\frac{1}{2}\))

Question 10.
If non-zero numbers a, b, c are in harmonic progression then show that the equation \(\frac{x}{a}+\frac{y}{b}+\frac{1}{c}\) = o represents a family of concurrent lines and find the point of concurrency.
Solution:
Given a, b, c are in H.P.
∴ \(\frac{2}{b}=\frac{1}{a}+\frac{1}{c}\)
\(\frac{1}{a}+\frac{(-2)}{b}+\frac{1}{c}\) = o
∴ For all values of a,b,c the equation \(\frac{x}{a}+\frac{y}{b}+\frac{1}{c}\) = o
represents a line passing through the point (1, -2)
∴ The equation, \(\frac{x}{a}+\frac{y}{b}+\frac{1}{c}\) = o represents a family of concurrent lines.
Point of concurrency is P (1, – 2)

III.

Question 1.
Find the point on the straight lines
3x + y 4- 4 = 0 which is equidistant from the points (-5, 6) and (3, 2).
Solution:
P(x₁, y₁) is the point on 3x + y + 4 = 0
3x₁ + y₁ + 4 = 0 …………. (1)
Given PA = PB ⇒ PA² = PB²

(x₁+ 5)² + (y₁ – 6)² = (x₁ – 3)² + (y₁ – 2)²
x²₁ + 10x₁ + 25 + y²₁ – 12y₁ + 36 = x²₁ -6x₁ + 9 + y²₁ – 4y₁ + 4
16x₁ – 8y₁ + 48 = 0
2x₁ – y₁+ 6 = 0 ……….. (2)
3x₁ + y₁ +4 = 0 ………… (1)
Adding 5x₁ + 10 = 0 ⇒ x₁ = – 2
From (1) – 6 + y₁ + 4 = 0
y₁ = 6 – 4 = 2
Co-ordinates of P are (-2, 2)

Question 2.
A straight line through P (3, 4) makes an angle of 60° with the positive direction of the X-axes. Find the co-ordinates of the points on that line which are 5 units away from P.
Solution:
Co-ordinates of any point on the line Q are (x₁ + r cos θ, y₁ + r sin θ)
Given (x₁, y₁) = (3, 4) i.e., x₁ = 3, y₁ = 4
θ = 60° ⇒ cos θ = cos 60° = \(\frac{1}{2}\), sin θ = sin 60° \(\frac{\sqrt{3}}{2}\)

Question 3.
A straight line through Q (√3, 2) makes an angle \(\frac{\pi}{6}\) with the positive direction of the X-axis. If the straight line intersects the line √3x – 4y + 8 = 0 at P; find the distance of PQ.
Note : AB and PQ are not perpendicular.
So we have to follow the first method only.
Solution:

PQ makes an angle \(\frac{\pi}{6}\) with the positive direction of X – axis.
m = Slope of PQ = tan 30° = \(\frac{1}{\sqrt{3}}\)
PQ passes through Q (√3, 2)
Equation of PQ is y – 2 = \(\frac{1}{\sqrt{3}}\)(x – √3)
√3y – 2√3 = x – √3
x – √3y = – √3 ………… (1)
Equation of AB is √3x – 4y + 8 = 0
√3x – 4y = -8 ……………. (2)
(1) × √3 ⇒ √3x – 3y = -3
Subtracting -y = -5
y = 5
From (1), x = √3y – √3
= 5√3 – √3 = 4√3
Equation of AB is
Co-ordinates of P are ( 4√3, 5)
Q are (√3, 2)
PQ²= (4√3 – √3)² +(5 – 2)²
= 27 + 9 = 36
PQ =6 units.

Question 4.
Show that the origin is with in the triangle whose angular points are (2,1), (3,-2) and (-4,-1).
Solution:
The vertices of the ∆ le ABC are
A = (2, 1), B = (3, -2), C = (-4, -1)

The equation of BC is

⇒ 7y + 7 = -x -4
L = x + 7y + 11 = 0 …………. (1)
The equation of CA is

⇒ 3y + 3 = x + 4
⇒ L’ = x – 3y + 1 = 0 ………. (2)

⇒ 3x – 6 = -y + 1
L” = – 3x + y – 7 = 0 …………… (3)
L” (- 4, -1)= 3 (- 4) – 1 – 7
= – 20 is negative.
L” (0,0) = 3(0) + 0 – 7
= – 7 is negative
So (- 4, -1), (0,0) lie on the same side of AB
Hence O (0,0) lies to the left of AB ………… (4)
L’ (3, -2) = 3 – 3 (- 2) + 1
= 10 is positive

L” (0, 0) =0 – 3 (0) + 1
= 1 is positive
So (0, 0), (3, -2) lie on the same side of AC
both down of AC ………… (5)
L (2, 1) = 2 + 7(1) + 11
= 20 is positive

L (0, 0) =0 + 7(0)+ 11
= 11 is positive
So (0, 0) and (2, 1) lie on the same side of BC.
So (0, 0) lie upwards from BC ……… (6)

From (4), (5), (6) it follows O (0,0) lies down-wards of AC, upwards of BC, and to the left of AB. So O (0, 0) will lie inside the ∆ ABC.

Question 5.
A straight line through Q(2, 3) makes an angle \(\frac{3 \pi}{4}\) with negative direction of the X – axis. If the straight line intersects the line x+ y – 7 = 0 at P, find the distance PQ.

Solution:
The line PQ makes an angle \(\frac{3 \pi}{4}\) with the negative direction of X – axis i.e., PQ makes and angle π – \(\frac{3 \pi}{4}\) = \(\frac{\pi}{4}\) with the positive direction of X – axis.
Co-ordinates of Q are (2, 3)
Co-ordinates of P are (x₁ + r cos θ, y₁ + r sin θ)

PQ = r = √2 units.

Question 6.
Show that the straight lines x + y = 0, 3x + y – 4 = 0 and x + 3y – 4 = 0 form an isosceles triangle.
Solution:
Given lines
x + y = 0 ………….. (1)
3x + y – 4 = 0 ………….. (2)
x + 3y – 4 = 0 ………….. (3)
Let θ₁ be the angle between (1), (2)

Let θ₂ be the angle between (2), (3)

Let 0, be the angle between (3), (1)

Since θ₁ = θ₃
∴ The ∆^le formed by the lines is an Isosceles triangle.

Question 7.
Find the area of the triangle formed by the straight lines 2x – y – 5 = 0, x – 5y + 11 = 0 and x + y – 1 = 0.
Solution:
Given lines
2x – y – 5 = 0 ………….. (1)
x – 5y + 11 =0 …………. (2)
x + y – 1 = 0 ………….. (3)
By solving (1), (2)

Let A = (4, 3)
By solving (2), (3)

∴ B = (- 1,2)
By solving (3), (1)

∴ C = (2, -1)

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B The Straight Line Solutions Exercise 3(b) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1B The Straight Line Solutions Exercise 3(b)

I.

Question 1.
Find the sum of the squares of the inter¬cepts of the line 4x – 3y = 12 on the axes, of co-ordinates.
Solution:
Equation of the given line is
\(\frac{4 x}{12}-\frac{3 y}{12}=1\)
\(\frac{x}{3}+\frac{y}{-4}=1\)
a = 3, b = -4
Sum of the squares = a² + b²
= 9 + 16 = 25

Question 2.
If the portion of a straight line inter¬cepted between the axes of co-ordinates is bisected at (2p, 2q), write the equation of the straight line.
Solution:
Equation of AB in the intercept form is
\(\frac{x}{a}+\frac{y}{b}=1\) …………… (1)

Co-ordinates of A are (a, 0) and B are (0, b)
M is the mid-point of AB
Co-ordinates of M are (\(\frac{a}{2}\), \(\frac{b}{2}\)) = (2p, 2q)
\(\frac{a}{2}\) = 2p, \(\frac{b}{2}\) = 2q
a = 4p, b = 4q
Substituting in (1), equation of AB is
\(\frac{x}{4 P}+\frac{y}{4 Q}=1\)
\(\frac{x}{P}+\frac{y}{Q}=4\)

Question 3.
If the linear equation ax + by + c = 0
(a,b,c ≠ 0) and lx + my + n = 0
represent the same line and r = \(\frac{l}{a}\) = \(\frac{n}{c}\)
write the value of r in terms of m and b.
Solution:
ax + by + c — 0 and
lx + my + n = 0 represent the same line
∴ \(\frac{1}{a}=\frac{m}{b}=\frac{n}{c}=r\)
\(\frac{m}{b}\) = r

Question 4.
Find the angle made by the straight line y = – √3x + 3 with the positive direction of the X-axis mea-sured in the counter clock-wise direction.
Solution:
Equation of the given line isy = -√3x + 3 Suppose a is the angle made by this line with positive X – axis in the counter clock – wise direction.
tan α = – √3 = tan \(\frac{2 \pi}{3}\)
α = \(\frac{2 \pi}{3}\)

Question 5.
The intercepts of a straight line on the axes of co-ordinates are a and b. If p is the length of the perpendicular drawn from the origin to this line. Write the value of p in terms of a and b.
Solution:
Equation of the line in the intercept form is

p = length of the perpendicular from origin

II.

Question 1.
In what follows, p denotes the distance of the straight line from the origin and a denotes the angle made by the normal ray drawn from the origin to the straight line with \(\stackrel{\leftrightarrow}{O X}\) measured in the anticlockwise sense. Find the equations of the straight lines with the following values of p and a.
i) p = 5, a = 60°
ii) p = 6, a = 150°
iii) p = 1, α = \(\frac{7 \pi}{4}\)
iv) p = 4, α = 90°
v) p = 0, α = 0
vi) p = 2√2, α = \(\frac{5 \pi}{4}\)
Solution:
Equation of the line in the normal form is x cos α + y sin α = p

i) p = 5, α = 60°
cos α = cos 60° = \(\frac{1}{2}\)
sin α = sin 60° = \(\frac{\sqrt{3}}{2}\)
Equation of the line is x. \(\frac{1}{2}\) + y. \(\frac{\sqrt{3}}{2}\) = 5
⇒ x + √3y= 10

ii) p = 6, α = 150°
cos α = cos 150° = cos (180° – 30°)
= -cos 30° = – \(\frac{\sqrt{3}}{2}\)
sin α = sin 150°
= sin (180° – 30°)
= sin 30° = \(\frac{1}{2}\)
Equation of the line is
x.(-\(\frac{\sqrt{3}}{2}\)) + y.\(\frac{1}{2}\) = 6
-√3x + y = 12
or √3x – y + 12 = 0

iii) p = 1, α = \(\frac{7 \pi}{4}\)
cos α = cos 315° = cos (360° – 45°)
= cos 45° = \(\frac{1}{\sqrt{2}}\)
sin α = sin 315° = sin (360° – 45°)
= -sin45° = \(\frac{1}{\sqrt{2}}\)
Equation of the line is
x.\(\frac{1}{\sqrt{2}}\) – y.\(\frac{1}{\sqrt{2}}\) = 1
x – y = √2
x – y – √2 = 0

iv) p = 4, α = 90°
cos α = cos 90° = 0, sin α = sin 90° = 1
Equation of the line is
x.0 + y.1 = 4
y = 4

v) p = 0, α = 0
cos α = cos 0 = 1, sin α = sin 0 = 0
Equation of the line is
x.1 + y.0 = 0
x = 0

vi) p = 2√2 , α = \(\frac{5 \pi}{4}\)
cos α = cos 225° = cos (180° + 45°)
= -cos 45° = – \(\frac{1}{\sqrt{2}}\)
sin α = sin 225° = sin (180° + 45°)
=-sin 45° = – \(\frac{1}{\sqrt{2}}\)
Equation of the line is
x(-\(\frac{1}{\sqrt{2}}\)) + y(-\(\frac{1}{\sqrt{2}}\)) = 2√2
– x – y = 4
or x + y + 4 = 0

Question 2.
Find the equations of the straight line in the symmetric form, given the slope and a point on the line in each part of the question.
i) √3, (2, 3)
ii) –\(\frac{1}{\sqrt{3}}\), (-2, 0)
iii) -1, (1,1)
Solution:
i) Equation of the line in the symmetric form is
\(\frac{x-x_{1}}{\cos \alpha}=\frac{y-y_{1}}{\sin \alpha}=r\)
(x₁, y₁) = (2, 3)
m = tan α = √3 ⇒ α = 60°
cos α = cos 60° = \(\frac{1}{2}\)
sin α = sm 60° = \(\frac{\sqrt{3}}{2}\)
Equation of the line in symmetric form is
\(\frac{x-2}{\cos \frac{\pi}{3}}=\frac{y-3}{\sin \frac{\pi}{3}}\)

ii) (x₁, y₁) = (-2, 0)
tan α = –\(\frac{1}{\sqrt{3}}\) ⇒ α = 180° – 30° = 150°
Equation of the line is \(\frac{x+2}{\cos 150^{\circ}}=\frac{y}{\sin 150^{\circ}}\)

iii) tan α = -1, α =180°- 45° = 135°
(x₁, y₁) = (1, 1)
Equation of the line is \(\frac{x-1}{\cos \left(\frac{3 \pi}{4}\right)}=\frac{y-1}{\sin \left(\frac{3 \pi}{4}\right)}\)

Question 3.
Transform the following equation into
a) Slope-intercept form
b) Intercept form and
c) Normal form
i) 3x + 4y = 5
ii) 4x – 3y + 12 = 0
iii) √3x + y = 4
iv) x + y + 2 = 0
v) x + y – 2 = 0
vi) √3x + y + 10 = 0
Solution:
i) 3x + 4y = 5
Slope-intercept form
4y = -3x + 5
\(y=\left(-\frac{3}{4}\right) x+\left(\frac{5}{4}\right)\)
Intercept form :
3x + 4y = 5

x cos α + y sin α = 1

ii) 4x – 3y + 12 = 0
Slope-intercept form :
3y = 4x + 12
y = (\(\frac{4}{3}\))x + 4
Intercept form:
4x – 3y+ 12 = 0
-4x + 3y = 12
\(\frac{-4 x}{12}+\frac{3 y}{12}=1\)
\(\frac{x}{(-3)}+\frac{y}{4}=1\)
Normal form :
4x – 3y + 12 = 0
– 4x + 3y = 12

x cos α + y sin α = \(\frac{12}{5}\)

iii) √3x + y = 4
Slope-intercept form :
√3x + y = 4
y = -√3x + 4
Intercept form :
√3x + y = 4

iv) x + y + 2 = 0
Slope-intercept form
x + y+.2 = 0
y = -x – 2
= (-1)x + (-2)

Intercept form:
x + y + 2 = 0
-x – y = 2
\(\frac{x}{(-2)}+\frac{y}{(-2)}=1\)
Normal form:
x + y + 2 = 0
-x – y = 2

v) x + y – 2 = 0
Slope-intercept form :
x + y – 2 = 0
y = – x + 2
Intercept form:
x + y – 2 = 0
x + y = 2
\(\frac{x}{2}+\frac{y}{2}\) = 1
Normal form :
x + y-2 = 0
x + y = 2

vi) √3x + y + 10 = 0
Slope-intercept form :
√3x + y + 10 = 0
y = -√3x – 10
Intercept form :
√3x + y = -10
Normal form :
√3x + y = -10
Dividing with \(\sqrt{3+1}=2\), we get
\(\frac{-\sqrt{3}}{2} \cdot x+\frac{-1}{2} \cdot y=5\)
x cos 30° + y sin 30° = 5

Question 4.
If the product of the intercepts made by the straight line x tan α + y sec α = 1 (0 ≤ α < \(\frac{\pi}{2}\)), on the co-ordinate axes is equal to sin α, find α.
Solution:
Equation of the line is x tan α + y sec α = 1
\(\frac{x}{\cot \alpha}+\frac{y}{\cos \alpha}=1\)
a = cot α, b = cos α
Given ab = sin α
cot α. cos α = sin α
\(\frac{\cos ^{2} \alpha}{\sin \alpha}\) = sin α ⇒ cos² α = sin α
tan² α = 1 ⇒ tan α = ±1
α = 45°

Question 5.
If the sum of the reciprocals of the intercepts made by a variable straight line on the axes of co-ordinate is a constant, then prove that the line always passes through a fixed point.
Solution:
Equation of the line in the intercept form is
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\) ………….. (1)
Sum of the reciprocals of the intercepts

The line (1) passes through the fixed point
\(\left(\frac{1}{k}, \frac{1}{k}, \frac{1}{k}\right)\)

Question 6.
Line L has intercepts a and b on the axes of coordinates. When the axes are rota¬ted through a given angle, keeping the origin fixed, the same line L has intercepts p and q on the transformed axes. Prove that \(\frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{P^{2}}+\frac{1}{Q^{2}}\).
Solution:
Equation of the line in the old system in the intercept form is
\(\frac{x}{a}+\frac{y}{b}=1 \Rightarrow \frac{x}{a}+\frac{y}{b}-1=0\)
Length of the perpendicular from origin

Equation of the line in the second system in the intercept form is
\(\frac{x}{p}+\frac{y}{Q}=1 \Rightarrow \frac{x}{p}+\frac{y}{Q}-1=0\)
Length of the perpendicular from origin

Since the origin and the given line remain unchanged we have from (1) and (2)

Question 7.
Transform the equation \(\frac{x}{a}+\frac{y}{b}=1\) into the normal form when a > 0 and b > 0. If the perpendicular distance of the straight line from the origin is p, deduce that \(\frac{1}{p^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}\).
Solution:

III.

Question 1.
A straight line passing through A(-2, 1) makes an angle 30° with \(\overrightarrow{O X}\) in the positive direction. Find the points on the straight line whose distance from A is 4 units.
Solution:
Co-ordinates of any point on the given line are (x₁ + r cos α, y₁ + r sin α)
α = 30° ⇒ cos α = cos 30° = \(\frac{\sqrt{3}}{2}\),
sin α = sin = 30° = \(\frac{1}{2}\)
(x₁, y₁) = (-2, 1) ⇒ x₁ = -2 y₁ = 1
Taking r = 4 ⇒ co-ordinates of P are

Question 2.
Find the points on the line 3x – 4y – 1 = 0 which are at a distance of 5 units from the point (3, 2).
Solution:
Equation of the line in the symmetric form is
\(\frac{x-3}{\cos \alpha}=\frac{y-2}{\sin \alpha}=r\)
Co-ordinates of the point P are
(3 + r cos α, 2 + r sin α) = (3 + 5 cos α, 2 + 5 sin α)
P is a point on 3x – 4y – 1 = 0
3(3 + 5 cos α) – 4(2 + 5 sin α) -1 = 0
9 + 15 cos α – 8 – 20 sin α – 1 = 0
15 cos α – 20 sin α = 0
15 cos α = + 20 sin α
tan α = + T
Case i) : cos α = +\(\frac{4}{5}\), sin α \(\frac{3}{5}\)
Case ii) : cos α = –\(\frac{4}{5}\), sin α \(\frac{3}{5}\)

Case i) : Co-ordinates of P are
\(\left(3+5 \cdot \frac{4}{5}, 2+5 \cdot \frac{3}{5}\right)=(7,5)\)

Case ii) : Co-ordinates of P are
\(\left(3-5 \cdot \frac{4}{5}, 2-5 \cdot \frac{3}{5}\right)=(-1,-1)\)

Question 3.
A straight line whose inclination with the positive direction of the X-axis measured in the antidock wise sense is it/3 makes positive intercept on the Y-axis. If the straight line is at a distance of 4 from the origin, find its equation.
Solution:
Given α = π/3, p = 4
m = tan α = tan 60° = √3
Equation of the line in the slope – intercept form is
y = √3 x + c
√3x – y + c = 0
Distance from the origin = 4
\(\frac{|0-0+c|}{\sqrt{3+1}}=4\)
|c| = 4 × 2 = 8
c = ± 8
Given c > 0 ∴ c = 8
Equation of the line is √3x – y + 8 = 0

Question 4.
A straight line L is drawn through the point A (2, 1) such that its point of intersection with the straight line x + y = 9 is at a distance of 3√2 from A. Find the angle which the line L makes With the positive direction of the X – axis.
Solution:
Suppose a is the angle made by L with the positive X – axis
Any point on the line is
(x₁ + r cos α₁, y₁ + r sin α) = (2 + 3√2 cos α 1 + 3√2 sin α)
This is a point on the line x + y = 9
2+ 3√2 cos α+ 1 + 3√2 sin α = 9
3√2 (cos α + sin α) = 6
cos α + sm α = \(\frac{6}{3 \sqrt{2}}\) = √2
\(\frac{1}{\sqrt{2}}\). cos α + \(\frac{1}{\sqrt{2}}\) sin α = 1
cos α. cos 45° + sin α. sin 45° = 1
cos (α – 45°) = cos 0°
α – 45° = 0 ⇒ α = 45° = \(\frac{\pi}{3}\)

Question 5.
A straight line L with negative slope passes through the point (8, 2) and cuts positive co-ordinate axes at the points P and Q. Find the minimum value of OP + OQ as L varies, when O is the origin.
Solution:
Equation of the line passing through A(8, 2) with negative slope ‘-m’ is y – 2 = -m(x – 8)
mx + y – (2 + 8m) = 0
mx + y = 2 + 8m

OP =X Intercept = \(\frac{2+8 m}{m}\)
OQ = Y – Intercept = 2 + 8m 2 + 8m

For f(m) to have minimum or maximum, we must have f'(m) = 0

and ‘f’ has minimum at m = \(\frac{1}{2}\)
∴ Minimum value of f(m) = Minimum Value I of OP + OQ at m = \(\frac{1}{2}\)

∴ Minimum value of OP + OQ as L varies, where ‘O’ is the origin is 18.

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B The Straight Line Solutions Exercise 3(a) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1B The Straight Line Solutions Exercise 3(a)

I.

Question 1.
Find the slope of the line x + y = 0 and x – y = 0.
Solution:
Slope of x + y = 0 is – \(\frac{a}{b}\) = -1
Slope of x – y = 0 is 1

Question 2.
Find the equation of the line containing the points (2, -3) and (0, -3).
Solution:
Equation of the line is
(y – y₁) (x₁ – x₂) = (x – x₁) (y₁ – y₂)
(y + 3)(2 – 0) = (x – 2)(-3 + 3)
2(y + 3) = 0
⇒ y + 3 = 0

Question 3.
Find the equation of the line containing the points (1, 2) and (1, -2).
Solution:
Equation of the line is
(y – y₁) (x₁ – x₂) = (x – x₁) (y₁ – y₂)
(y – 2)(1 – 1) = (x – 1) (2 + 2)
0 = 4(x – 1) ⇒ x – 1 =0

Question 4.
Find the angle which the straight line y = √3x – 4 makes with the Y-axis.
Solution:
Equation of the line is y = √3x – 4
Slope = m = √3 = tan \(\frac{\pi}{6}\)
Angle made with X-axis = \(\frac{\pi}{6}\)
Angle made with Y – axis = \(\frac{\pi}{2}-\frac{\pi}{6}=\frac{\pi}{3}\)

Question 5.
Write the equation of the reflection of the line x = 1 in the Y-axis.
Solution:

Equation of PQ is x = 1
Reflection about Y – axis is x = – 1
i.e., x + 1 = 0

Question 6.
Find the condition for the points (a, 0), (h, k) and (0, b) when ab ≠ 0 to be collinear.
Solution:
A(a, 0), B(h, k), C(0, b) are collinear.
⇒ Slope of AB = Slope of AC
\(\frac{k-0}{h-a}=\frac{-b}{a}\)
ak = -bh + ab
bh + ak = ab
\(\frac{h}{a}+\frac{k}{b}=1\)

Question 7.
Write the equations of the straight lines parallel to X-axis is
i) at a distance of 3 units above the X-axis and ii)at a distance of 4 units below the X-axis.
Solution:
I)

Equation of the required line AB is y = 3

ii)

Equation of A’B’ is y = — 4 ; y + 4 = 0

8. Write the equations of the straight line parallel to Y – axis and
i) at a distance of 2 units from the Y-axis to the right of it.
ii) at a distance of 5 units from the Y-axis to the left of it.
Solution:
i)

Equation of the required line AB is x = 2

ii)

Equation of the required line A’B’
x = -5
x + 5 = 0

II.

Question 1.
Find the slopes of the straight line passing through the following pairs of points.
i) (-3, 8) (10, 5)
ii) (3, 4) (7, -6)
iii) (8,1), (-1, 7)
iv) (-P, q) (q, -p) (pq ≠ 0)
Solution:

Question 2.
Find the value of x, if the slope of the line passing through (2, 5) and (x, 3) is 2.
Solution:
Slope = \(\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{5-3}{2-x}=2\)
2 = 2(2 – x)
x = 2 – 1 = 1

Question 3.
Find the value of y if the line joining the points (3, y) and (2, 7) is parallel to the line joining the points (-1, 4) and (0,6).
Solution:
A(3, y), B(2, 7), P(-1, 4) and Q(0, 6) are the given points.
m₁ = Slope of AB = \(\frac{y-7}{3-2}\) = y – 7
m₂ = Slope of PQ = \(\frac{4-6}{-1-0}=\frac{-2}{-1}=2\)

AB and PQ are parallel
m₁ = m₂ ⇒ y – 7 = 2
y = 2 + 7 = 9

Question 4.
Find the slopes of the lines i) parallel to and ii) perpendicular to the line passing through (6, 3) and (- 4,5).
Solution:
A(6, 3) and B(-4, 5) are the given points.
m = Slope of AB = \(\frac{3-5}{6+4}=\frac{-2}{10}=-\frac{1}{5}\)

PQ is parallel to AB
(i) Slope of PQ = m = – \(\frac{1}{5}\)
RS is perpendicular to AB

(ii)Slope of RS = – \(\frac{1}{m}\) =5 m

Question 5.
Find the equation of the straight line which makes the following angles with the positive X-axis in the positive direction and which pass through the points given below
i) \(\frac{\pi}{4}\) and (0,0)
ii) \(\frac{\pi}{3}\) and(1, 2)
iii) 135° and (3, -2)
iv) 150° and (-2, -1)
Solution:
i) m = Slope = tan 45° = 1
Equation of the line is y – y₁ = m(x – x₁)
y – 0 = 1(x – 0)
i.e., y = x
or x – y = 0

ii) m = tan 60° – √3
Equation of the line is
y – 2 = √3 (x – 1)
= √3x – √3
√3x – y+(2 – √3) = 0

iii) m = tan 135° = tan (180° -45°)
= – tan 45° = -1
Equation of the line is y + 2 = -1(x – 3)
= – x + 3
i.e., x + y – 1 = 0

iv) m = tan 150° = tan (180° – 30°)
= – tan 30° = – \(\frac{1}{\sqrt{3}}\)
Equation of the line is
y + 1 = – \(\frac{1}{\sqrt{3}}\) (x + 2)
√3y + √3 = – x – 2
x + √3y + (2 + √3) = 0

Question 6.
Find the equations of the straight lines passing through the origin and making equal angles with the co-ordinate axes.
Solution:
Case (i) : PP1 makes an angle 45° with positive X-axis
m = tan 45° = 1
PP’ passes through 0(0, 0)
Equation of PP’ is y – 0 = 1 (x – 0)

Case ii) : QQ’ makes an angle 135° with positive X-axis
m = tan 135° = tan (180° – 45°) = -tan 45°
Equation of QQ’ is y – 0 = -1 (x – 0)
y = -x

Question 7.
The angle made by a straight line with the positive X-axis in the positive direction and the Y-intercept cut off by it are given below. Find the equation of the straight line.
i) 60°, 3
ii) 150°, 2
iii) 45°, -2
iv) Tan^-1(\(\frac{2}{3}\)), 3
Solution:
i) Equation of the line is y = mx + c
m = tan 60° = √3, c = 3
Equation of the line is y = √3x + 3
√3x – y + 3 = 0

ii) m = tan 150° = tan (180°-30°)
= – tan 30° = \(\frac{-1}{\sqrt{3}}\), c = 2
Equation of the line is y =- \(\frac{1}{\sqrt{3}}\)x + 2
√3 y = -x + 2 √3x
x + √3y – 2 – √3 =0

iii) m = tan 45° = 1
c = -2
Equation of the line is
y = x – 2
x – y – 2=0

iv) θ = tan^-1(\(\frac{2}{3}\)) ⇒ m = tan θ = \(\frac{2}{3}\),c = 3
Equatidn of the line is y = \(\frac{2}{3}\) x + 3
3y = 2x + 9
2x – 3y + 9 = 0

Question 8.
Find the equation of the straight line passing through (-4, 5) and cutting off equal and non-zero intercepts on the coordinate axes.
Solution:
Equation of the line in the intercept form is
\(\frac{x}{a}+\frac{y}{b}\) = 1
Given a = b
Equation of the line is \(\frac{x}{a}+\frac{y}{b}\) = 1
⇒ x + y = a
This line passes through P(- 4, 5)
-4 + 5 = a ⇒ a = 1
Equation of the required line is x + y = 1 or x + y – 1 = 0

Question 9.
Find the equation of the straight line passing through (-2, 4) and making non¬zero intercepts whose sum is zero.
Solution:
Equation of the line in the intercept form is
\(\frac{x}{a}+\frac{y}{b}\) = 1
Given a + b = 0 ⇒ b = -a
Equation of the line is \(\frac{x}{a}-\frac{y}{b}\)
⇒ x – y = a
This line passes through P(-2,4)
∴ -2 – 4 = a ⇒ a = -6
Equation of the required line is x – y = -6
⇒ x – y + 6 = 0

III.

Question 1.
Find the equation of the straight line passing through the point (3, -4) and making X and Y-intercepts which are in the ratio 2 : 3.
Solution:
Equation of the line in the intercept form is x y
\(\frac{x}{a}+\frac{y}{b}\) = 1
Given \(\frac{a}{b}\) = \(\frac{2}{3}\) ⇒ b = \(\frac{3a}{2}\)
Equation of the line is \(\frac{x}{a}+\frac{2 y}{3 a}=1\)
⇒ 3x + 2y = 3a
This line passes through P(3, – 4)
9 – 8 = 3a ⇒ 3a = 1

Equation of the required line is 3x + 2y = 1
⇒ 3x + 2y – 1 = 0

Question 2.
Find the equation of the straight line passing through the point (4, -3) and perpendicular to the line passing through the points (1, 1) and (2, 3).
Solution:
A(1, 1), B(2, 3) are the given points.
m = Slope of AB = \(\frac{1-3}{1-2}=\frac{-2}{-1}=2\)

PQ is perpendicular to AB
Slope of PQ = –\(\frac{1}{m}\) = – \(\frac{1}{2}\)
PQ passes through P(4, -3)
Equation of PQ is y – y₁ = m(x – x₁)
y + 3 = –\(\frac{1}{2}\)(x – 4)
2y + 6 = -x + 4 ⇒ x + 2y + 2 = 0

Question 3.
Show that the following sets of points are collinear and find the equation of the line L containing them.
i) (-5, 1), (5, 5), (10, 7)
ii) (1, 3), (-2, – 6), (2, 6)
iii) (a, b + c), (b, c + a), (c, a + b)
Solution:
i) A(-5, 1), B(5, 5), C(10, 7) are the given points.
Equation of AB is
(y – y₁) (x₁ – x₂) = (x – x₁) (y₁ – y₂)
(y- 1) (-5 – 5) = (x + 5) (1 – 5)
– 10y + 10 = -4x – 20
4x – 10y + 30 = 0
or 2x – 5y + 15 = 0

C(10, 7)
2x – 5y + 15 = 2.10 – 5.7 + 15
= 20 – 35 + 15 = 0

A, B, C are collinear.
Equation of the line containing them is 2x – 5y + 15 = 0

ii) A(1, 3), B(-2, -6), C(2, 6)
Equation of AB is
(y – 3) (1 + 2) = (x – 1) (3 + 6)
3(y – 3) = 9(x – 1)
y – 3 = 3x – 3
3x – y = 0

C(2, 6)
3x – y = 3.2 – 6 = 6 – 6 = 0

∴ The given points A, B, C are collinear.
Equation of the line containing A,B,C is 3x – y = 0

iii) A(a, b f c), B(b, c + a), C(c, a + b)
Equation of AB is
(y – (b + c)) (a-b) = (x – a)(b + c – c – a)
(y – b – c) (a – b) = -(a – b) (x – a)
y – b – c = -x + a
or x + y – (a + b + c) = 0
C (c, a + b)
c + a + b – a – b – c = 0
C lies on AB
A, B, C are collinear.
Equation of the line containing them is x + y = a + b + c

Question 4.
A(10, 4), B(-4, 9) and C(-2, -1) are the vertices of a triangle. Find the equations of
i) \(\stackrel{\leftrightarrow}{A B}\)
ii) the median through A
iii) the altitude through B
iv) the perpendicular bisector the side of \(\stackrel{\leftrightarrow}{A B}\)
Solution:
i) A(10,4), B(-4, 9) are the given points.
Equation of AB is
(y – 4) (10 + 4) = (x – 10) (4 – 9)
14y – 56 = -5x + 50
5x + 14y – 106 = 0

ii) D is the mid-point of BC

A (10,4) is the other vertex Equation of AD is
(y – 4) (10 + 3) = (x + 3) (4 – 4)
13(y – 4) = 0 ⇒ y – 4 = 0 (or) y = 4

iii)

Slope of AC = \(\frac{4+1}{10+2}=\frac{5}{12}\)
BE is perpendicular to AC
Slope of BE = \(\frac{-1}{m}=\frac{-12}{5}\)

BE passes through B(-4, 9)
Equation of the altitude BE is
y – 9 = \(\frac{-12}{5}\)(x + 4)
5y – 45 = -12x – 48
12x + 5y + 3 = 0

iv) O is the mid-point of AB

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Transformation of Axes Solutions Exercise 2(a) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1B Transformation of Axes Solutions Exercise 2(a)

I.

Question 1.
When the origin is shifted to (4, -5) by the translation of axes, find the coordinates of the following points with reference to new axes.
i) (0, 3), ii) (-2, 4) iii) (4, -5)
Solution:
i) New origin = (4, -5); h = 4, k = -5
Old co-ordinates are (0, 3)
x = 0, y = 3
x’ = x – h = 0 – 4 = -4
y’=y-k = 3 + 5 = 8
New co-ordinates are (—4, 8)

ii) Old co-ordinates are (-2, 4)
x = -2, y = 4
x’ = x- h = -2 – 4 = -6
y’=y-k=4 + 5 = 9
New co-ordinates are (-6, 9)

iii) Old co-ordinates are (4, -5)
x = 4, y = -5
x’ = x – h = 4- 4 = 0
y = y – k = -5 + 5 = 0
New co-ordinates are (0,0)

Question 2.
The origin is shifted to (2, 3) by the translation of axes. If the coordinates of a point P changes as follows, find the coordinates of P in the original system.
i) (4, 5) ii) (-4, 3), iii) (0, 0)
Solution:
i) New co-ordinates are (4, 5)
x’ = 4, y’ = 5
x = x’ + h = 4 + 2 = 6
y = y’ + k = 5 + 3 = 8
Old co-ordinates are (6, 8)

ii) New co-ordinates are (-4, 3)
x’ = – 4, y’ = 3
x = x’ + h = -4 + 2 = -2
y = y’ + k = 3 + 3 = 6
Old co-ordinates are (-2, 6)

iii) New co-ordinates are (0, 0)
x’ = 0, y’ = 0
x = x’ + h = 0 + 2 = 2
y = y’ + k = 0 + 3 = 3
Old co-ordinates are (2, 3)

Question 3.
Find the, point to which the origin is to be shifted so that the point (3, 0) may change to (2, -3).
Solution:
(x, y) = (3, 0)
(x’, y’) = (2, -3)
Let (h, k) be the shifting origin.
h = x – x’= 3- 2 = 1
k = y – y’ = 0 + 3 = 3
∴ (h, k) = (1, 3)

Question 4.
When the origin is shifted to (-1, 2) by the translation of axes, find the transformed equations of the following.
i) x² + y² + 2x – 4y.+ 1 = 0
ii) 2x² + y² – 4x + 4y = 0
Solution:
i) The given equation is
x² + y² + 2x – 4y + 1 = 0
Origin is shifted to (-1, 2)
h = -1, k = 2
Equation of transformations are
x = x’ + h, y = y’ + k
i.e., x = x’ – 1, y = y’ + 2
The new equation is
(x’ – 1)² + (y’ + 2)² + 2(x’ – 1) – 4(y’ + 2) + 1 = 0
⇒ (x’)² + 1 – 2x’ + (y’)² + 4 + 4y’ + 2x’ – 2 -4y’ – 8 + 1 = 0
(x’)² + (y’)² -4 = 0
The transformed equation is x² + y² – 4 = 0

ii) Old equation is
2x² + y² – 4x + 4y = 0
New equation is 2(x’ – 1)² + (y’ + 2)² —4(x’ – 1) + 4(y’ + 2) = 0
2[(x’)² + 1 – 2x’] + (y’)² + 4 + 4y’ – 4x’ + 4 + 4y’ + 8 = 0
2(x’)² + 2 – 4x’ + (y’)² + 4 + 4y’ – 4x’ + 4 + 4y’ + 8 = 0
2(x’)² + (y’)² – 8x’ + 8y’ + 18 = 0
The transformed equation is
2x² + y² – 8x + 8y+18 = 0

Question 5.
The point to which the origin is shifted and the transformed equation are given below. Find the original equation.
i) (3,-4);x² + y² = 4
ii) (-1, 2); x² + 2y² + 16 = 0
Solution:
i) Given shifting origin = (3, – 4) = (h, k)
x’ = x – h,
= x – 3

y’ = y – k
= y + 4

The original equation of (x’)² + (y’)² = 4 is
(x – 3)² + (y + 4)2 = 4
x² – 6x + 9 + y² + 8y + 16 = 4
x² + y² – 6x + 8y + 21 =0

ii) Given shifting origin = (h, k) = (-1,2)
x’ = x – h,
= x + 1

y’ = y – k
y = y – 2

The original equation of
x’² + 2y’² + 16 = 0 is
(x + 1)² + 2(y – 2)² + 16 = 0
x² + 2x + 1 + 2y² – 8y + 8 + 16 = 0
x² + 2y² + 2x – 8y + 25 = 0

Question 6.
Find the point to which the origin is to be shifted so as to remove the first degree terms from the equation.
4x² + 9y² – 8x + 36y + 4 = 0
Solution:
The given equation is
4x² + 9y² – 8x + 36y + 4 = 0
a = 4 g = -4
b = 9 f = 18

Origin should be shifted to (1, -2)

Question 7.
When the axes are rotated through an angle 30°, find the new coordinates of the following points,
i) (0, 5) ii) (-2, 4) hi) (0, 0)
Solution:
i) Given 0 = 30°
Old co-ordinates are (0, 5)
i.e., x = 0, y = 5
x‘ = x. cos θ + y. sin θ
= 0. cos 30° + 5. sin 30° = \(\frac{5}{2}\)
= – x sin θ + y cos θ
= – 0. sin 30° + 5 cos 30° =
New co-ordinates are \(\left(\frac{5}{2}, \frac{5 \sqrt{3}}{2}\right)\)

ii) Old co-ordinates are (-2, 4)
x= -2, y = 4
x’ = x cos θ + y sin θ
= (-2). cos 30°+ 4. sin 30°
= -2. \(\frac{\sqrt{3}}{2}\) + 4. \(\frac{1}{2}\) = – √3 +2
y’ = -x sin θ + y cos θ
= – (-2) sin 30° + 4 cos 30°
= 2 . \(\frac{1}{2}\) + 4. \(\frac{\sqrt{3}}{2}\)
= 1 + 2 √3
New co-ordinates are (- √3 + 2, 1 + 2√3)

iii) Given (x, y) = (0,0) and 0 = 30°
x = (0, y) ⇒ x = x’. cos 30° – y’ sin 30°
= 0. \(\frac{\sqrt{3}}{2}\) – 0. \(\frac{1}{2}\) =0
y = x’. sin 30° + y’.cos 30°
= 0.\(\frac{1}{2}\) + 0.\(\frac{\sqrt{3}}{2}\) = 0
New co-ordinates of the point are (0, 0)

Question 8.
When the axes are rotated through an angle 60°, the new co-ordinates of three points are the following
i) (3, 4) ii) (-7, 2) iii) (2, 0) Find their original coordinates.
Solution:
i) Given 0 = 60°
New co-ordinates are (3, 4)
x’ = 3, y’ = 4
x = x’ cos θ – y’ sin θ
= 3. cos 60° – 4. sin 60°
\(=3 \cdot \frac{1}{2}-\frac{4 \cdot \sqrt{3}}{2}=\frac{3-4 \sqrt{3}}{2}\)
y = x’ sin θ + y’ cos θ
= 3 sin 60° + 4. cos 60°

ii) New co-ordinates are (-7, 2)
x’= -7, y’ = 2
x = x’ cos θ – y’ sin θ
= (-7) cos 60° – 2. sin 60°
\(=-7 \cdot \frac{1}{2}-2 \cdot \frac{\sqrt{3}}{2}=\frac{-7-2 \sqrt{3}}{2}\)
y = x’ sin θ + y’. cos θ
= – 7. sin 60° + 2. cos 60°

iii) New co-ordinates are (2, 0)
x’ = 2, y’ = 0
x = x’ cos θ – y’ sin θ
= 2. cos 60° – 0. sin 60°
= 2.\(\frac{1}{2}\) – 0.\(\frac{\sqrt{3}}{2}\) =1 – 0 = 1
y = x’ sin θ + y’ cos θ
= 2. sin 60° + 0. cos 60°
= 2.\(\frac{\sqrt{3}}{2}\) + 0.\(\frac{1}{2}\) = √3
Co-ordinates of R are (1, √3)

Question 9.
Find the angle through which the axes are to be rotated so as to remove the xy term in the equation.
x² + 4xy + y² – 2x + 2y – 6 = 0.
Solution:
Compare the equation
x² + 4xy + y² – 2x + 2y – 6 = 0
with ax² + 2hxy + by² + 2gx + 2fy + c = 0
a = 1, h = 2, b = 1, g = -1, f = 1, c = -6
Let ‘θ’ be the angle of rotation of axes, then

II.

Question 1.
When the origin is shifted to the point (2, 3), the transformed equation of a curve is x² + 3xy – 2y² + 17x – 7y – 11= 0. Find the original equation of the curve.
Solution:
Equations of transformation are
x = x’ + h, y = y’ + k
x’ = x – h = x – 2, y’ = y – 3
Transformed equation is
x² + 3xy – 2y² + 17x – 7y – 11 = 0
Original equation is
(x – 2)² + 3(x – 2) (y – 3) – 2(y – 3)² + 17(x – 2) – 7(y – 3) – 11 = 0
⇒ x² – 4x + 4 + 3xy – 9x – 6y + 18 – 2y² + 12y – 18 + 17x – 34 – 7y + 21 -11 = 0
⇒ x² + 3xy – 2y² + 4x – y – 20 = 0
This is the required original equation.

Question 2.
When the axes are rotated through an angle 45°, the transformed equation of acurveis 17x² – 16xy + 17y² = 225. Find the original equation of the curve.
Solution:
Angle of rotation = θ = 45
x’ = x cos θ + y sin θ = x cos 45 + y sin 45 = \(\frac{x+y}{\sqrt{2}}\)
y’ = – x sin θ + y cos θ = – x sin 45 + y cos 45 = \(\frac{-x+y}{\sqrt{2}}\)

The original equation of
17x’²- 16x’y’ + 17y’² = 225 is

⇒ 17x² + 17y² + 34xy – 16y² + 16x² + 17x² + 17y² – 34xy = 450
⇒ 50x² + 18y² = 450
∴ x² + y² = 9 is the original equation.

Question 3.
When the axes are rotated through an angle a, find the transformed equation of x cos a + y sin a = p. iMM&mmn
Solution:
The given equation is x cos α + y sin α = p
∵ The axes are rotated through an angle α
x = x’ cos α – y’ sin α
y = x’ sin α + y’ cos α

The given equation transformed to
(x’ cos α – y’ sin α) cos α +
(x’ sin α + y’ cos α) sin α = p
⇒ x’ (cos² α + sin² α) = p
⇒ x’ = p
The equation transformed to x = p

Question 4.
When the axes are rotated through an angle n/6. Find the transformed equation of x² + 2 √3xy – y² = 2a².
Solution:

⇒ 3X² – 2√3 XY + Y² + 2√3|√3X² +2XY – √3Y²|- (x² +3Y² +2√3XY) =8a²
⇒ 3X² -2V3XY +Y² +6X² + W3XY – 6Y² – X² – 3Y² – 2√3 XY = 8a²
⇒ 8X² – 8Y² = 8a² ⇒ X² – Y² = a²
The transformed equation is x² – y² = a²

Question 5.
When the axes are rotated through an angle \(\frac{\pi}{4}\), find the transformed equation of 3x² + 10xy + 3y² = 9.
Solution:
Given equation is
3x² + 10xy + 3y² – 9 = 0 ………….. (1)
Angle of rotation of axes = θ = \(\frac{\pi}{4}\)
Let (X, Y) be the new co-ordinates of (x, y)
x = X cos θ – Y sin θ

Transformed equation of (1) is

⇒ 3X² – 6XY + 3Y² + 10X² – 10Y² + 3X² + 6XY + 3Y² – 18 = 0
∴ 16X² – 4Y² -18 = 0
∴ 8X² – 2Y² = 9
∴ 8X² – 2Y² = 9
The transformed equation is 8x² – 2y² = 9

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Locus Solutions Exercise 1(a) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1B Locus Solutions Exercise 1(a)

I.

Question 1.
Find the equation of the locus of a point that is at a distance 5 from A (4, – 3).
Solution:
A (4, -3) is the given point P(x, y) is any point on the locus.

Given condition is CP = 5
CP² = 25
(x – 4)² + (y + 3)² = 25
x² – 8x + 16 + y² + 6y + 9 – 25 = 0
Equation of the locus of P is x²+ y² – 8x + 6y = 0.

Question 2.
Find the equation of locus of a point which is equidistant from the points A(-3, 2) and B(0, 4).
Solution:
A (-3, 2), B (0,4) are the given points
P (x, y) is any point on the locus

Given condition is PA = PB
PA² = PB²
(x+3)² + (y – 2)² = (x – 0)² + (y – 4)²
x² + 6x + 9 + y² – 4y + 4 = x² + y² – 8y +16
6x + 4y = 3 is the equation of the locus.

Question 3.
Find the equation of locus of a point P such that the distance of P from the origin is twice the distance of P from A (1, 2).
Solution:
O(0,0), A (1,2) are the given points
P (x, y) is any point on the locus

Given condition is OP = 2AP
OP² = 4 AP²
x² + y² = 4 [(x – 1)² + (y – 2)²]
= 4 (x² – 2x + 1 + y² – 4y + 4)
x² + y² = 4x² + 4y² – 8x – 16y + 20
Equation to the locus of P is
3x² + 3y² – 8x – 16y + 20 = 0.

Question 4.
Find the equation of locus of a point which is equidistant from the coordi¬nate axes.
Solution:
P(x, y) is any point on the locus.
PM and PN are perpendiculars from P on X and Y – axes.

Given PM = PN ⇒ PM² = PN²
y² = x²
Locus of P is x² – y² = 0

Question 5.
Find the equation of locus of a point equidistant from A (2, 0) and the Y- axis.
Solution:
A (2,0) is the given point.
P (x, y) is any point on the locus.

Draw PN perpendicular to Y – axis.
Given condition is PA = PN
PA² = PN²
(x – 2)² + (y – 0)² = x²
x² – 4x + 4 + y² = x²
Locus of P is y² – 4x + 4 = 0

Question 6.
Find the equation of locus of a point P, the square of whose distance from the origin is 4 times its y coordinate.

Solution:
P(x, y) is any point on the locus
Given condition is OP² = 4y ⇒ x² + y² = 4y
Equation of the locus of P is x² + y² – 4y = 0

Question 7.
Find the equation of locus of a point ‘P’ such that PA² + PB² = 2c², where A = (a, 0), B = (-a, 0) and 0 < |a| < |c|.
Solution:
Let P (x, y) be a point in locus.
A = (a, 0); B = (-a, 0)
Given condition is PA² + PB² = 2c²
(x – a)² + (y – 0)² + (x + a)² + (y – 0)² = 2c²
x² – 2ax + a² + y² + x² + 2ax + a² + y² = 2c²
2x² + 2y² = 2c² – 2a²
∴ x² + y² = c² – a² is the locus.

II.

Question 1.
Find the equation of locus of P, if the line segment joining (2, 3) and (-1, 5) subtends a right angle at P.
Solution:
A(2,3), B (-1, 5) are the given points.
P(x, y) is any point on the locus.

Given condition is, ∠APB = 90°
AP² + PB² = AB²
(x – 2)² + (y – 3)² + (x + 1)² + (y – 5)² = (2 + 1)² + (3 – 5)²
x² – 4x + 4 + y² – 6y + 9 + x² + 2x + 1+ y² – 10y + 25 = 9 + 4
2x² + 2y² – 2x – 16y + 26 = 0
Locus of P is x² + y² – x – 8y + 13 = 0
(x, y) ≠ (2,3) and (x, y) ≠ (-1, 5)

Question 2.
The ends of the hypotenuse of a right angled triangle are (0, 6) and (6, 0). Find the equation of locus of its third vertex.
Solution:
A(0,6), B (6,0) are the ends of the hypotenuse,
P (x, y) is the third vertex.
∴ The given condition is ∠APB = 90°

AP² + PB² = AB²
(x – 0)² + (y – 6)² + (x – 6)² + (y – 0)² = (6 – 0)² + (0 – 6)²
x² + y² – 12y + 36 + x² – 12x + 36 + y² = 36 + 36
2x² + 2y² -12x -12y = 0
Locus of P is x² + y² – 6x – 6y = 0
(x, y) ≠ (0,6) and (x, y) ≠ (6,0).

Question 3.
Find the equation of the locus of a point, the difference of whose distances from (-5, 0) and (5, 0) is 8.
Solution:
A(5,0), B(-5,0) are the given points.

P(x, y) is any point on the locus.
Given condition is |PA – PB| = 8
PA – PB = 8 ………..(1)
PA² – PB² = [(x – 5)² + (y – 0)²] – [(x + 5)² + (y – 0)²]
= x² – 10x + 25 + y² – x² – 10x – 25 – y²
= – 20x
(PA + PB) (PA – PB) = -20x
(PA + PB) 8 = -20x
PA + PB = –\(\frac{5}{2}\)x …………..(2)
Adding (1) and (2),
2PA = –\(\frac{5x}{2}\) + 8 = \(\frac{-5 x+16}{2}\)
4PA = – 5x + 16
16PA² = (- 5x + 16)²
16 [(x – 5)² + y²] = (- 5x + 16)²
16 [x² – 10x + 25 + y²] = [- 5x + 16]²
16x² + 16y²2 – 160x + 400 = 25x² + 256 – 160x
9x² – 16y² = 144
Dividing with 144, locus of P is
\(\frac{9 x^{2}}{144}-\frac{16 y^{2}}{144}=1\) i.e., \(\frac{x^{2}}{16}-\frac{y^{2}}{9}=1\)

Question 4.
Find the equation of locus Of P, if A=(4,0), B = (- 4,0) and |PA- PB| =4.
Solution:
A = (4, 0), B = (-4, 0) are the given points.
P (x, y) is any point on the locus.
The given condition is |PA – PB| = 4 …………… (1)
PA² – PB²= [(x- 4)² + (y – 0)²] – [(x + 4)² + y²]
= x² – 8x+ 16 + y² – x² – 8x – 16 – y²
= -16x

(PA + PB) (PA – PB) = -16x
(PA + PB) 4 = – 16x
PA + PB = – 4x ………….. (2)
Adding (1) and (2),
2PA = 4 – 4x
PA = 2 – 2x
PA² = (2 – 2x)²
(x – 4)² + (y – 0)² = (2 – 2x)²
x² – 8x + 16 + y² = 4 + 4x² – 8x
3x² – y² = 12
Dividing with 12, locus of P is \(\frac{3 x^{2}}{12}-\frac{y^{2}}{12}=1\)
i.e., \(\frac{x^{2}}{4}-\frac{y^{2}}{13}=1\)

Question 5.
Find the equation of the locus of a point, the sum of whose distances from (0, 2) and (0, -2) is 6.
Solution:
A (0,2), B (0, -2) are the given points.

P(x, y) is any point on the locus.
Given condition is PA + PB = 6 ……….. (1)
PA² – PB² = [(x – 0)² + (y – 2)²] – [(x – 0)² + (y + 2)²]
= x² + y² – 4y + 4 – x² – y² – 4y – 4 = – 8y
(PA + PB) (PA – PB) = -8y
6(PA – PB) = – 8y
PA – PB = –\(\frac{8y}{6}\)
PA-PB = –\(\frac{4y}{3}\) ……. (2)
Adding (1) and (2), 2PA = 6 –\(\frac{4y}{3}\)

9x² + 9y² + 36 = 81 + 4y²
9x² + 5y² = 45
Dividing with 45,

Question 6.
Find the equation (rf the locus of P, if A= (2,3), B = (2, -3) and PA + PB = 8.
Solution:
A (2, 3), B (2, -3) are the given points.

P(x, y) is any point on the locus.
Given condition is PA + PB = 8 ……….. (1)
PA² – PB² = [(x – 2)² + (y – 3)²] – [(x – 2)² + (y + 3)²]
= (x – 2)² + (y – 3)² – (x – 2)² – (y + 3)² = (y – 3)² – (y + 3)² = -12y
(PA + PB) (PA – PB) = -12y
8(PA – PB) = -12y
PA – PB = \(\frac{-12 y}{8}\)
PA – PB = \(\frac{-3 y}{2}\) ……….. (2)
Adding (1) and (2),
2PA = 8 – \(\frac{3 y}{2}\) = \(\frac{16 -3y}{2}\)
4PA = 16 – 3y
16PA² = (16 – 3y)²
16 [(x – 2)² + (y – 3)²] = (16 – 3y)²
16(x² – 4x + 4 + y² – 6y + 9) = (16 – 3y)²
16x² + 16y² – 64x – 96y + 208 = 256 + 9y² – 96y
16x² + 7y² – 64x – 48 = 0
Locus of P is 16x² + 7y² – 64x – 48 = 0.

Question 7.
A(5, 3) and B (3, -2) are two fixed points. Find the equation of the locus of P, so that the area of triangle PAB is 9.
Solution:
A(5, 3), B(3, -2) are the given points.
P(x, y) is any point on the locus.

Given condition is ∆PAB = 9
\(\frac{1}{2}\)|5(-2, -y) + 3(y – 3) + x(3 + 2)|=9
|-10 – 5y + 3y – 9 + 5x| = 18
5x – 2y – 19 = ± 18
5x – 2y – 19 = 18 or 5x – 2y – 19 = – 18
5x – 2y – 37 = 0 or 5x – 2y – 1 = 0
Locus of P is (5x- 2y- 37) (5x- 2y- 1) = 0

Question 8.
Find the equation of the locus of a point, which forms a triangle of ar5a 2 with the points A(l, 1) and B (-2, 3).
Solution:
A (1, 1), B(-2, 3) are the given points.

P(x, y) is any point on the locus.
Given condition is ∆PAB = 2
\(\frac{1}{2}\)|1(3 – y) – 2 (y – 1) + x (1 – 3)| = 2
|3 – y – 2y + 2 – 2x| = 4
-2x – 3y + 5 = ± 4
-2x – 3y + 5 = 4 or – 2x – 3y + 5 = – 4
2x + 3y – 1 = 0 or 2x + 3y – 9 = 0
Locus of P is (2x + 3y – 1) (2x + 3y – 9) = 0.

Question 9.
If the distance from ‘P’ to the points (2, 3) and (2, -3) are in the ratio 2 : 3, then find the equation of locus of P.
Solution:
Let P (x, y) be a point on locus.
Given points A = (2, 3), B = (2, -3)
Given condition is
PA : PB = 2 : 3
⇒ 3PA = 2PB
⇒ 9PA² = 4PB²
⇒ 9[(x – 2)² + (y – 3)²] = 4[(x – 2)² + (y + 3)²]
⇒ 9[x² – 4x + 4 + y² – 6y + 9] = 4 [x² – 4x + 4 + y² + 6y + 9]
∴ 5x² + 5y² – 20 x – 78 y + 65 = 0 is the equation of locus.

Question 10.
A (1, 2), B (2, -3) and C (-2, 3) are three points. A point ‘P’ moves such that PA² + PB² = 2PC². Show that the equation to the locus of ‘P’ is 7x – 7y + 4 = 0.
Solution:
Let P (x, y) be a point on locus.
Given points A = (1, 2), B = (2) -3) and C = (-2, 3)
Given condition is PA² + PB² = 2 PC²
⇒ (x – 1)² + (y – 2)² + (x – 2)² + (y + 3)² = 2 [(x + 2)² + (y-3)²]
⇒ 2x² + 2y² – 6x + 2y + 18 = 2x² + 2y² + 8x – 12y + 26
⇒ 14x – 14y + 8 = 0
∴ 7x – 7y + 4 = 0 is the equation of locus.

SOLVED PROBLEMS

Question 1.
Find the equation of the locus of a point which is at a distance 5 from (-2, 3) in the xoy plane.
Solution:
Let the given point be A = (-2, 3) and P(x, y) be a point on the plane.
The geometric condition to be satisfied by P to be on the locus is that
AP = 5 …………… (1)
Expressing this condition algebraically, we get
\(\sqrt{(x+2)^{2}+(y-3)^{2}}=5\)
i.e., x² + 4x + 4 + y² – 6y + 9 = 25
i.e., x² + y² + 4x – 6y – 12 = 0 ……………. (2)
Let Q(x₁, y₁) satisfy (2).

Hence AQ = 5.
This means that Q(x₁, y₁) satisfies the geometric condition (1).
∴ The required equation of locus is
x² + y² + 4x – 6y – 12 = 0.

Question 2.
Find the equation of locus of a point P, if the distance of P from A(3,0) is twice the distance of P from B(-3, 0).
Solution:
Let P(x, y) be a point on the locus. Then the geometric condition to be satisfied by P is
PA = 2PB …………. (1)
i.e., PA² = 4PB²
i.e., (x – 3)² + y² = 4 [(x + 3)² + y²]
i.e., x² – 6x + 9 + y² = 4 [x² + 6x + 9 + y²]
i.e., 3x² + 3y² + 30x + 27 = 0
i.e., x2 + y2 + lOx + 9 = 0 ………….. (2)
Let Q(x₁, y₁) satisfy (2).

∴ QA = 2QB.
This means that Q(x₁, y₁) satisfies (1).
Hence, the required equation of locus is
x² + y² + 10x + 9 = 0.

Question 3.
Find the locus of the third vertex of a right angled triangle, the ends of whose hypotenuse are (4, 0) and (0, 4).
Solution:
Let A = (4, 0) and B = (0, 4).
Let P(x, y) be a point such that PA and PB are
perpendicular. Then PA² + PB² = AB².
i.e., (x – 4)² + y² + x² + (y – 4)² = 16 + 16
i.e., 2x² + 2y² – 8x – 8y = 0
or x² + y² – 4x – 4y = 0
Let Q(x₁, y₁) satisfy (2) and Q be different from A and B.

Hence QA² + QB² = AB², Q ≠ A and Q ≠ B.
This means that Q(x₁, y₁) satisfies (1).
∴ The required equation of locus is (2), which is the circle with \(\overline{\mathrm{AB}}\) as diameter, deleting the points A and B.
Though A and B satisfy equation (2), they do not satisfy the required geometric condition.

Question 4.
Find the equation of the locus of P, if the ratio of the distances from P to A(5, -4) and B(7, 6) is 2 : 3.
Solution:
Let P(x, y) be any point on the locus.
The geometric condition to be satisfied by P
is\(\frac{AP}{PB}\) = \(\frac{2}{3}\).
i.e., 3AP = 2PB ………. (1)
i.e., 9AP² = 4PB²
i.e., 9[(x – 5)² + (y + 4)²] = 4[(x – 7)² + (y – 6)²]
i.e., 9[x² + 25 – 10x + y² + 16 + 8y] = 4[x² + 49 – 14x +y² + 36 – 12y]
i.e, 5x² + 5y² – 34x + 120y + 29 = 0 ……… (2)
Let Q(x₁ y₁) satisfy (2). Then

(by using (3))
– 4 [(x₁ – 7)² + (y₁ – 6)²] = 4PB²
Thus 3AQ = 2PB. This means that Q(x₁, y₁) satisfies (1).
Hence, the required equation of locus is
5(x² + y²) – 34x + 120y + 29 = 0.

Question 5.
A(2, 3) and B(-3, 4) are two given points. Find the equation of locus of P so that the area of the triangle PAB is 8.5.
Solution:
Let P(x, y) be a point on the locus.
The geometric condition to be satisfied by P is that.
area of ∆PAB = 8.5 ……….. (1)
i.e., \(\frac{1}{2}\)|x(3 – 4) + 2(4 – y) – 3(y – 3)| = 8.5
i.e., |-x + 8-2y-3y+9| = 17
i.e., |-x- 5y + 17| = 17
i.e., -x- 5y + 17 = 17 or-x-5y + 17 = -17
i.e., x + 5y = 0 or x + 5y = 34
∴ (x + 5y) (x + 5y – 34) = 0
i.e., x² + 10xy + 25y² – 34x – 170y = 0 …………. (2)
Let Q(x₁, y₁) satisfy (2). Then
x₁ + 5y₁ = 0 or x₁ + 5y₁ = 34 ………… (3)
Now, area of ∆QAB
= \(\frac{1}{2}\)|x₁(3-4) + 2(4-y₁) – 3(y₁ – 3)|
= \(\frac{1}{2}\) |-x₁ + 8 – 2y₁ – 3y₁ + 9|
= \(\frac{1}{2}\) |-x₁ – 5y₁ + 17|
= \(\frac{17}{2}\) =8.5 (by using (3))
This means that Q(x₁, y₁) satisfies (1).
Hence, the required equation of locus is
(x + 5y) (x + 5y – 34) = 0 or
x² + 10xy + 25 – y² – 34x- 170y = 0.

Students get through Maths 1A Important Questions Inter 1st Year Maths 1A Properties of Triangles Important Questions which are most likely to be asked in the exam.

Intermediate 1st Year Maths 1A Properties of Triangles Important Questions

Question 1.
In ∆ABC, if a = 3, b = 4 and sinA = \(\frac{3}{4}\), find angle B.
Solution:
By sine Rule \(\frac{a}{\sin A}\) = \(\frac{b}{\sin B}\)
⇒ sin B = \(\frac{\text { b. } \sin A}{a}\) = \(\frac{4}{3}\) (\(\frac{3}{4}\)) = 1
⇒ sin B = 1 ⇒ B = 90°

Question 2.
If the lengths of the sides of a triangle are 3, 4, 5 find the circumradius of the triangle.
Solution:
∴ 3² + 4² = 5²
∴ The triangle is right angled and its hypotenuse = 5 = circum diameter.
∴ Circum radius = \(\frac{1}{2}\) (hypotenu) = \(\frac{5}{2}\) cms.

Question 3.
If a = 6, b = 5, c = 9, then find angle A.
Solution:
∵ cos A = \(\frac{\mathrm{b}^{2}+c^{2}-a^{2}}{2 b c}\)
= \(\frac{5^{2}+9^{2}-6^{2}}{2(5)(9)}\) = \(\frac{25+81-36}{2(5)(9)}\)
= \(\frac{70}{90}\) = \(\frac{7}{9}\)
∴ A = cos^-1(\(\frac{7}{9}\))

Question 4.
If ∆ABC, show that ∆ (b + c) cos A = 2s.
Solution:
L.H.S. = (b + c) cos A + (c + a) cos B + (a + b) cos C
= (b cos A + a cos B) + (c cos B + b cos C) + (a cos C + c cos A)
= c + a + b = 2S = R.H.S.

Question 5.
If the sides of a triangle are 13, 14, 15, then find the circum diameter.
Solution:
Let a = 13, b = 14, c = 15
Then 2s = a + b + c = \(\frac{13+14+15}{2}\) = \(\frac{42}{2}\)
∴ s = 21
s – a = 21 – 13 = 8
s – b = 21 – 14 = 7
s – c = 21 – 15 = 6
Now ∆ = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{21 \times 8 \times 7 \times 6}\)
= \(\sqrt{21 \times 21 \times 16}\) = 21 × 4 = 84
∵ ∆ = \(\frac{a b c}{4 R}\) ⇒ 4R = \(\frac{a b c}{\Delta}\)
= \(\frac{13 \times 14 \times 15}{84}\) = \(\frac{65}{2}\)
∴ R = \(\frac{65}{8}\)
∴ Circum diameter(2R) = 2 × \(\frac{65}{8}\) = \(\frac{65}{4}\)cms.

Question 6.
In a ∆A B C, if (a + b + c) (b + c – a) = 3 be, find A.
Solution:
(2s – sa) = 3bc ⇒ \(\frac{s(s-a)}{b c}\) = \(\frac{3}{4}\)
⇒ cos² \(\frac{A}{2}\) = \(\frac{3}{4}\) ⇒ cos \(\frac{A}{2}\) = \(\frac{1}{2}\) = cos 30°
∴ \(\frac{A}{2}\) = 30° = A = 60°

Question 7.
If a = 4, b = 5, c = 7, find cos \(\frac{B}{2}\).
Solution:
2s = a + b + c = 4 + 5 + 7 = 16
⇒ s = 8 and s – b = 8 – 5 = 3
Now cos \(\frac{B}{2}\) = \(\sqrt{\frac{s(s-b)}{a c}}\) = \(\sqrt{\frac{8 \times 3}{4 \times 7}}\) = \(\sqrt{\frac{6}{7}}\)

Question 8.
In ∆ ABC, find b cos² \(\frac{C}{2}\) + c cos² \(\frac{B}{2}\).
Solution:
b cos² \(\frac{C}{2}\) + c cos ² \(\frac{B}{2}\) = b[latex]\frac{s(s-c)}{a b}[/latex] + c[latex]\frac{s(s-b)}{c a}[/latex]
= \(\frac{s(s-c)}{a}\) + \(\frac{s(s-b)}{a}\) = \(\frac{\mathrm{s}}{\mathrm{a}}\)[s – c + s – b]
= \(\frac{\mathrm{s}}{\mathrm{a}}\) • a = s

Question 9.
If tan \(\frac{A}{2}\) = \(\frac{5}{6}\) and tan \(\frac{C}{2}\) = \(\frac{2}{5}\), determine the relation between a, b, c.    [Mar 05]
Solution:
tan \(\frac{A}{2}\) • tan \(\frac{C}{2}\) = \(\frac{5}{6}\) • \(\frac{2}{5}\)
\(\sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}\) = \(\frac{2}{6}\)
⇒ \(\frac{s-b}{s}\) = \(\frac{1}{3}\) ⇒ 3s – 3b = s ⇒ 2s = 3b
⇒ a + b + c = 3b ⇒ a + c = 2b Hence a, b, c are in A.P.

Question 10.
If cot \(\frac{A}{2}\) = \(\frac{b+c}{a}\), find angle B.
Solution:
cot \(\frac{A}{2}\) = \(\frac{b+c}{a}\) ⇒ \(\frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}\) = \(\frac{\cos \left(\frac{B-C}{2}\right)}{\sin \frac{A}{2}}\) (by Mollweide rule)
⇒ \(\frac{A}{2}\) = \(\frac{B-C}{2}\)
⇒ A = B – C ⇒ A + C = B ⇒ A + B + C = 2B
∴ 2B = 180° ⇒ B = 90°

Question 11.
If tan (\(\frac{C-A}{2}\)) = k cot \(\frac{B}{2}\), find k.
Solution:
Comparing with tan (\(\frac{C-A}{2}\)) = \(\frac{c-a}{c+a}\) cot \(\frac{B}{2}\), (by tangent law)
we get that k = \(\frac{c-a}{c+a}\)

Question 12.
In ∆ABC, show that \(\frac{b^{2}-c^{2}}{a^{2}}\) = \(\frac{\sin (B-C)}{\sin (B+C)}\).
Solution:

Question 13.
Show that a² cot A + b² cot B + c² cot C = \(\frac{a b c}{R}\).    [Mar 14]
Solution:
L.H.S. = a² cot A + b² cot B + c² cot C
= 4R² sin² A. \(\frac{\cos A}{\sin A}\) + 4R² sin²B. \(\frac{\cos B}{\sin B}\) + 4R² sin²C. \(\frac{\cos C}{\sin C}\) (by sine rule)
= 2R² (2 sin A cos A + 2 sin B cos B + 2 sin C cos C)
= 2R² (sin 2A + sin 2B + sin 2C) .
= 2R² (4 sin A sin B sin C) (from transformations)
= \(\frac{1}{R}\) (2R sin A) (2R sin B) (2R sin C)
= \(\frac{abc}{R}\) = R.H.S

Question 14.
Show that (b – c)² cos² \(\frac{A}{2}\) + (b + c)² sin² \(\frac{A}{2}\) = a².
Solution:
L.H.S. = (b² + c² – 2bc) cos² \(\frac{A}{2}\) + (b² + c² + 2bc) sin² \(\frac{A}{2}\)
= (b² + c²) [cos² \(\frac{A}{2}\) + sin² \(\frac{A}{2}\)] – 2bc (cos² \(\frac{A}{2}\) – sin² \(\frac{A}{2}\))
= b² + c² – 2bc cos A = a²

Question 15.
Prove that a (b cos C – c cos B) = b² – c²     [Mar 07]
Solution:
L.H.S. = ab cos C – ca cos B
= (\(\frac{a^{2}+b^{2}-c^{2}}{2}\)) – (\(\frac{c^{2}+a^{2}-b^{2}}{2}\)) (by cosine rule)
= \(\frac{1}{2}\)[a² + b² – c² – c² – a² + b²]
= b² – c² = R.H.S.

Question 16.
Show that \(\frac{c-b \cos A}{b-c \cos A}\) = \(\frac{\cos B}{\cos C}\).
Solution:
From projection rule c = a cos B – b cos A and b = c cos A + a cos C

Question 17.
In ∆ABC, if \(\frac{1}{a+c}\) + \(\frac{1}{b+c}\) = \(\frac{3}{a+b+c}\) show that C = 60°.
Solution:
\(\frac{1}{a+c}\) + \(\frac{1}{b+c}\) = \(\frac{3}{a+b+c}\)
⇒ \(\frac{b+c+a+c}{(a+c)(b+c)}\) = \(\frac{3}{a+b+c}\)
⇒ 3(a + c) (b + c) = (a + b + 2c) (a + b + c)
⇒ 3(ab + ac + bc + c²) = (a² + b² + 2ab) + 3c(a + b) + 2c²
⇒ ab = a² + b² – c²
⇒ 2ab cos C (from cosine rule)
⇒ cos C = \(\frac{1}{2}\) ⇒ C = 60°

Question 18.
If a = (b – c) sec θ, prove that tan θ = \(\frac{2 \sqrt{b c}}{b-c}\) sin \(\frac{A}{2}\).    [Mar 16]
Solution:
a = (b – c) sec θ ⇒ sec θ = \(\frac{a}{b-c}\)
tan² θ = sec² θ – 1

Question 19.
In ∆ABC, show that (a + b + c) (tan \(\frac{A}{2}\) + tan \(\frac{B}{2}\)) = 2c cot \(\frac{C}{2}\).
Solution:

Question 20.
Show that b² sin 2C + c² sin 2B = 2bc sin A.
Solution:
L.H.S. = b² sin 2C + c² sin 2B
= 4R² sin² B (2 sin C cos C) + 4R² sin² C (2 sin B cos B)
= 8R² sin B sin C (sin B cos C + cos B sin C)
= 8R² sin B sin C sin (B + C)
= 2(2R sin B) (2R sin C) sin A
= 2bc sin A = R.H.S.

Question 21.
Prove that cot A + cot B + cot C = \(\frac{a^{2}+b^{2}+c^{2}}{4 \Delta}\)    [Mar 15]
Solution:

Question 22.
Show that a cos² \(\frac{A}{2}\) + b cos² \(\frac{B}{2}\) + c cos ² \(\frac{C}{2}\) = s + \(\frac{\Delta}{R}\).
Solution:
L.H.S. = Σ a cos² \(\frac{A}{2}\) = \(\frac{1}{2}\) Σ a (1 + cos A)
= \(\frac{1}{2}\) Σ (a + a cos A) = \(\frac{1}{2}\) (a + b + c) + \(\frac{1}{2}\) Σ (2R sin A cos A)
= \(\frac{1}{2}\) (2s) + \(\frac{R}{2}\) Σ sin 2A
= s + \(\frac{R}{2}\) (sin 2A + sin 2B + sin 2C)
= s + \(\frac{R}{2}\) (4 sin A sin B sin C)
= s + \(\frac{1}{R}\) (2R² sin A sin B sin C)
= s + \(\frac{\Delta}{R}\) (∵ ∆ = 2R² sin A sin B sin C)
= R.H.S.

Question 23.
In ∆ ABC, if a cos A = b cos B, prove that the triangle is either isosceles or right angled.
Solution:
a cos A = b cos B
⇒ 2R sin A cos A = 2R sin B cos B
⇒ sin 2A = sin 2B (or) = sin (180° – 2B)
Hence 2A = 2B or 2A = 180° – 2B
⇒ A = B or A = 90° – B ⇒ A = B or A + B = 90°
⇒ C = 90°
∴ The triangle is isosceles or right angled.

Question 24.
If cot \(\frac{A}{2}\) : cot \(\frac{B}{2}\) : cot \(\frac{C}{2}\) = 3 : 5 : 7, show that a : b : c = 6 : 5 : 4.
Solution:

Then s – a = 3k, s – b = 5k, s – c = 7k
Adding 3s – (a + b + c) = 3k + 5k+ 7k
⇒ 3s – 2s = 15k ⇒ s = 15k
Now a = 12k, b = 10k, c = 8k
∴ a : b : c = 12k : 10k : 8k = 6 : 5 : 4

Question 25.
Prove that a³ cos (B – C) + b³ cos (C – A) + c³ cos(A – B) = 3abc.
Solution:
L.H.S. = Σ a³ cos (B – C)
= Σ a² (2R sin A) cos (B – C)
= R Σ a² [2 sin (B + C) cos (B- C)]
= R Σ a² (sin 2B + sin 2C)
= R Σ a² (2 sin B cos B + 2 sin C cos C)
= Σ [a²(2R sin B) cos B + a²(2R sin C) cos C] Σ (a² b cos B + a²c cos C)
= (a²b cos B + a²c cos C) + (b²c cos C + b² a cos A) + (c² a cos A + c²b cos B)
= ab (a cos B + b cos A) + bc (b cos C + c cos B) + ca (c cos A + a cos C)
= ab(c) + bc(a) + ca(b) = 3 abc = R.H.S

Question 26.
If p₁, p₂, p₃ are the altitudes of the ∆ A, B, C show that \(\frac{1}{p_{1}^{2}}+\frac{1}{p_{2}^{2}}+\frac{1}{p_{3}^{2}}\) = \(\frac{\cot A+\cot B+\cot C}{\Delta}\)
Solution:
Since p₁, p₂, p₃ are the altitudes of ∆ ABC,

Question 27.
The angle of elevation of the top point P of the vertical tower PQ of height h from a point A is 45° and from a point B is 60°, where B is a point at a distance 30 meters from the point A measured along the line AB which makes an angle 30° with AQ. Find the height of the tower.
Solution:

PQ = h, ∠PAQ = 45°
∠BAQ = 30° and ∠PBC = 60°
Also AB = 30 mts.
∴ ∠BAP = ∠APB = 15°.
This gives BP = AB = 30 and h = PC + CQ = BP sin 60° + AB sin 30°
= 15 \(\sqrt{3}\) + 15 = 15(\(\sqrt{3}\) +1) metres.

Question 28.
Two trees A and B are on the same side of a river. From a point C in the river the distances of the trees A and B are 250m and 300m respectively. If the angle C is 45°, find the distance between the trees (use \(\sqrt{2}\) = 1.414).
Solution:

From the triangle ABC, using the cosine rule
AB² = 250² + 300² – 2(250) (300) cos 45°
= 100 (625 + 900 – 750\(\sqrt{2}\)) = 46450.
∴ AB = 215.5m. (approximately).

Question 29.
In A ABC, prove that \(\frac{1}{r_{1}}\) + \(\frac{1}{r_{2}}\) + \(\frac{1}{r_{3}}\) = \(\frac{1}{r}\).
Solution:
L.H.S. = \(\frac{1}{r_{1}}\) + \(\frac{1}{r_{2}}\) + \(\frac{1}{r_{3}}\) = \(\frac{s-a}{\Delta}\) + \(\frac{s-b}{\Delta}\) + \(\frac{s-c}{\Delta}\)
= \(\frac{3 s-(a+b+c)}{\Delta}\) = \(\frac{3 s-2 s}{\Delta}\) = \(\frac{s}{\Delta}\) = \(\frac{1}{r}\)
= R.H.S

Question 30.
Show that rr₁r ₂r₃ = ∆²
Solution:
L.H.S. = rr₁r ₂r₃ = \(\frac{\Delta}{s} \cdot \frac{\Delta}{s-a} \cdot \frac{\Delta}{s-b} \cdot \frac{\Delta}{s-c}\)
= \(\frac{\Delta^{4}}{\Delta^{2}}\) = ∆² = R.H.S

Question 31.
In an equilateral triangle, find the value of \(\frac{r}{R}\).
Solution:
\(\frac{r}{R}\) = \(\frac{4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}}{R}\)
= 4 sin³ 30° (∵ A = B = C = 60°)
= 4 . (\(\frac{1}{2}\))³ = \(\frac{1}{2}\)

Question 32.
The perimeter of ∆ ABC is 12 cm. and its in radius is 1 cm. Then find the area of the triangle.
Solution:
Given that 2s = 12 ⇒ s = 6 cm.
and r = 1 cm
Area of ∆ ABC = ∆ = rs = (1) (6) = 6 sq.cms.

Question 33.
Show that rr₁ = (s – b) (s – c).
Solution:
L.H.S. = rr₁
= [(s – b) tan \(\frac{B}{2}\)] [(s – c) cot \(\frac{B}{2}\)]
= (s – b) (s – c) = R.H.S.
ans:

Question 34.
Express \(\frac{a \cos \mathbf{A}+\mathbf{b} \cos \mathbf{B}+\cos \mathbf{C}}{\mathbf{a}+\mathbf{b}+\mathbf{c}}\) in terms of R and r.
Solution:

Question 35.
In ∆ABC, ∆ = 6sq.cm and s = 1.5 cm, find r.
Solution:
r = \(\frac{\Delta}{s}\) = \(\frac{6}{1.5}\) = 4 cm.

Question 36.
Show that rr₁ cot\(\frac{A}{2}\) = ∆.
Solution:
rr₁ cot\(\frac{A}{2}\) = \(\frac{\Delta}{s}\) (s tan \(\frac{A}{2}\)) cot \(\frac{A}{2}\) = ∆

Question 37.
If a = 13, b = 14, c = 15, find r₁.
Solution:
2s = a + b +c = 42
⇒ s = 21
s – a = 8, s – b = 7, s – c = 6
∆² = 21 × 8 × 7 × 6
⇒ ∆ = 7 × 12 = 84 sq. units
∴ r₁ = \(\frac{\Delta}{s-b}\) = \(\frac{84}{8}\) = 10.5 units

Question 38.
If rr₂ = r₁r₃, then find B.
Solution:

Question 39.
In a ∆ABC, show that the sides a, b, c are in A.P if and only if r₁, r₂, r₃ are in H.P.
Solution:
r₁, r₂, r₃ are in A.P.
⇔ \(\frac{1}{r_{1}}\), \(\frac{1}{r_{2}}\), \(\frac{1}{r_{3}}\) are in A.P.
⇔ \(\frac{s-a}{\Delta}\), \(\frac{s-b}{\Delta}\), \(\frac{s-c}{\Delta}\) are in A.P.
⇔ s – a, s – b, s – c are in A.P.
⇔ -a, -b, -c are in A.P.
⇔ a, b. c, are in A.R

Question 40.
If A = 90°, show that 2(r + R) = b + c.
Solution:
L.H.S = 2r + 2R
= 2(s – a) tan \(\frac{A}{2}\) + 2R.1
= 2(s – a) tan 45°+ 2RsinA
(∵A = 90°)
= (2s – 2a). 1 + a
= b + c = R.H.S.

Question 41.
If (r₂ – r₁) (r₃ – r₁) = 2r₂r₃, show that A = 90°.
Solution:
(r₂ – r₁) (r₃ – r₁) = 2r₂r₃

⇒ 2(bc – ca – ab + a²) = b² + c² + a² + 2bc – 2ca – 2ab
⇒ 2a² = b² + c² + a²
⇒ b² + c² = a²
Hence △ABC is right angled and A = 90°.

Question 42.
Prove that \(\frac{r_{1}\left(r_{2}+r_{3}\right)}{\sqrt{r_{1} r_{2}+r_{2} r_{3}+r_{3} r_{1}}}\) = a
Solution:

Question 43.
Show that \(\frac{1}{r^{2}}+\frac{1}{r_{1}^{2}}+\frac{1}{r_{2}^{2}}+\frac{1}{r_{3}^{2}}\) = \(\frac{a^{2}+b^{2}+c^{2}}{\Delta^{2}}\)
Solution:

Question 44.
Prove that Σ(r + r₁) tan \(\left(\frac{B-C}{2}\right)\) = 0
Solution:
r + r₁ = 4R sin \(\frac{A}{2}\) sin \(\frac{B}{2}\) sin \(\frac{C}{2}\) + 4R sin \(\frac{A}{2}\) cos \(\frac{B}{2}\) cos \(\frac{C}{2}\)
= 4R sin\(\frac{A}{2}\)[sin \(\frac{B}{2}\) sin \(\frac{C}{2}\) + cos \(\frac{B}{2}\) cos \(\frac{C}{2}\)]

Question 45.
Show that \(\frac{r_{1}}{b c}+\frac{r_{2}}{c a}+\frac{r_{3}}{a b}=\frac{1}{r}-\frac{1}{2 R}\) .    [May 07]
Solution:

Question 46.
If r: R: r₁ = 2 : 5 : 12. then prove that the triangle ¡s right angled at A.
Solution:
r : R : r₁ = 2 : 5 : 12 then r = 2k. R = 5k and r₁ = 12K
r₁ – r = 12k – 2k = 10k = 2(5k) = 2R
⇒ 4R sin \(\frac{A}{2}\)[cos \(\frac{B}{2}\) cos \(\frac{C}{2}\) – sin \(\frac{B}{2}\) sin \(\frac{C}{2}\)] = 2R
⇒ 2 sin\(\frac{A}{2}\) cos\(\left(\frac{B+C}{2}\right)\) = 1
⇒ 2 sin \(\frac{A}{2}\) = \(\frac{1}{2}\) [cos \(\left(\frac{B+C}{2}\right)\) = sin \(\frac{A}{2}\)]
⇒ sin \(\frac{A}{2}\) = \(\frac{1}{\sqrt{2}}\) = sin 45°
⇒ \(\frac{A}{2}\) = 45° ⇒ A = 90°
Hence the triangle is right angled at A.

Question 47.
Show that r + r₃ + r₁ – r₂ = 4R cos B.
Solution:
r + r₃

Question 48.
If A, A₁, A₂, A₃ are the areas of incircle and excircles of a triangle respectively, then prove that \(\frac{1}{\sqrt{A_{1}}}+\frac{1}{\sqrt{A_{2}}}+\frac{1}{\sqrt{A_{3}}}=\frac{1}{\sqrt{A}}\) .
Solution:
r, r₁, r₂, r₃ are the in radius and ex-radii of the circles whose areas are A, A₁, A₂, A₃ respectively, then
A = itr², A₁ = πr₁², A₂ = πr₂², A₃ = πr₃²

Question 49.
Show that (r₁ + r₂) sec² \(\frac{C}{2}\) = (r₂ + r₃) sec² \(\frac{A}{2}\) = (r₃ + r₁) sec² \(\frac{B}{2}\).
Solution:
r₁ + r₂ = 4R cos \(\frac{C}{2}\) [sin\(\frac{A}{2}\) cos\(\frac{B}{2}\) + cos \(\frac{A}{2}\) sin \(\frac{B}{2}\)]
= 4R cos\(\frac{C}{2}\) sin\(\left(\frac{A+B}{2}\right)\)
= 4R cos²\(\frac{C}{2}\) ___________ (1)
⇒ (r₁ + r₂) sec²\(\frac{C}{2}\) = 4R
Similarly, we can show that
(r₂ + r₃) sec²\(\frac{A}{2}\) = (r₃ + r₁) sec²\(\frac{B}{2}\) = 4R
∴ (r₁ + r₂)sec²\(\frac{C}{2}\) = (r₂ + r₃)
sec²\(\frac{A}{2}\) = (r₃ + r₁) sec²\(\frac{B}{2}\) = 4R

Question 50.
In ∆ABC, if AD, BE, CF are the perpendiculars drawn from the vertices A, B, C to the opposite sides, show that
i) \(\frac{1}{A D}+\frac{1}{B E}+\frac{1}{C F}=\frac{1}{r}\) and
ii) AD. BE. CF = \(\frac{(a b c)^{2}}{8 R^{3}}\)
Solution:
Since AD ⊥r to BC,

Question 51.
In ∆ ABC, if r₁ = 8, r₂ = 12, r₃ = 24, find a, b, c.
Solution:
∵ \(\frac{1}{r}\) = \(\frac{1}{r_{1}}\) + \(\frac{1}{r_{2}}\) + \(\frac{1}{r_{3}}\)
⇒ \(\frac{1}{r}\) = \(\frac{1}{8}\) + \(\frac{1}{12}\) + \(\frac{1}{24}\)
⇒ \(\frac{1}{r}\) = \(\frac{3+2+1}{24}\) ⇒ r = 4
But ∆² = rr₁r₂r₃ = 4 × 8 × 12 × 24
= (8 × 12)²
⇒ ∆ = 96 sq. cm
∵ r = \(\frac{\Delta}{\mathrm{s}}\) ⇒ 4 = \(\frac{96}{s}\) ⇒ s = 24
∵ r₁ = \(\frac{\Delta}{s-a}\)
⇒ δ = \(\frac{96}{s-a}\) ⇒ s – a = \(\frac{96}{8}\) ⇒ 24 – a
= 12 ⇒ a = 12
Similarly s – b = \(\frac{96}{12}\) ⇒ 24 – b = 8 ⇒ b = 16
s – c = \(\frac{96}{24}\) ⇒ 24 – c = 4 ⇒ c = 20

Question 52.
Show that \(\frac{a b-r_{1} r_{2}}{r_{3}}\) = \(\frac{b c-r_{2} r_{3}}{r_{1}}\) = \(\frac{c a-r_{3} r_{1}}{r_{2}}\)   [May 08; Mar 08]
Solution:

Students get through Maths 1A Important Questions Inter 1st Year Maths 1A Hyperbolic Functions Important Questions which are most likely to be asked in the exam.

Intermediate 1st Year Maths 1A Hyperbolic Functions Important Questions

Question 1.
Prove that for any x ∈ R, sinh (3x) = 3 sinh x + 4sinh x
Answer:
LHS = sinh (3x)
= sinh (2x + x)
= sinh (2x) . cosh(x) + cosh (2x) . sinh (x) = (2sinh x cosh x)cosh x (1 +2sinh² x)sinh x
= 2 sinh x (cosh² x) + (1 + 2 sinh² x) sinh x
= 2 sinh x (1 + sinh² x) + (1 + 2 sinh² x) sinh x
∵ cosh² x – sinh² x = 1
= 3 sinh x + 4 sinh³ x
∴ sinh (3x) = 3 sinh x + 4 sinh³ x

Question 2.
Prove that for any x ∈ R, tanh 3x = \(\frac{3 \tanh x+\tanh ^{3} x}{1+3 \tanh ^{2} x}\)
Answer:
tanh 3x = tan (2x + x)

Question 3.
If cosh x = \(\frac{5}{2}\), find the values of
i) cosh (2x) and
ii) sinh (2x)
Answer:
cosh (x) = \(\frac{5}{2}\)
(i) cosh (2x) = 2 cosh² (x) – 1
= 2(\(\frac{5}{2}\))² – 1 = \(\frac{25}{2}\) – 1 = \(\frac{23}{2}\)

ii) sinh² (2x) = cosh² (2x) – 1
= (\(\frac{23}{2}\))² – 1 = \(\frac{529-4}{2}\) = \(\frac{525}{4}\)
∴ sinh (2x) = ±\(\sqrt{\frac{525}{4}}\) = ±\(\frac{5 \sqrt{21}}{2}\)

Question 4.
If coshx = sec θ then prove that tan h²\(\frac{x}{2}\) = tan²\(\frac{\theta}{2}\)
Answer:
tan h²\(\frac{x}{2}\) = \(\frac{\cosh x-1}{\cosh x+1}\)
= \(\frac{\sec \theta-1}{\sec \theta+1}\) = \(\frac{1-\cos \theta}{1+\cos \theta}\) = tan²\(\frac{\theta}{2}\)

Question 5.
If θ ∈ (-\(\frac{\pi}{4}\), \(\frac{\pi}{4}\)) and x = log_e(cot(\(\frac{\pi}{4}\) + θ) then prove that
i) cosh x = sec 2θ and
ii) sinh x = -tan 2θ
Answer:

Question 6.
If sinh x = 5, show that x = log_e (5 + \(\sqrt{26}\))
Answer:
∴ sinh (x) = 5
⇒ x = sinh^-1 (5)
= log_e (5 + \(\sqrt{5^{2}+1}\))
= log_e (5 + \(\sqrt{26}\))
[sin^-1 (x) = log_e (x + \(\sqrt{x^{2}+1}\)) for all x ∈ R]

Question 7.
Show that tanh^-1(\(\frac{1}{2}\)) = \(\frac{1}{2}\) log_e3 (A.P) [Mar 15; May 07, 05; Mar 08, 05]
Answer:
∵tanh^-1 (x) = \(\frac{1}{2}\)log_e(\(\frac{1+x}{1-x}\)) for all x ∈ (-1, 1)
∵ tanh^-1 (\(\frac{1}{2}\)) = \(\frac{1}{2}\)log_e(\(\frac{1+\frac{1}{2}}{1-\frac{1}{2}}\))
= \(\frac{1}{2}\)log_e(3)

Students get through Maths 1A Important Questions Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions which are most likely to be asked in the exam.

Intermediate 1st Year Maths 1A Inverse Trigonometric Functions Important Questions

Question 1.
Find the values of the following.
i) sin^-1(-\(\frac{1}{2}\))
Solution:
sin^-1(-\(\frac{1}{2}\)) = -sin^-1(\(\frac{1}{2}\)) = –\(\frac{\pi}{6}\)

ii) cos^-1(-\(\frac{\sqrt{3}}{2}\))
Solution:
cos^-1(-\(\frac{\sqrt{3}}{2}\)) = π – cos^-1(\(\frac{\sqrt{3}}{2}\))
= π – \(\frac{\pi}{6}\) = \(\frac{5\pi}{6}\)

iii) tan^-1(\(\frac{1}{\sqrt{3}}\))
Solution:
tan^-1(\(\frac{1}{\sqrt{3}}\)) = \(\frac{\pi}{6}\)

iv) cot^-1(-1)
Solution:
cot^-1(-1) = π – cot^-1 (1) = π – \(\frac{\pi}{4}\)
= \(\frac{3\pi}{4}\)

v) sec^-1(-\(\sqrt{2}\))
Solution:
sec^-1(-\(\sqrt{2}\)) = π – sec^-1(\(\sqrt{2}\))
= π – \(\frac{\pi}{4}\) = \(\frac{3\pi}{4}\)

vi) cosec^-1(\(\frac{2}{\sqrt{3}}\))
Solution:
cosec^-1(\(\frac{2}{\sqrt{3}}\)) = sin^-1(\(\frac{\sqrt{3}}{2}\)) = \(\frac{\pi}{3}\)

Question 2.
Find the values of the following.
i) sin^-1(sin \(\frac{4\pi}{3}\))
Solution:

ii) cos^-1(cos \(\frac{4\pi}{3}\))
Solution:
cos^-1(cos \(\frac{4\pi}{3}\))
= cos^-1 (cos (π + \(\frac{\pi}{3}\)))
= cos^-1 (-cos \(\frac{\pi}{3}\))
= π – cos^-1 (cos \(\frac{\pi}{3}\)) = π – \(\frac{\pi}{3}\) = \(\frac{2\pi}{3}\);
∵ \(\frac{2\pi}{3}\) ∈ (0, π)

iii) tan^-1(tan \(\frac{4\pi}{3}\))
Solution:
tan^-1(tan \(\frac{4\pi}{3}\)) = tan^-1(tan(π + \(\frac{\pi}{3}\)))
= tan^-1(tan \(\frac{\pi}{3}\)) = \(\frac{\pi}{3}\);
∵ \(\frac{\pi}{3}\) ∈ (-\(\frac{\pi}{2}\), \(\frac{\pi}{2}\))

Question 3.
Find the values of the following
i) sin(cos^-1 \(\frac{5}{13}\))
Solution:
sin(cos^-1 \(\frac{5}{13}\)) = sin (sin^-1 \(\frac{12}{13}\)) = \(\frac{12}{13}\)

ii) tan (sec^-1 \(\frac{25}{7}\))
Solution:
tan (sec^-1 \(\frac{25}{7}\)) = tan (tan^-1 \(\frac{24}{7}\)) = \(\frac{24}{7}\)

iii) cos (tan^-1 \(\frac{24}{7}\))
Solution:
cos (tan^-1 \(\frac{24}{7}\)) = cos (cos^-1 \(\frac{7}{25}\)) = \(\frac{7}{25}\)

Question 4.
Find the values of the following
i) sin² (tan^-1 \(\frac{3}{4}\))
Solution:
sin (tan^-1 \(\frac{3}{4}\)) = sin (sin^-1 \(\frac{3}{5}\)) = \(\frac{3}{5}\)
∴ sin² (tan^-1 \(\frac{3}{4}\)) = (\(\frac{3}{5}\))² = \(\frac{9}{25}\)

ii) sin (\(\frac{\pi}{2}\) – sin^-1(-\(\frac{4}{5}\)))
Solution:
sin (\(\frac{\pi}{2}\) – sin^-1(-\(\frac{4}{5}\))
= sin (\(\frac{\pi}{2}\) – sin^-1(\(\frac{4}{5}\))
= cos (sin^-1 \(\frac{4}{5}\))
= cos (cos^-1 \(\frac{3}{5}\)) = \(\frac{3}{5}\)

iii) cos (cos^-1(-\(\frac{2}{3}\)) – sin^-1(\(\frac{2}{3}\)))
Solution:
cos (cos^-1(-\(\frac{2}{3}\)) – sin^-1(\(\frac{2}{3}\)))
= cos (π – cos^-1\(\frac{2}{3}\) – sin^-1(\(\frac{2}{3}\)))
= cos (π – (cos^-1\(\frac{2}{3}\) + sin^-1\(\frac{2}{3}\)))
= cos (π – \(\frac{\pi}{2}\)) = cos (\(\frac{\pi}{2}\)) = 0

iv) sec²(cot^-1 3) + cosec² (tan^-1 2)
Solution:
Let cot^-1 (3) = α and tan^-1 (2) = β
Then cot α = 3 and tan β = 2
⇒ tan α = \(\frac{1}{3}\) and cot β = \(\frac{1}{2}\)
Now sec²(cot^-1 3) + cosec² (tan^-1 2)
= sec² α + cosec² β
= (1 + tan²α) + (1 + cot² β)
= 1 + (\(\frac{1}{3}\))² + 1 + (\(\frac{1}{2}\))²
= 2 + \(\frac{1}{9}\) + \(\frac{1}{4}\)
= \(\frac{72+4+9}{36}\) = \(\frac{85}{36}\)

Question 5.
Find the value of cot^-1 \(\frac{1}{2}\) + cot^-1 \(\frac{1}{3}\)
Solution:
cot^-1 \(\frac{1}{2}\) + cot^-1 \(\frac{1}{3}\)
= tan^-1 (2) + tan^-1 (3)
∵ x = 3, y = 2, xy > 1
= π + tan^-1 (\(\frac{2+3}{1-(2)(3)}\))
= π + tan^-1 (\(\frac{5}{-5}\))
= π + tan^-1 (-1)
= π – \(\frac{\pi}{4}\) = \(\frac{3\pi}{4}\)

Question 6.
Prove that sin^-1 \(\frac{4}{5}\) + sin^-1 \(\frac{7}{25}\) = sin^-1 \(\frac{117}{125}\) [Mar 16]
Solution:
Method (i):
Let sin^-1(\(\frac{4}{5}\)) = α and sin^-1 \(\frac{7}{25}\) = β
Then sin α = \(\frac{4}{5}\) and sin β = \(\frac{7}{25}\) and α, β ∈ (0, \(\frac{\pi}{2}\))
So that cos α = \(\frac{3}{5}\) and cos β = \(\frac{24}{25}\) and α + β ∈ (0, π)
Now
cos(α + β) = cos α cos β – sin α sin β
= \(\frac{3}{5}\) . \(\frac{24}{25}\) – \(\frac{4}{5}\) . \(\frac{7}{25}\)
= \(\frac{72-28}{125}\) = \(\frac{44}{125}\) > 0
⇒ (α + β) ∈ (0, \(\frac{\pi}{2}\))
Now sin(α + β) = sin α cos β – cos α sin β
= \(\frac{4}{5}\) . \(\frac{24}{25}\) – \(\frac{3}{5}\) . \(\frac{7}{25}\)
= \(\frac{96+21}{125}\) = \(\frac{117}{125}\)
⇒ (α + β) = sin (\(\frac{117}{125}\))
∴ sin^-1 (\(\frac{4}{5}\)) + sin^-1 (\(\frac{7}{25}\)) = sin^-1 (\(\frac{117}{125}\)

Method (ii) :
We know that

Question 7.
If x ∈ (-1, 1), prove that 2 tan^-1x = tan^-1(\(\frac{2 x}{1-x^{2}}\))
Solution:
∵ x ∈ (-1, 1) and let tan^-1 x = α
Then tan α = x and \(\frac{-\pi}{4}\) < \(\frac{\pi}{4}\)
Now tan^-1 (\(\frac{2 x}{1-x^{2}}\)) = tan^-1 (\(\frac{2 \tan \alpha}{1-\tan ^{2} \alpha}\))
= tan^-1 (tan 2α)
= 2α,
since 2α ∈ (-\(\frac{-\pi}{2}\), \(\frac{-\pi}{2}\))
∴ tan^-1(\(\frac{2 x}{1-x^{2}}\)) = 2 tan^-1 (x)

Question 8.
Prove that sin^-1 \(\frac{4}{5}\) + sin^-1 \(\frac{5}{13}\) + sin^-1 \(\frac{16}{25}\) = \(\frac{\pi}{2}\)
Solution:
Let sin^-1 \(\frac{4}{5}\) = α and sin^-1 \(\frac{5}{13}\) = β
Then α, β are acute angles and sin α = \(\frac{4}{5}\), sin β = \(\frac{5}{13}\)
So that cos α = \(\frac{3}{5}\) and cos β = \(\frac{12}{13}\)
Now
cos (α + β) = cos α cos β – sin α sin β
= \(\frac{3}{5}\) . \(\frac{12}{13}\) – \(\frac{4}{5}\) . \(\frac{5}{13}\)
= \(\frac{16}{65}\)
∴ α + β = cos^-1 (\(\frac{16}{65}\))
⇒ sin^-1 \(\frac{4}{5}\) + sin^-1 \(\frac{5}{13}\) = cos^-1(\(\frac{16}{65}\)) __________ (1)
LHS
= (sin^-1\(\frac{4}{5}\) + sin^-1 \(\frac{5}{13}\)) + sin^-1(\(\frac{16}{65}\))
= cos^-1 \(\frac{16}{65}\) + sin^-1 \(\frac{16}{65}\) = \(\frac{\pi}{2}\) [By (1)]
LHS = RHS

Question 9.
Prove that cot^-1 9 + cosec^-1 \(\frac{\sqrt{41}}{4}\) = \(\frac{\pi}{4}\)
Solution:
Let cot^-1 (9) = α and cosec^-1 \(\frac{\sqrt{41}}{4}\) = β
⇒ cot α = 9 and cosec β = \(\frac{\sqrt{41}}{4}\)
⇒ tan α = \(\frac{1}{9}\) and cot β =
= \(\sqrt{\frac{41}{16}-1}\) = \(\sqrt{\frac{25}{16}}\) = \(\frac{5}{4}\)
∴tan α = \(\frac{1}{9}\) and tan β = \(\frac{4}{5}\)
Now tan(α + β) = \(\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}\)
= \(\frac{\frac{1}{9}+\frac{4}{5}}{1-\left(\frac{1}{9}\right)\left(\frac{4}{5}\right)}\)
= \(\left(\frac{5+36}{45-4}\right)\) = 1
⇒ tan (α + β) = tan \(\frac{\pi}{4}\)
⇒ α + β = \(\frac{\pi}{4}\)
⇒ cot^-1 (9) + cosec^-1 (\(\frac{\sqrt{41}}{4}\)) = \(\frac{\pi}{4}\)

Question 10.
Show that cot (sin^-1 \(\sqrt{\frac{13}{17}}\)) = sin (tan^-1 \(\frac{2}{3}\))
Solution:

Question 11.
Find the value of tan (2 tan^-1(\(\frac{1}{5}\)) – \(\frac{\pi}{4}\))
Solution:

Question 12.
Prove that sin^-1 \(\frac{4}{5}\) + 2 tan^-1 \(\frac{1}{3}\) = \(\frac{\pi}{2}\)
Solution:

Question 13.
Prove that cos (2 tan^-1 \(\frac{1}{7}\)) = sin (4 tan^-1 \(\frac{1}{3}\))
Solution:
Let tan^-1 \(\frac{1}{7}\)) = α and tan^-1 \(\frac{1}{3}\) = β
Then tan α = \(\frac{1}{7}\) and tan β = \(\frac{1}{3}\)
Now

Question 14.
If sin^-1 x + sin^-1y + sin^-1z = π, then prove that x⁴ + y⁴ + z⁴ + 4x²y²z² = 2 (x²y² + y²z² + z²x²)
Solution:
Let sin^-1x = A, sin^-1 y = B and sin^-1(z) = c
then A + B + C = π ________(1) and
sin A = x, sin B = y and sin C = z
Now A + B = π – C’
= cos (A + B) = cos( π – C’)
= cos A cos B – sin A sin B = -cos C
⇒ \(\sqrt{1-x^{2}} \sqrt{1-y^{2}}\) – xy = – \(\sqrt{1-z^{2}}\)
⇒ \(\sqrt{1-x^{2}} \sqrt{1-y^{2}}\) = xy – \(\sqrt{1-z^{2}}\)
On squaring both sides we get
(1 – x²)(1 – y²) = x²y² + (1 – z²) – 2xy \(\sqrt{1-z^{2}}\)
⇒ 2xy\(\sqrt{1-z^{2}}\) = x² + y² – z²
Again on squaring both sides, we get
(2xy \(\sqrt{1-z^{2}}\))² (x² + y² – z²)²
⇒ 4x²y²(1 – z²) = x⁴ + y⁴ + z⁴ + 2x²y² – 2y²z² – 2z²x²
⇒ 4x²y² – 4x²y²z² = x⁴ + y⁴ + z⁴ + 2x²y² – 2y²z² — 2z²x²
⇒ x⁴ + y⁴ + z⁴ + 4x²y²z² = 2x²y² + 2y²z² + 2z²x²

Question 15.
If cos^-1\(\frac{p}{a}\) + cos^-1\(\frac{q}{b}\) = α, then prove that \(\frac{p^{2}}{a^{2}}\) – \(\frac{2 p q}{a b}\) cos α + \(\frac{q^{2}}{b^{2}}\) = sin² α
Solution:
Let cos^-1\(\frac{p}{a}\) = α and cos^-1\(\frac{q}{b}\) = β
then cosα = \(\frac{p}{a}\) and cos β = \(\frac{q}{b}\) and
A + B = α (given)
Now
cos α = cos(A + B)
= cosA cosB – sinA sinB

Question 16.
Solve are sin(\(\frac{5}{x}\)) + arc sin \(\frac{12}{x}\) = \(\frac{\pi}{2}\); (x > 0)
Solution:
Given that
sin^-1 (\(\frac{5}{x}\)) + sin^-1 \(\frac{12}{x}\) = \(\frac{\pi}{2}\); x > 0
Let sin^-1 \(\frac{5}{x}\) = α and sin^-1 \(\frac{12}{x}\) = β
then sin α = \(\frac{5}{x}\) and sin β = \(\frac{12}{x}\), x > 0
Now α + β = \(\frac{\pi}{2}\)
⇒ α = \(\frac{\pi}{2}\) – β
sin α = sin(\(\frac{\pi}{2}\) – β) ⇒ sin α = cos β
= \(\frac{5}{x}\) = \(\sqrt{1-\left(\frac{12}{x}\right)^{2}}\)
⇒ \(\frac{25}{x^{2}}\) = 1 – \(\frac{144}{x^{2}}\)
⇒ \(\frac{169}{x^{2}}\) = 1 ⇒ x² = 169 ⇒ x = ± 13
⇒ x = 13 (∵ x > 0)

Question 17.
Solve sin^-1 (\(\frac{3x}{5}\)) + sin^-1 (\(\frac{4x}{5}\)) = sin^-1(x)
Solution:
Let sin^-1 (\(\frac{3x}{5}\)) = α, sin^-1 (\(\frac{4x}{5}\)) = β and sin^-1(x) = γ
Then sin α = \(\frac{3x}{5}\), sin β = \(\frac{3x}{5}\) and sin γ = x
⇒ cos α = \(\sqrt{1-\frac{9 x^{2}}{25}}\), cos β = \(\sqrt{1-\frac{16 x^{2}}{25}}\) and cos γ = \(\sqrt{1-x^{2}}\)
Now α + β = γ
⇒ sin (α + β) = sin γ
⇒ sinα cos β + cos α sin β = sin γ

Squaring on both sides
16(25 – 9x²) = 625 – 150\(\sqrt{25-16 x^{2}}\) + 9(25 – 16x²)
⇒ 400 – 144x² = 625 – 150\(\sqrt{25-16 x^{2}}\) + 225 – 144x²
⇒ 150\(\sqrt{25-16 x^{2}}\) = 225 + 225
⇒ \(\sqrt{25-16 x^{2}}\) = 3
⇒ 25 – 16x² = 9
⇒ 16x² = 16 ⇒ x = ± 1
∴ x = 0, + 1, – 1.
All these values of x satisfy the given equation.

Question 18.
Solve sin^-1 x + sin^-1 2x = \(\frac{\pi}{3}\)
Solution:
Given that sin^-1 x + sin^-1 2x = \(\frac{\pi}{3}\)
⇒ cos (sin^-1x + sin^-1 2x) = cos(\(\frac{\pi}{3}\))

But when x = –\(\frac{\sqrt{3}}{2 \sqrt{7}}\),. both sin^-1x and sin^-1(2x) are negative. The given equation does not satisfy.
Hence x = \(\frac{\sqrt{3}}{2 \sqrt{7}}\) is the only solution.

Question 19.
If sin [2 cos^-1 {cot (2 tan^-1 x)}] = 0, find x
Solution:
sin [2cos^-1 {cot (2 tan^-1 x)}] = 0
⇔ 2 cos^-1 [cot (2tan^-1 x)] = 0 or π or 2π
(since the range of cos^-1 is (0, π)
⇔ cos^-1 [cot (2 tan^-1 x)] = 0 or \(\frac{\pi}{2}\) or π
⇒ cot (2 tan^-1 x) = 1 or 0 or -1
⇒ 2 tan^-1 x = ±\(\frac{\pi}{4}\) or ±\(\frac{\pi}{2}\) (or) ±\(\frac{3\pi}{4}\)
∴ tan^-1 (x) = ±\(\frac{\pi}{8}\) (or) ±\(\frac{\pi}{4}\) (or) ± \(\frac{3\pi}{8}\)
x = ± (\(\sqrt{2}\) – 1) (or) ±1 (or) ± (\(\sqrt{2}\) + 1)

Question 20.
Prove that cos^-1 [tan^-1 (sin (cot^-1x)}] = \(\sqrt{\frac{x^{2}+1}{x^{2}+2}}\)
Solution:
Let cot^-1 (x) = θ,
then cot θ = x and 0 < x < π

Students get through Maths 1A Important Questions Inter 1st Year Maths 1A Products of Vectors Important Questions which are most likely to be asked in the exam.

Intermediate 1st Year Maths 1A Products of Vectors Important Questions

Question 1.
If \(\bar{a}\) = 6\(\bar{i}\) + 2\(\bar{j}\) + 3\(\bar{k}\) and \(\bar{b}\) = 2\(\bar{i}\) – 9\(\bar{j}\) + 6\(\bar{k}\), then find \(\bar{a}\) . \(\bar{b}\) and the angle between \(\bar{a}\) and \(\bar{b}\).
Solution:

Question 2.
If a = i + 2j – 3k, b = 3i – j + 2k then show that a + b and a – b are perpendicular to each other. (A.P) (Mar. ’15, May ’11)
Solution:
a + b = (1 + 3)i + (2 – 1)j+ (-3 + 2) k
= 4i + j – k
Similarly a – b = – 2i + 3j – 5k
(a + b). (a – b) = 4(-2) + 1(3) + (-1)(-5) = 0
Hence (a + b) and (a – b) are mutually perpendicular.

Question 3.
Let \(\bar{a}\) and \(\bar{b}\) be non- zero, non-collinear vectors. If |\(\bar{a}\) + \(\bar{b}\) | = | \(\bar{a}\) – \(\bar{b}\) |, then find the angle between \(\bar{a}\) and \(\bar{b}\).
Solution:
| \(\bar{a}\) + \(\bar{b}\) |² = | a – b |².
⇒ |\(\bar{a}\)|² + |\(\bar{b}\)|² + 2 (\(\bar{a}\). \(\bar{b}\))
⇒ |\(\bar{a}\)|² + |\(\bar{b}\)|² – 2 (\(\bar{a}\). \(\bar{b}\))
⇒ 4(\(\bar{a}\) . \(\bar{b}\)) = 0 ⇒ \(\bar{a}\) . \(\bar{b}\) = 0
⇒ Angle between \(\bar{a}\) and \(\bar{b}\) is 90°.

Question 4.
If |\(\bar{a}\)| = 11, |\(\bar{b}\)| = 23 and | \(\bar{a}\) – \(\bar{b}\) | = 30, then find the angle between the vectors \(\bar{a}\), \(\bar{b}\) and also find | \(\bar{a}\) + \(\bar{b}\) |.
Solution:
Given that | \(\bar{a}\) | = 11, | \(\bar{b}\) | = 23 and | \(\bar{a}\) – \(\bar{b}\) | = 30, Let (\(\bar{a}\), \(\bar{b}\)) = θ.

Question 5.
If \(\bar{a}\) = \(\bar{i}\) – \(\bar{j}\) – \(\bar{k}\) and \(\bar{b}\) = 2\(\bar{i}\) – 3\(\bar{j}\) + \(\bar{k}\), then find the projection vector of \(\bar{b}\) on \(\bar{a}\) and its magnitude.
Solution:

Question 6.
If P, Q, R and S are points whose position vectors are \(\bar{i}\) – \(\bar{k}\), –\(\bar{i}\) +2\(\bar{j}\), 2\(\bar{i}\) – 3\(\bar{k}\) and 3\(\bar{i}\) – 2\(\bar{j}\) – \(\bar{k}\) respectively, then find the component of \(\overline{\mathbf{R S}}\) on \(\overline{\mathbf{P Q}}\).
Solution:
Let ‘O’ be the origin.
\(\overline{\mathrm{OP}}\) = \(\overline{\mathbf{i}}\) – \(\overline{\mathbf{k}}\), \(\overline{\mathrm{OQ}}\) = –\(\overline{\mathbf{i}}\) + 2\(\overline{\mathbf{j}}\), \(\overline{\mathrm{OR}}\) = 2\(\overline{\mathbf{i}}\) – 3\(\overline{\mathbf{k}}\) and \(\overline{\mathrm{OR}}\) = 3\(\overline{\mathbf{i}}\) – 2\(\overline{\mathbf{j}}\) – \(\overline{\mathbf{k}}\)

The component of \(\overline{\mathrm{RS}}\) and \(\overline{\mathrm{PQ}}\)

Question 7.
If the vectors λ\(\overline{\mathbf{i}}\) – 3\(\overline{\mathbf{j}}\) + 5\(\overline{\mathbf{k}}\) and 2 λ\(\overline{\mathbf{i}}\) – λ\(\overline{\mathbf{j}}\) + \(\overline{\mathbf{k}}\) are perpendicular to each other, find λ. (T.S) (Mar. ’16)
Solution:
Since the vectors are perpendicular
⇒ (λ\(\overline{\mathbf{i}}\) – 3\(\overline{\mathbf{j}}\) + 5\(\overline{\mathbf{k}}\)) . (2λ\(\overline{\mathbf{i}}\) – λ\(\overline{\mathbf{j}}\) – \(\overline{\mathbf{k}}\)) = 0
⇒ λ(2λ) + (-3)(-λ) + 5(-1) = 0
⇒ 2λ² + 3λ – 5 = 0
⇒ (2λ + 5)(λ – 1) = 0
∴ λ = 1 (or) \(\frac{-5}{2}\)

Question 8.
Let \(\overline{\mathbf{a}}\) = 2\(\overline{\mathbf{i}}\) + 3\(\overline{\mathbf{j}}\) + \(\overline{\mathbf{k}}\), \(\overline{\mathbf{b}}\) = 4\(\overline{\mathbf{i}}\) + \(\overline{\mathbf{j}}\) and \(\overline{\mathbf{c}}\) = \(\overline{\mathbf{i}}\) – 3\(\overline{\mathbf{j}}\) – 7\(\overline{\mathbf{k}}\). Find the vector \(\overline{\mathbf{r}}\) such that \(\overline{\mathbf{r}}\) . \(\overline{\mathbf{a}}\) = 9, \(\overline{\mathbf{r}}\) . \(\overline{\mathbf{b}}\) = 7 and \(\overline{\mathbf{r}}\) . \(\overline{\mathbf{c}}\) = 6.
Solution:
Let \(\overline{\mathbf{r}}\) = x\(\overline{\mathbf{i}}\) + y\(\overline{\mathbf{j}}\) + z\(\overline{\mathbf{k}}\)
∴ \(\overline{\mathbf{r}}\) . \(\overline{\mathbf{a}}\) = 9 ⇒ 2x + 3y + z = 9
\(\overline{\mathbf{r}}\) . \(\overline{\mathbf{b}}\) = 7 ⇒ 4x + y = 7 and \(\overline{\mathbf{r}}\) . \(\overline{\mathbf{c}}\) = 6 ⇒ x – 3y – 7z = 6
By solving these equations, we get x = 1, y = 3 and z = -2
∴ \(\overline{\mathbf{r}}\) = \(\overline{\mathbf{i}}\) + 3\(\overline{\mathbf{j}}\) – 2\(\overline{\mathbf{k}}\)

Question 9.
Show that the points 2\(\overline{\mathbf{i}}\) – \(\overline{\mathbf{j}}\) + \(\overline{\mathbf{k}}\), \(\overline{\mathbf{i}}\) – 3\(\overline{\mathbf{j}}\) – 5\(\overline{\mathbf{k}}\) and 3\(\overline{\mathbf{i}}\) – 4\(\overline{\mathbf{j}}\) – 4\(\overline{\mathbf{k}}\) are the vertices of a right angle triangle. Also, find the other angles.
Solution:
Let ‘O’ be the origin and A, B, C be the given points, then

Question 10.
Prove that the angle θ between any two diagonals of a cube is given by cos θ = \(\frac{1}{3}\). (May ’11)
Solution:
Let us suppose that the cube is a unit cube

Question 11.
Let \(\overline{\mathbf{a}}\), \(\overline{\mathbf{b}}\), \(\overline{\mathbf{c}}\) be non—zero mutually orthogonal vectors, if x \(\overline{\mathbf{a}}\) + y\(\overline{\mathbf{b}}\) + z = 0, then x = y = z = 0.
Solution:
x\(\overline{\mathbf{a}}\) + y\(\overline{\mathbf{b}}\) + z\(\overline{\mathbf{c}}\) = 0
⇒ \(\overline{\mathbf{a}}\) • (x\(\overline{\mathbf{a}}\) + y\(\overline{\mathbf{b}}\) + z\(\overline{\mathbf{c}}\)) = 0
⇒ x(\(\overline{\mathbf{a}}\) • \(\overline{\mathbf{a}}\)) = 0
⇒ x = 0 (∵ \(\overline{\mathbf{a}}\) • \(\overline{\mathbf{a}}\) ≠ 0)
Similarly y = 0, z = 0.

Question 12.
Let \(\overline{\mathbf{a}}\), \(\overline{\mathbf{b}}\) and \(\overline{\mathbf{c}}\) be mutually orthogonal vectors of equal magnitudes. Prove that the vector \(\overline{\mathbf{a}}\) + \(\overline{\mathbf{b}}\) + \(\overline{\mathbf{c}}\) is equally inclined to each of \(\overline{\mathbf{a}}\), \(\overline{\mathbf{b}}\) and \(\overline{\mathbf{c}}\), the angle of inclination being cos^-1\(\left(\frac{1}{\sqrt{3}}\right)\).
Solution:

Similarly, it can be proved that \(\overline{\mathbf{a}}\) + \(\overline{\mathbf{b}}\) + \(\overline{\mathbf{c}}\) inclines at an angle of cos^-1\(\left(\frac{1}{\sqrt{3}}\right)\) with \(\overline{\mathbf{b}}\) and \(\overline{\mathbf{c}}\).

Question 13.
The vectors \(\overline{\mathbf{A B}}\) = 3\(\bar{i}\) – 2\(\bar{j}\) + 2\(\bar{k}\) and \(\overline{\mathbf{A D}}\) = \(\bar{i}\) – 2\(\bar{k}\) represent the adjacent sides of a parallelogram ABCD. Find the angle between the diagonals.
Solution:

Question 14.
For any two vectors a and b, show that
i) |a • b| (Cauchy – Schwartz inequality)
ii) || a + b | ≤ ||a|| + ||b| (triangle inequality
Solution:
i)
If a = 0 or b = 0, the inequalities hold trivially.

ii)

Question 15.
Find the cartesian equation of the plane passing through the point (-2, 1, 3) and perpendicular to the vector 3\(\bar{i}\) + \(\bar{j}\) + 5\(\bar{k}\).
Solution:
Let A = -2\(\bar{i}\) + \(\bar{j}\) + 3\(\bar{k}\) and \(\bar{r}\) = x\(\bar{i}\) + y\(\bar{j}\) + z\(\bar{k}\) be any point in the plane
∴ \(\overline{\mathrm{AP}}\) = (x + 2)\(\bar{i}\) + (y – 1)\(\bar{j}\) + (z – 3)\(\bar{k}\)
∴ \(\overline{\mathrm{AP}}\) is perpendicular to 3\(\bar{i}\) + \(\bar{j}\) + 5\(\bar{k}\)
⇒ (x + 2)3 + (y – 1)1 + (z – 3)5 = 0
⇒ 3x + y + z – 10 = 0

Question 16.
Find the cartesian equation of the plane through the point A (2, -1, -4) and parallel to the plane 4x – 12y – 3z – 7 = 0.
Solution:
The equation of the plane parallel to
4x – 12y – 3z – 7 = 0 be
4x -12y – 3z = p
∴ If passes through the point A(2, -1, -4)
⇒ 4(2) – 12(-1) – 3(-4) = p
⇒ p = 32
∴ The equation of the required plane is
4x – 12y – 3z = 32

Question 17.
Find the angle between the planes 2x – 3y – 6z = 5 and 6x + 2y – 9z = 4.
Solution:
Equation to the planes is 2x – 3y – 6z = 5
⇒ \(\overline{\mathbf{r}}\) .(2\(\overline{\mathbf{i}}\) – 3\(\overline{\mathbf{j}}\) – 6\(\overline{\mathbf{k}}\)) = 5 ———- (1)
and 6x + 2y – 9z = 4 —— (2)
∴ The angle between the planes \(\overline{\mathbf{r}} \cdot \overline{\mathbf{n}}_{1}\) = p, and \(\overline{\mathbf{r}} \cdot \overline{\mathbf{n}}_{2}\) = q is θ, then

Question 18.
Find unit vector orthogonal to the vector 3\(\bar{i}\) + 2\(\bar{j}\) +6\(\bar{k}\) and coplanar with thevectors 2\(\bar{i}\) + \(\bar{j}\) + \(\bar{k}\) and \(\bar{i}\) – \(\bar{j}\) + \(\bar{k}\).
Solution:

⇒ (2x + y)3 + (x – y)2 + (x + y)6 = 0
⇒ 6x + 3y + 2x – 2y + 6x + 6y = 0
⇒ 14x + 7y = 0
⇒ y = -2x —— (1)
∵ |\(\overline{\mathbf{r}}\)| = 1
⇒ (2x + y)² + (x – y)² + (x + y)² = 1
⇒ 4x² + 4xy + 9 + 2(x² + y²) = 1
⇒ 6x² + 4x (-2x) + 3(-2x)² = 1 (By using (1))
⇒ 10x² = 1

Question 19.
If \(\overline{\mathbf{a}}\) = 2\(\overline{\mathbf{i}}\) – 3\(\overline{\mathbf{j}}\) + 5\(\overline{\mathbf{k}}\), \(\overline{\mathbf{b}}\) = –\(\overline{\mathbf{i}}\) + 4\(\overline{\mathbf{j}}\) + 2\(\overline{\mathbf{k}}\), then find \(\overline{\mathbf{a}}\) × \(\overline{\mathbf{b}}\) and unit vector perpendicular to both \(\overline{\mathbf{a}}\) and \(\overline{\mathbf{b}}\).
Solution:
\(\overline{\mathbf{a}}\) = 2\(\overline{\mathbf{i}}\) – 3\(\overline{\mathbf{j}}\) + 5\(\overline{\mathbf{k}}\) and \(\overline{\mathbf{b}}\) = –\(\overline{\mathbf{i}}\) + 4\(\overline{\mathbf{j}}\) + 2\(\overline{\mathbf{k}}\)

Question 20.
If \(\bar{a}\) = 2\(\bar{i}\) – 3\(\bar{j}\) + 5\(\bar{k}\), \(\bar{b}\) = –\(\bar{i}\) + 4\(\bar{j}\) + 2\(\bar{k}\), then find (\(\bar{a}\) + \(\bar{b}\)) × (\(\bar{a}\) – \(\bar{b}\)) and unit vector perpendicular to both \(\bar{a}\) + \(\bar{b}\) and \(\bar{a}\) – \(\bar{b}\).
Solution:

Question 21.
Find the area of the parallelogram for which the vectors \(\bar{a}\) = 2\(\bar{i}\) – 3\(\bar{j}\) and \(\bar{b}\) = 3\(\bar{i}\) – \(\bar{k}\) are adjacent sides. (Mar. ; May ‘08)
Solution:
The vector area of the given parallelogram

Question 22.
If \(\overline{\mathbf{a}}\), \(\overline{\mathbf{b}}\), \(\overline{\mathbf{c}}\) and \(\overline{\mathbf{d}}\) are vectors such that \(\overline{\mathbf{a}}\) × \(\overline{\mathbf{b}}\) = \(\overline{\mathbf{c}}\) × \(\overline{\mathbf{d}}\) and \(\overline{\mathbf{a}}\) × \(\overline{\mathbf{c}}\) = \(\overline{\mathbf{b}}\) × \(\overline{\mathbf{d}}\) then show that the vectors \(\overline{\mathbf{a}}\) – \(\overline{\mathbf{d}}\) and \(\overline{\mathbf{b}}\) – \(\overline{\mathbf{c}}\) are parallel.
Solution:

Question 23.
If \(\overline{\mathbf{a}}\) = \(\overline{\mathbf{i}}\) + 2\(\overline{\mathbf{j}}\) + 3\(\overline{\mathbf{k}}\) and \(\overline{\mathbf{b}}\) = 3\(\overline{\mathbf{i}}\) + 5\(\overline{\mathbf{j}}\) – \(\overline{\mathbf{k}}\) are two sides of a triangle, then find
its area.
Solution:
Area of the triangle = \(\frac{1}{2}|\bar{a} \times \bar{b}|\)

Question 24.
In Δ ABC, if \(\overline{\mathrm{BC}}\) = \(\bar{a}\), \(\overline{\mathrm{CA}}\) = \(\bar{b}\), \(\overline{\mathrm{AB}}\) = \(\bar{c}\) then show that \(\bar{a}\) × \(\bar{b}\) = \(\bar{b}\) × \(\bar{c}\) = \(\bar{c}\) × \(\bar{a}\).
Solution:

Question 25.
Let \(\bar{a}\) = 2\(\bar{i}\) – \(\bar{j}\) + \(\bar{k}\) and \(\bar{b}\) = 3\(\bar{i}\) + 4\(\bar{j}\) – \(\bar{k}\). If θ is the angle between \(\bar{a}\) and \(\bar{b}\) then find sin θ.
Solution:

Question 26.
Let \(\bar{a}\), \(\bar{b}\) and \(\bar{c}\) be such that \(\bar{c}\) ≠ 0,
\(\bar{a}\) × \(\bar{b}\) = \(\bar{c}\), \(\bar{b}\) × \(\bar{c}\) = \(\bar{a}\). Show that \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) are pair wise orthogonal vectors and |\(\bar{b}\)| = 1, |\(\bar{c}\)| = |\(\bar{a}\)|.
Solution:

Question 27.
Let \(\overline{\mathbf{a}}\) = 2\(\overline{\mathbf{i}}\) + \(\overline{\mathbf{j}}\) – 2\(\overline{\mathbf{k}}\) ; \(\overline{\mathbf{b}}\) = \(\overline{\mathbf{i}}\) + \(\overline{\mathbf{j}}\). If \(\overline{\mathbf{c}}\) is a vector such that \(\overline{\mathbf{a}}\) . \(\overline{\mathbf{c}}\) = |\(\overline{\mathbf{c}}\)|, |\(\overline{\mathbf{c}}\) – \(\overline{\mathbf{a}}\)| = \(2 \sqrt{2}\) and the angle between \(\overline{\mathbf{a}}\) × \(\overline{\mathbf{b}}\) = \(2 \sqrt{2}\) and the angle between \(\overline{\mathbf{a}}\) × \(\overline{\mathbf{b}}\) and \(\overline{\mathbf{c}}\) is 30°, then find the value of |(\(\overline{\mathbf{a}}\) × \(\overline{\mathbf{b}}\)) × \(\overline{\mathbf{c}}\)|
Solution:

Question 28.
Let \(\overline{\mathbf{a}}\), \(\overline{\mathbf{b}}\) be two non-collinear unit vectors. If \(\bar{\alpha}\) = \(\overline{\mathbf{a}}\) – (\(\overline{\mathbf{a}}\) • \(\overline{\mathbf{b}}\))\(\overline{\mathbf{b}}\) and \(\bar{\beta}\) =
= \(\overline{\mathbf{a}}\) × \(\overline{\mathbf{b}}\), then show that |\(\overline{\boldsymbol{\beta}}\)| = |\(\bar{\alpha}\)|.
Solution:

Question 29.
A non-zero vector \(\bar{a}\) is parallel to the line of intersection of the plane determined by the vectors \(\bar{i}\), \(\bar{i}\) + \(\bar{j}\) and the plane determined by the vectors \(\bar{i}\) – \(\bar{j}\), \(\bar{i}\) + \(\bar{k}\). Find the angle between \(\bar{a}\) and the vector \(\bar{i}\) – 2\(\bar{j}\) + 2\(\bar{k}\).
Solution:
Let l be the line of intersection of the planes determined by the pairs \(\bar{i}\), \(\bar{i}\) + \(\bar{j}\) and \(\bar{i}\) – \(\bar{j}\), \(\bar{i}\) + \(\bar{k}\)

∴ \(\bar{n}_{1}\) is perpendicular to l and \(\bar{n}_{2}\) is also perpendicular to l.
∴ \(\bar{a}\) is parallel to the line l, follows that \(\bar{a}\) is perpendicularto both \(\bar{n}_{1}\), and \(\bar{n}_{2}\)

Question 30.
Let \(\overline{\mathbf{a}}\) = 4\(\overline{\mathbf{i}}\) + 5\(\overline{\mathbf{j}}\) – \(\overline{\mathbf{k}}\), \(\overline{\mathbf{b}}\) = \(\overline{\mathbf{i}}\) – 4\(\overline{\mathbf{j}}\) + 5\(\overline{\mathbf{k}}\) and \(\overline{\mathbf{c}}\) = 3\(\overline{\mathbf{i}}\) + \(\overline{\mathbf{j}}\) – \(\overline{\mathbf{k}}\). Find the vector \(\bar{\alpha}\) which is perpendicular to both \(\overline{\mathbf{a}}\) and \(\overline{\mathbf{b}}\) and \(\bar{\alpha} \cdot \mathbf{c}\) = 21.
Solution:

Question 31.
For any vector a, show that |a × i|² + |a × j|² + |a × k|² = 2|a|².
Solution:
Let a = x i + y j + z k.
Then a × i = -yk + zj.
∴ |a × i|² = y² + z²
Similarly |a × j|² = z² + x² and |a × k|² = x²
+ y²
∴ |a × i|² + |a × j|² + |a × k|² = 2(x² + y² + z²) = 2|a|²

Question 32.
If \(\overline{\mathbf{a}}\) is a non-zero vector and \(\overline{\mathbf{b}}\), \(\overline{\mathbf{c}}\) are two vector such that \(\overline{\mathbf{a}}\) × \(\overline{\mathbf{b}}\) = \(\overline{\mathbf{a}}\) × \(\overline{\mathbf{c}}\) and \(\overline{\mathbf{a}}\) • \(\overline{\mathbf{b}}\) = \(\overline{\mathbf{a}}\) • \(\overline{\mathbf{c}}\), then prove that \(\overline{\mathbf{b}}\) = \(\overline{\mathbf{c}}\).
Solution:

Question 33.
Prove that the vectors \(\bar{a}\) = 2\(\bar{i}\) – \(\bar{j}\) + \(\bar{k}\), \(\bar{b}\) = \(\bar{i}\) – 3\(\bar{j}\) – 5\(\bar{k}\) and \(\bar{c}\) = 3\(\bar{i}\) – 4\(\bar{j}\) – 4\(\bar{k}\) are coplanar.
Solution:

Question 34.
Find the volume of the parallelopiped whose cotersminus edges are represented by the vectors 2\(\bar{i}\) – 3\(\bar{j}\) + \(\bar{k}\), \(\bar{i}\) – \(\bar{j}\) + 2\(\bar{k}\) and 2\(\bar{i}\) + \(\bar{j}\) – \(\bar{k}\).
Solution:
Let \(\bar{a}\) = 2\(\bar{i}\) – 3\(\bar{j}\) + \(\bar{k}\)

Question 35.
If the vectors \(\bar{a}\) = 2\(\bar{i}\) – \(\bar{j}\) + \(\bar{k}\), \(\bar{b}\) = \(\bar{i}\) + 2\(\bar{j}\) – 3\(\bar{k}\) and \(\bar{c}\) = 3\(\bar{i}\) + p\(\bar{j}\) + 5\(\bar{k}\) are coplanar, then find p.
Solution:

Question 36.
Show that
\(\overline{\mathbf{i}}\) × (\(\overline{\mathbf{a}}\) × \(\overline{\mathbf{i}}\)) + \(\overline{\mathbf{j}}\) × (\(\overline{\mathbf{a}}\) × \(\overline{\mathbf{j}}\)) + k × (\(\overline{\mathbf{a}}\) × \(\overline{\mathbf{k}}\)) = 2\(\overline{\mathbf{a}}\) for any vector \(\overline{\mathbf{a}}\).
Solution:

Question 37.
Prove that for any three vectors \(\overline{\mathbf{a}}\), \(\overline{\mathbf{b}}\), \(\overline{\mathbf{c}}\), [\(\overline{\mathbf{b}}\) + \(\overline{\mathbf{c}}\) \(\overline{\mathbf{c}}\) + \(\overline{\mathbf{a}}\) \(\overline{\mathbf{a}}\) + \(\overline{\mathbf{b}}\)] = 2[\(\overline{\mathbf{a}}\) \(\overline{\mathbf{b}}\) \(\overline{\mathbf{c}}\)]
Solution:

Question 38
For any three vectors \(\overline{\mathbf{a}}\), \(\overline{\mathbf{b}}\), \(\overline{\mathbf{c}}\) prove that [\(\overline{\mathbf{b}}\) × \(\overline{\mathbf{c}}\) \(\overline{\mathbf{c}}\) × \(\overline{\mathbf{a}}\) \(\overline{\mathbf{a}}\) × \(\overline{\mathbf{b}}\)] = [\(\overline{\mathbf{a}}\) \(\overline{\mathbf{b}}\) \(\overline{\mathbf{c}}\)]²
Solution:

Question 39.
Let \(\overline{\mathbf{a}}\), \(\overline{\mathbf{b}}\) and \(\overline{\mathbf{c}}\) be unit vectors such that \(\overline{\mathbf{b}}\) is not parallel to \(\overline{\mathbf{c}}\) and \(\overline{\mathbf{a}}\) × (\(\overline{\mathbf{b}}\) × \(\overline{\mathbf{c}}\)) = \(\frac{1}{2} \bar{b}\). Find the angles made by \(\overline{\mathbf{a}}\) with each of \(\overline{\mathbf{b}}\) and \(\overline{\mathbf{c}}\).
Solution:

Since \(\overline{\mathbf{b}}\) and \(\overline{\mathbf{c}}\) are non—collinear vectors. Equating corresponding coeffs. on both sides

Question 40.
Let \(\overline{\mathbf{a}}\) = \(\overline{\mathbf{i}}\) + \(\overline{\mathbf{j}}\) + \(\overline{\mathbf{k}}\), \(\overline{\mathbf{b}}\) = 2\(\overline{\mathbf{i}}\) – \(\overline{\mathbf{j}}\) + 3\(\overline{\mathbf{k}}\), \(\overline{\mathbf{c}}\) = \(\overline{\mathbf{i}}\) – \(\overline{\mathbf{j}}\) and \(\overline{\mathbf{d}}\) = 6\(\overline{\mathbf{i}}\) + 2\(\overline{\mathbf{j}}\) + 3\(\overline{\mathbf{k}}\). Express \(\overline{\mathbf{d}}\) interms of \(\overline{\mathbf{b}}\) × \(\overline{\mathbf{c}}\), \(\overline{\mathbf{c}}\) × \(\overline{\mathbf{a}}\) and \(\overline{\mathbf{a}}\) × \(\overline{\mathbf{b}}\)  (May. ’12)
Solution:

Then from the above problem,

Question 41.
For any four vectors \(\overline{\mathbf{a}}\), \(\overline{\mathbf{b}}\), \(\overline{\mathbf{c}}\) and \(\overline{\mathbf{d}}\) prove that
(\(\overline{\mathbf{b}}\) × \(\overline{\mathbf{c}}\)) . (\(\overline{\mathbf{a}}\) × \(\overline{\mathbf{d}}\)) + (\(\overline{\mathbf{c}}\) × \(\overline{\mathbf{a}}\)) . (\(\overline{\mathbf{b}}\) × \(\overline{\mathbf{d}}\)) + (\(\overline{\mathbf{a}}\) × \(\overline{\mathbf{b}}\)) . (\(\overline{\mathbf{c}}\) × \(\overline{\mathbf{d}}\)) = 0.
Solution:

Question 42.
Find the equation of the plane passing through the points A = (2, 3, -1), B = (4, 5, 2) and C = (3, 6, 5).
Solution:
Let ‘O’ be the origin

Let P be any point on the plane passing through the points A, B, C.

Hence, equation of the required plane is

Question 43.
Find the equation of the plane passing through the point A (3, -2, -1) and parallel to the vector \(\bar{b}\) = \(\bar{i}\) – 2\(\bar{j}\) + 4\(\bar{k}\) and \(\bar{c}\) = 3\(\bar{i}\) + 2\(\bar{j}\) – 5\(\bar{k}\).
Solution:
The required plane passes through A = (3, -2, -1) and is parallel to the vectors
\(\bar{b}\) = \(\bar{i}\) – 2\(\bar{j}\) + 4\(\bar{k}\) and \(\bar{c}\) = 3\(\bar{i}\) + 2\(\bar{j}\) – 5\(\bar{k}\). Hence if P is a point in the plane, then AP, b and C are coplanar and [AP b c] = 0.

i.e., 2x + 17y + 8z + 36 = 0 is the equation of the required plane.

Question 44.
Find the vector equation of the plane passing through the intersection of the planes r. (i + j + k) = 6 and r. (2i + 3j+ 4k) = -5 and the point (1, 1, 1).
Solution:

Substituting this value of λ in equation (1), we get

Question 45.
Find the distance of a point (2, 5, -3) from the plane r. (6i – 3j + 2k) = 4.
Solution:
Here a = 2i + 5j – 3k, N = 6i – 3j + 2k and d = 4.
∴ The distance of the point (2, 5, -3) from the given plane is

Question 46.
Find the angle between the line \(\frac{x+1}{2}\) = \(\frac{y}{3}\) = \(\frac{z-3}{6}\) and the plane 10x + 2y – 11z = 3.
Solution:
Let φ be the angle between the given line and the normal to the plane.
Converting the given equations into vector form, we have
r = (-i + 3k) + λ(2i + 3j + 6k)
and r. (10i + 2j – 11k) = 3
Here b = 2i + 3j + 6k and n = 10i + 2j – 11k

Question 47.
For any four vectors a, b, c and d, (a × b) × (c × d) = [a c d]b – [b c d]a and (a × b) × (c × d) = [a b d]c – [a b c]d.
Solution:
Let m = c × d
∴ (a × b) × (c × d) = (a × b) × m
= (a.m)b – (b.m)a
= (a. (c × d))b – (b. (c × d))a
= [a c d]b – [b c d]a.
Again Let a × b = n.
Then(a × b) × (c × d) = n × (c × d)
= (n .d)c – (n.c)d
= ((a × b). d) c – ((a × b). c)d
= [a b d]c – [a b c]d.

Question 48.
Find the shortest distance between the skew line r = (6i + 2j + 2k) + t(i – 2j + 2k) and r = (-4i – k) + s (3i – 2j – 2k).
Solution:
The first line passes through the point
A(6, 2, 2) and is parallel to the vector b = i – 2j + 2k. Second line passes through the point C(-4, 0, -1) and is parallel to the vector d = 3i – 2j – 2k

Question 49.
The angle ¡n a semi circle is a right angle. (May ’08)
Solution:
Let O be the centre and AOB be the diameter of the given šemi circle. Let P be a point on the semi circle. With reference to O as the origin, if the position vectors of A and P are taken as \(\overline{\mathrm{a}}\) and \(\overline{\mathrm{p}}\),then

Question 50.
The altitude of a triangle are concurrent.
ii) The perpendicular bisectors of the sides of a triangle are concurrent.
Solution:
i) In the given triangle ABC, let the altitudes from A, B drawn to BC. CA respectively intersect them at D, E. Assume that AD and BE intersect at O, join CO and produce it to meet AB at F. With reference to O, let the position vectors of A, B and C be a, b and c respectively.

From Fig. we have

ii) In the given ABC, let the mid-points of BC, CA and AB be D, E and F respectively. Let the perpendicular bisectors drawn to BC and CA at D and E meet at O. Join OF. With respect to O, let the position vectors of A, B and C be \(\bar{a}\), \(\bar{b}\) and \(\bar{c}\) respectively.


On adding equations (1) and (2), we obtain

Question 51.
Show that the vector area of the quadrilateral ABCD having diagonals \(\overline{\mathbf{A C}}\), \(\overline{\mathbf{B D}}\) is \(\frac{1}{2}(\overline{A C} \times \overline{B D})\)
Solution:
If the point of intersection of the diagonals AC, BD of the given quadrilateral ABCD is assumed as Q, the vector area of the quadrilateral ABCD.

= Sum of the vector areas of ΔQAB, ΔQBC, ΔQCD and ΔQDA.

Question 52.
For any vectors a, b and c (a × b) × c = (a.c) b – (b.c) a (AP) (Mar. 15 May ‘06; June ’04)
Solution:
Equation (1) is evidently true if a and b are parallel.
Now suppose that a and b are non-parallel.
Let O denote the origin. Choose points A and B such that \(\overline{\mathrm{OA}}\) = \(\overline{\mathbf{a}}\) and \(\overline{\mathrm{OB}}\) = \(\overline{\mathbf{b}}\). Since \(\overline{\mathbf{a}}\) and \(\overline{\mathbf{b}}\) are non-parallel, the points O, A and B are non-collinear. Hence they determine a plane.
Let i denote the unit vector along \(\overline{\mathrm{OA}}\).
Let j be a unit vector in the OAB plane perpendicular to i. Let k = i × j. Then

Question 53.
For any four vectors a, b, c and d(a × b). (c × d) = \(\left|\begin{array}{cc}
a \cdot c & a \cdot d \\
b . c & b . d
\end{array}\right|\) and in particular (a × b)² = a²b² – (a. b)².
Solution:

Question 54.
In the above fou rmula if \(\overline{\mathbf{a}}\), \(\overline{\mathbf{b}}\), \(\overline{\mathbf{c}}\) and \(\overline{\mathbf{d}}\)(\(\overline{\mathbf{a}}\) × \(\overline{\mathbf{b}}\)). (\(\overline{\mathbf{c}}\) × \(\overline{\mathbf{d}}\)) = (\(\overline{\mathbf{a}}\) . \(\overline{\mathbf{c}}\))(\(\overline{\mathbf{b}}\) . \(\overline{\mathbf{d}}\)) – (\(\overline{\mathbf{a}}\) . \(\overline{\mathbf{d}}\))(\(\overline{\mathbf{b}}\) . \(\overline{\mathbf{c}}\)) = \(\left|\begin{array}{ll}
\overline{\mathbf{a}} \cdot \overline{\mathbf{c}} & \overline{\mathbf{a}} \cdot \overline{\mathbf{d}} \\
\overline{\mathbf{b}} \cdot \overline{\mathbf{c}} & \bar{b} \cdot \overline{\mathbf{d}}
\end{array}\right|\).
Solution:

Since this is a scalar, this is also called the scalar product of four vectors.

Students get through Maths 1A Important Questions Inter 1st Year Maths 1A Addition of Vectors Important Questions which are most likely to be asked in the exam.

Intermediate 1st Year Maths 1A Addition of Vectors Important Questions

Question 1.
Find unit vector in the direction of vector a = 2i + 3j + k. (Mar. 14)
Solution:
The unit vector in the direction of a vector a is given by \(\hat{a}\) = \(\frac{1}{|a|} a\)
Now |a| = \(\sqrt{2^{2}+3^{2}+1^{2}}\) = \(\sqrt{14}\).
∴ \(\hat{a}\) = \(\frac{1}{\sqrt{14}}\)(2i + 3j + k)
= \(\frac{2}{\sqrt{14}} i\) + \(\frac{3}{\sqrt{14}} \mathrm{j}\) + \(\frac{1}{\sqrt{14}} k\)

Question 2.
Find a vector in the direction of vector a = i – 2j that has magnitude 7 units.
Solution:
The unit vector in the direction of the given vector a is
\(\hat{a}\) = \(\frac{1}{|\mathrm{a}|} \mathrm{a}\) = \(\frac{1}{\sqrt{5}}(i-2 j)\) = \(\frac{1}{\sqrt{5}} \mathrm{i}\) – \(\frac{2}{\sqrt{5}} \mathrm{j}\)
Therefore, the vector having a magnitude equal to 7 and in the direction of a is
7a = 7\(\left(\frac{1}{\sqrt{5}} \mathrm{i}-\frac{2}{\sqrt{5}} \mathrm{j}\right)\) = \(\frac{7}{\sqrt{5}} \mathrm{i}\) – \(\frac{14}{\sqrt{5}} j\)

Question 3.
Find the unit vector in the direction of the sum of the vectors.
a = 2i + 2j – 5k and b = 2i + j + 3k.
Solution:
The sum of the given vectors is
a + b (= c, say) = 4i + 3j – 2k and
|c| = \(\sqrt{4^{2}+3^{2}+(-2)^{2}}\) = \(\sqrt{29}\).
∴ \(\hat{c}\) = \(\frac{a+b}{|a+b|}\) = \(\frac{4 i+3 j-2 k}{\sqrt{29}}\)

Question 4.
Write direction ratios of the vector a = i + j – 2k and hence calculate its direction cosines.
Solution:
Note that direction ratios a, b, c of a vector r = xi + yj + zk are just the respective components x. y and z of the vector. So, for the given vector, we have a = 1, b = 1, c = – 2, Further, if l, m and n the direction cosines of the given vector, then
l = \(\frac{a}{|r|}\) = \(\frac{1}{\sqrt{6}}\), m = \(\frac{b}{|r|}\) = \(\frac{1}{\sqrt{6}}\),
n = \(]\frac{c}{|r|}\) = \(\frac{2}{\sqrt{6}}\) as |r| = \(\sqrt{6}\).
Thus, the direction cosines are
(\(\frac{1}{\sqrt{6}}\), \(\frac{1}{\sqrt{6}}\), \(-\frac{2}{\sqrt{6}}\)).

Question 5.
Consider two points P and Q with position vectors OP = 3a – 2b and OQ = a + b. Find the position vector of a point R which divides the line joining P and Q in the ratio 2 :1,
(i) internally and
(ii) externally.
Solution:
i) The position vector of the point R dividing the join of P and Q internally in the ratio 2 : 1 is
OR = \(\frac{2(a+b)+(3 a-2 b)}{2+1}\) = \(\frac{5 a}{3}\)

ii) The position vector of the point R dividing the join of P and Q externally in the ratio 2 : 1 is
OR = \(\frac{2(a+b)-(3 a-2 b)}{2-1}\) = 4b – a.

Question 6.
Show that the points A (2i – j + k), B(i – 3j – 5k), C(3i – 4j – 4k) are the vertices of a right angled triangle.
Solution:
We have AB = (1 – 2)i + (-3 + 1)j + (-5 – 1)
k = -i – 2j – 6k
BC = (3 – 1)i + (-4 + 3)j + (-4 + 5)k
= 2i – j + k. and CA = (2 – 3)i + (-1 + 4)j + (1 +4)k = -i + 3j + 5k.
we have |AB|² = |BC|² + |CA|².

Question 7.
Let A. B, C, D be four points with position vectors \(\overline{\mathbf{a}}\) + 2\(\overline{\mathbf{b}}\), 2\(\overline{\mathbf{a}}\) – \(\overline{\mathbf{b}}\), \(\overline{\mathbf{a}}\) and 3\(\overline{\mathbf{a}}\) + \(\overline{\mathbf{b}}\) respectively. Express the vectors \(\overline{\mathbf{A C}}\), \(\overline{\mathbf{D A}}\), \(\overline{\mathbf{B A}}\) and \(\overline{\mathbf{B C}}\) interms of \(\overline{\mathbf{a}}\) and \(\overline{\mathbf{b}}\).
Solution:
The position vectors of A, B, C, D with respect to the origin O are

Question 8.
Let ABCDEF be a regular hexagon with centre O. Show that \(\overline{\mathbf{A B}}\) + \(\overline{\mathbf{A C}}\) + \(\overline{\mathbf{A D}}\) + \(\overline{\mathbf{A E}}\) + \(\overline{\mathbf{A F}}\) = 3\(\overline{\mathbf{A D}}\) = 6\(\overline{\mathbf{A O}}\) (Mar. ’16, ’15)
Solution:
Let ABCDEF be a regular hexagon with centre O.
Then \(\overline{\mathbf{A B}}\) = \(\overline{\mathbf{B C}}\) = \(\overline{\mathbf{C D}}\) = \(\overline{\mathbf{D E}}\) = \(\overline{\mathbf{E F}}\) = \(\overline{\mathbf{F A}}\)

Question 9.
In ΔABC, If \(\overline{\mathbf{a}}\), \(\overline{\mathbf{b}}\), \(\overline{\mathbf{c}}\) are position vectors of the vertices A, B and C respectively, then prove that the position vector of the centroid G is \(\frac{1}{3}(\bar{a}+\bar{b}+\bar{c})\).
Solution:
Let G be the centroid of ΔABC and AD is the median through the vector A.
Then AG : GD = 2 : 1. Let ‘O’ be the origin \(\overline{\mathrm{OA}}\) = \(\bar{a}\), \(\overline{\mathrm{OB}}\) = \(\bar{b}\), \(\overline{\mathrm{OC}}\) = \(\bar{c}\) ∵ D is Mid point of BC, the P.V. of D is \(\overline{O D}\) = \(\frac{1}{2}(\bar{b}+\bar{c})\)

G divides the median AD in the ratio 2 : 1
∴ The P.V. of G is

Question 10.
In ΔABC, if ‘O’ is the circum centre and H is the orthocentre, then show that
i) \(\overline{O A}\) + \(\overline{O B}\) + \(\overline{O C}\) = \(\overline{O H}\)
ii) \(\overline{H A}\) + \(\overline{H B}\) + \(\overline{H C}\) = 2\(\overline{H O}\)
Solution:
Let D be the midpoint of BC

i) Take’O’as the origin, let \(\overline{O A}\) = \(\bar{a}\) \(\overline{O B}\) = \(\bar{b}\) and \(\overline{O C}\) = \(\bar{c}\)

(∵ AH =2 R cos A, OD = R cos A, R is the circum radius of Δ ABC and hence AH = 2 (OD))

ii) \(\overline{\mathrm{HA}}\) + \(\overline{\mathrm{HB}}\) + \(\overline{\mathrm{HC}}\) = \(\overline{\mathrm{HA}}\) +2\(\overline{\mathrm{HD}}\)
= \(\overline{\mathrm{HA}}\) + 2(\(\overline{\mathrm{HO}}\) + \(\overline{\mathrm{OD}}\))
= \(\overline{\mathrm{HA}}\) + 2\(\overline{\mathrm{HO}}\) + 2\(\overline{\mathrm{OD}}\)
= \(\overline{\mathrm{HA}}\) + 2\(\overline{\mathrm{HO}}\) + \(\overline{\mathrm{AH}}\) = 2\(\overline{\mathrm{HO}}\)

Question 11.
Let \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) and \(\bar{d}\) be the position vectors of A, B,C and D respectively which are the vertices of a tetrahedron. Then Prove that the lines joining the vertices to the centroids of the opposite faces are concurrent (this point is called the centroid or the centre of the tetrahe dron)
Solution:
Let ‘O’ be the origin of reference
Let G₁, G₂, G₃ and G₄ be the centroids of ΔBCD, ΔCAD, ΔABD and ΔABC, respectively then OG₁ = \(\frac{\overline{\mathrm{b}}+\overline{\mathrm{c}}+\overline{\mathrm{d}}}{3}\)

Consider the point P that divides AG₁ in the ratio 3 : 1

Similarly we can show that the P.vs. of the points dividing BG₂, CG₃ and DG₄ in the ratio 3 : 1 are equal to \(\frac{1}{4}\) (\(\overline{\mathrm{a}}\) + \(\overline{\mathrm{b}}\) + \(\overline{\mathrm{c}}\) + \(\overline{\mathrm{d}}\)). Therefore P lies on each of AG₁, BG₂, CG₃ and DG₄.

Question 12.
Let OABC be a parallelogram and D the mid point of OA. Prove that the segment CD trisects the diagonal OB and is trisected by the diagonal OB.
Solution:
Let \(\overline{\mathrm{OA}}\) = \(\bar{a}\), \(\overline{\mathrm{OC}}\) = \(\bar{b}\), So that
\(\overline{\mathrm{OB}}\) = \(\bar{a}\) + \(\bar{b}\), \(\overline{\mathrm{OD}}\) = \(\frac{1}{2}\)(\(\bar{a}\))

Let M be the intersection of OB and CD.
Let OM : MB = k : 1 and CM : MD = l : 1

∴ l = 2 and k = \(\frac{1}{2}\)
∴ CD trisects OB and OB trisects CD

Question 13.
Let \(\bar{a}\), \(\bar{b}\) be non – collinear vectors, if \(\bar{\alpha}\) = (x + 4y)\(\bar{a}\) + (2x + y + 1)\(\bar{b}\) and \(\bar{\beta}\) = (y – 2x + 2)\(\bar{a}\) + (2x – 3y – 1)\(\bar{b}\) are such that 3\(\bar{\alpha}\) = \(\bar{\beta}\), then find x andy.
Solution:
Given 3\(\bar{\alpha}\) = 2\(\bar{\beta}\)
⇒ 3(x + 4y)\(\bar{a}\) + 3(2x + y + 1)\(\bar{b}\) = 2(y – 2x + 2)\(\bar{a}\) + 2(2x – 3y – 1)\(\bar{b}\)
On comparing the coeffs. of \(\bar{a}\) and \(\bar{b}\), we get
3x + 12y = 2y – 4x + 4 ⇒ 7x + 10y = 4 ——–— (1)
and 6x + 3y + 3 = 4x – 6y – 2 ⇒
2x + 9y = -5 ——–—(2)
Solve (1) and (2), we get
x = 2 and y = -1

Question 14.
Show that the points whose position vectors are
-2\(\bar{a}\) + 3\(\bar{b}\) + 5\(\bar{c}\), \(\bar{a}\) + 2\(\bar{b}\) + 3\(\bar{c}\), 7\(\overline{\mathrm{a}}\) + \(\overline{\mathrm{c}}\) are non – collinear when \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) are non – coplanar vectors.
Solution:
Let O be the origin and let P, Q, R be the P.Vs. of the given points

⇒ The points P, Q, R are collinear.

Question 15.
If the points whose position vectors are 3\(\bar{i}\) – 2\(\bar{j}\) – \(\bar{k}\), 2\(\bar{i}\) + 3\(\bar{j}\) – 4\(\bar{k}\), –\(\bar{i}\) + \(\bar{j}\) + 2\(\bar{k}\) and 4\(\bar{i}\) + 5\(\bar{j}\) + λ\(\bar{k}\) are coplanar, then show that λ = \(\frac{-146}{7}\)
Solution:
Let ‘O’ be the origin and Let A, B, C and D be the given points. Then

Equating the corresponding coeffs.
-x – 4y = 1, 5x + 3y = 7, -3x + 3y = λ + 1
Solving x + 4y + 1 = 0 and 5x + 3y – 7 = 0

Question 16.
In the two dimensional plane, Prove by using vector methods, the equation of the line whose intercepts on the
axes are ‘a’ and ’b’ is \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1.
Solution:
Let A = (a, 0), B = (0, b)
∴ A = a\(\overline{\mathbf{i}}\), B = b\(\overline{\mathbf{j}}\)
The equation of the line is
\(\overline{\mathbf{r}}\) = (1 – t)a\(\overline{\mathbf{i}}\) + t(b\(\overline{\mathbf{j}}\))
If \(\overline{\mathbf{r}}\) = x\(\overline{\mathbf{i}}\) + y\(\overline{\mathbf{j}}\), then x = (1 – t)a and y = t b
By eliminating ‘t’
x = \(\left(1-\frac{y}{b}\right) a\)
⇒ \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1

Question 17.
Using the vector equation of the straight line passing through two points, prove that the points whose position vectors are \(\overline{\mathbf{a}}\), \(\overline{\mathbf{b}}\) and (3\(\overline{\mathbf{a}}\) – 2\(\overline{\mathbf{b}}\)) are collinear.
Solution:
The vector equation of the st. line passing through two pt.s \(\overline{\mathbf{a}}\) and \(\overline{\mathbf{b}}\) is
\(\overline{\mathbf{r}}\) = (1 – t) \(\overline{\mathbf{a}}\) + t\(\overline{\mathbf{b}}\) ; for some t ∈ R
This line also passes through the pt. which
P.v. is 3\(\overline{\mathbf{a}}\) – 2\(\overline{\mathbf{b}}\) then
3\(\overline{\mathbf{a}}\) – 2\(\overline{\mathbf{b}}\) = (1 – t)\(\overline{\mathbf{a}}\) + t\(\overline{\mathbf{b}}\)
Equating the corresponding coeffs,
1 – t = 3 and t = -2
∴ The three given points are collinear.

Question 18.
Find the equation of the line parallel to the vector 2\(\overline{\mathbf{i}}\) – \(\overline{\mathbf{j}}\) + 2\(\overline{\mathbf{k}}\) and which passes through the point A whose position vector is 3\(\overline{\mathbf{i}}\) + \(\overline{\mathbf{j}}\) – \(\overline{\mathbf{k}}\). If P is a point on this line such that AP = 15, find the position vector of R
Solution:
The vector equation of the line passing through the point A whose P.v is

Question 19.
Show that the line joining the pair of points 6\(\overline{\mathbf{a}}\) – 4\(\overline{\mathbf{b}}\) + 4\(\overline{\mathbf{c}}\) and the line joining the pair of points –\(\overline{\mathbf{a}}\) – 2\(\overline{\mathbf{b}}\) – 3\(\overline{\mathbf{c}}\), \(\overline{\mathbf{a}}\) + 2\(\overline{\mathbf{b}}\) – 5\(\overline{\mathbf{c}}\) intersect at the point -4\(\overline{\mathbf{c}}\) when \(\overline{\mathbf{a}}\), \(\overline{\mathbf{b}}\), \(\overline{\mathbf{c}}\) are non-coplanar vectors. (T.S) (Mar. ’16)
Solution:
The vector equation o+ the-line joining the pair of points \(\bar{q}\) = 6\(\bar{a}\) – 4\(\bar{b}\) + 4\(\bar{c}\), and \(\bar{p}\) = -4\(\bar{c}\) is \(\bar{r}\) = (1 – t)\(\bar{p}\) + t\(\bar{q}\) for t ∈ R

Let \(\bar{r}\) be the point of intersection of (1) and (2) then equating the coeffs of \(\bar{a}\), \(\bar{b}\) and \(\bar{c}\), we get
6t = 2s – 1 ⇒ 6t – s = -1 ——- (3)
– 4t = 4s – 2 ⇒ 4t + 4s = 2 —— (4)
-4 + 8t = -2s – 3 ⇒ 8t + 2s = 1 —— (5)
From (3) and (4)

From (1) s = \(\frac{1}{2}\)
Substitute t = 0 and s = \(\frac{1}{2}\) in (5)
8(0) + 2(\(\frac{1}{2}\)) = 1 ⇒ 1 = 1
Hence the lines (1) and (2) point of intersection put t = 0 in (1)
(i.e) \(\bar{r}\) = -4\(\bar{c}\)
∴ The two lines intersect at the point whose P.v. is -4\(\bar{c}\)

Question 20.
Find the point of intersection of the line \(\bar{r}\) = 2\(\bar{a}\) + \(\bar{b}\) + t(\(\bar{b}\) – \(\bar{c}\)) and the plane \(\bar{r}\) = \(\bar{a}\) + x(\(\bar{b}\) + \(\bar{c}\)) + y(\(\bar{a}\) + 2\(\bar{b}\) – \(\bar{c}\)) where \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) are non — coplanar vectors.
Solution:
Let \(\bar{r}\) be the P.v. of the point P, the intersection of the line and the plane.
Then
2\(\bar{a}\) + \(\bar{b}\) + t(\(\bar{b}\) – \(\bar{c}\)) = \(\bar{a}\) + x(\(\bar{b}\) + \(\bar{c}\)) + y(\(\bar{a}\) + 2\(\bar{b}\) – \(\bar{c}\))
∵ \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) are non – coplanar vectors.
Equation th coeffs. of \(\bar{a}\), \(\bar{b}\), \(\bar{c}\)
2 = 1 + y ⇒ y = 1
1 + t = x + 2y
⇒ 1 + t = x + 2(1)
⇒ t – x = 1 ——— (1)
and -t = x – y
x + t = y
⇒ x + t = 1 ——— (2)
From (1),(2) t = 1, x = 0, y = 1
∴ Point of intersection is
\(\bar{r}\) = 2\(\bar{a}\) + \(\bar{b}\) + 1(\(\bar{b}\) – \(\bar{c}\))
⇒ \(\bar{r}\) = 2\(\bar{a}\) + \(\bar{b}\) + 1(\(\bar{b}\) – \(\bar{c}\))
⇒ \(\bar{r}\) = 2\(\bar{a}\) + 2\(\bar{b}\) – \(\bar{c}\)
∴ P.v. of the point of intersection is 2\(\bar{a}\) + 2\(\bar{b}\) – \(\bar{c}\)

Students get through Maths 1A Important Questions Inter 1st Year Maths 1A Matrices Important Questions which are most likely to be asked in the exam.

Intermediate 1st Year Maths 1A Matrices Important Questions

Question 1.
If A = \(\left[\begin{array}{ccc}
2 & 3 & -1 \\
7 & 8 & 5
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
1 & 0 & 1 \\
2 & -4 & -1
\end{array}\right]\) then find A + B.
Solution:
The matrices A and B both are of the same order 2 × 3. Hence A + B is defined. Now

Question 2.
If \(\left[\begin{array}{ccc}
x-1 & 2 & y-5 \\
z & 0 & 2 \\
1 & -1 & 1+a
\end{array}\right]\) = \(\left[\begin{array}{ccc}
1-x & 2 & -y \\
2 & 0 & 2 \\
1 & -1 & 1
\end{array}\right]\) then find the values of x, y, z and a.
Solution:
From the equality of matrices
x – 1 = 1 – x ⇒ 2x = 2 ⇒ x = 1
y – 5 = -y ⇒ 2y = 5 ⇒ y = \(\frac{5}{2}\) ⇒ z = 2
z = 2
1 + a = 1 ⇒ a = 1 – 1 ⇒ a = 0

Question 3.
Find the trace of A if A = \(\left[\begin{array}{ccc}
1 & 2 & -\frac{1}{2} \\
0 & -1 & 2 \\
-\frac{1}{2} & 2 & 1
\end{array}\right]\) (Mar. ’04)
Solution:
The elements of the principal diagonal of A are 1,-1, 1.
Hence the trace of A is 1 + (-1) + 1 =1

Question 4.
If A = \(\left[\begin{array}{cc}
4 & -5 \\
-2 & 3
\end{array}\right]\) then find -5A.
Solution:
-5A = -5\(\left[\begin{array}{cc}
4 & -5 \\
-2 & 3
\end{array}\right]\) = \(\left[\begin{array}{cc}
-20 & 25 \\
10 & -15
\end{array}\right]\)

Question 5.
Find the additive inverse of A, where
A = \(\left[\begin{array}{ccc}
i & 0 & 1 \\
0 & -i & 2 \\
-1 & 1 & 5
\end{array}\right]\)
Solution:
The additive inverse of A is -A = (-1)A
∴ Additive inverse of A

Question 6.
If A = \(\left[\begin{array}{ccc}
2 & 3 & 1 \\
6 & -1 & 5
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
1 & 2 & -1 \\
0 & -1 & 3
\end{array}\right]\) then find the matrix X such that A + B – X = 0. What is the order of the matrix X ?
Solution:
Given, A + B – X = 0 ⇒ X = A + B
= \(\left[\begin{array}{ccc}
2 & 3 & 1 \\
6 & -1 & 5
\end{array}\right]\) + \(\left[\begin{array}{ccc}
1 & 2 & -1 \\
0 & -1 & 3
\end{array}\right]\)
∴ X = \(\left[\begin{array}{ccc}
3 & 5 & 0 \\
6 & -2 & 8
\end{array}\right]\) ∴ Order of X is 2 × 3.

Question 7.
If A = \(\left[\begin{array}{lll}
0 & 1 & 2 \\
2 & 3 & 4 \\
4 & 5 & 6
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
1 & -2 & 0 \\
0 & 1 & -1 \\
-1 & 0 & 3
\end{array}\right]\) then find A – B and 4B – 3A.
Solution:

Question 8.
If A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\), B = \(\left[\begin{array}{ll}
3 & 8 \\
7 & 2
\end{array}\right]\) and 2X + A = B then find ‘X’. (A.P.) (Mar. ’15, ’11)
Solution:
2X + A = B ⇒ 2X = B – A

Question 9.
Two factories I and II produce three varieties of pens namely. Gel, Ball and Ink pens. The sale in rupees of these varieties of pens by both the factories in the month of September and October in a year are given by the following matrices A and B.
September sales (in Rupees)

i) Find the combined sales in September and October for each factory in each variety.
ii) Find the decrease in sales from September to October.
Solution:
i) Combined sales in September and October for each factory in each variety is given by

ii) Decrease in sales from September to October is given by

Question 10.
Construct a 3 × 2 matrix whose elements are defined by a_ij = \(\frac{1}{2}|\mathbf{i}-3 \mathbf{j}|\). (T.S) (Mar. ’15)
Solution:
In general 3 × 2 matrix is given by

Question 11.
If A = \(\left[\begin{array}{lll}
0 & 1 & 2 \\
1 & 2 & 3 \\
2 & 3 & 4
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
1 & -2 \\
-1 & 0 \\
2 & -1
\end{array}\right]\) then find AB and BA.
Solution:
The number of columns of A = 3 = the number of rows of B.
Hence AB is defined and

Since the number of columns of B = 2 ≠ 3 = the number of rows of A.
∴ BA is not defined.

Question 12.
If A = \(\left[\begin{array}{ccc}
1 & -2 & 3 \\
2 & 3 & -1 \\
-3 & 1 & 2
\end{array}\right]\) and B = \(\left[\begin{array}{lll}
1 & 0 & 2 \\
0 & 1 & 2 \\
1 & 2 & 0
\end{array}\right]\) then examine whether A and B commute with respect to multiplication of matrices.
Solution:
Both A and B are square matrices of order 3. Hence both AB and BA are define and are matrices of order 3.

which shows that AB ≠ BA
Therefore A and B do not commute with respect to multiplication of matrices.

Question 13.
If A = \(\left[\begin{array}{cc}
\mathbf{i} & 0 \\
\mathbf{0} & -\mathbf{i}
\end{array}\right]\) then show that A² = -I where i² = -1.
Solution:

Question 14.
If A = \(\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\) then show that A² – 4A – 5I = 0 then show that for all the positive integers n.
Aⁿ = \(\left[\begin{array}{cc}
\cos n \theta & \sin n \theta \\
-\sin n \theta & \cos n \theta
\end{array}\right]\)
Solution:
We solve this problem by using the principle of mathematical induction consider the statement P(n) : Aⁿ = \(\left[\begin{array}{cc}
\cos n \theta & -\sin n \theta \\
-\sin n \theta & \cos n \theta
\end{array}\right]\)
Since A = \(\left[\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right]\)
⇒ A¹ = \(\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\), P(n) is true for n = 1
Suppose that the given statement P(n) is true for n = k, k ≥ 1.
∴ Then A^k = \(\left[\begin{array}{cc}
\cos k \theta & \sin k \theta \\
-\sin k \theta & \cos k \theta
\end{array}\right]\) .
Consider A^k+1 = A^k.A

∴ Therefore P (n) is true for n = k + 1.
Hence, by mathematical induction, P(n) is true for all positive integral values of n.

Question 15.
If A = \(\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]\) then show that A² – 4A – 5I = 0. (A.P.) (Mar. ’16)
Solution:

Question 16.
If A = \(\left[\begin{array}{ccc}
-2 & 1 & 0 \\
3 & 4 & -5
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
1 & 2 \\
4 & 3 \\
-1 & 5
\end{array}\right]\) then find A + B.
Solution:
A + B = \(\left[\begin{array}{ccc}
-2 & 1 & 0 \\
3 & 4 & -5
\end{array}\right]\) + \(\left[\begin{array}{ccc}
1 & 4 & -1 \\
2 & 3 & 5
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
-1 & 5 & -1 \\
5 & 7 & 0
\end{array}\right]\)

Question 17.
If A = \(\left[\begin{array}{cc}
-1 & 2 \\
0 & 1
\end{array}\right]\) then find AA’. Do A and A’ commute with respect to multiplication of matrices ?
Solution:

Since AA’ ≠ A’A, A and A’ do not commute with respect to multiplication of matrices.

Question 18.
If A = \(\left[\begin{array}{ccc}
0 & 4 & -2 \\
-4 & 0 & 8 \\
2 & -8 & x
\end{array}\right]\) is a skew symmetric matrix, find the value of x.
Solution:
A is a skew symmetric matrix and x is an element of the diagonal. Hence x = 0.

Question 19.
For any n × n matrix A, prove that A can be uniquely expressed as a sum of a symmetric matrix and a skew symmetric matrix.
Solution:
For A square matrix of order n, A + A’ is symmetric and A – A’ is a skew symmetric matrix and
A = \(\frac{1}{2}\)(A + A’) + \(\frac{1}{2}\)(A – A’)
To prove uniqueness, let B be a symmetric matrix and C be a skew-symmetric matrix, such that
A = B + C
Then A’ = (B + C)’ = B’ + C’
= B + (-C) = B – C
and hence B = \(\frac{1}{2}\)(A + A’)
C = \(\frac{1}{2}\)(A – A’)

Question 20.
Show that
(Mar. ’05)
Solution:

Question 21.
Without expanding the determinant show that (AP) (Mar. ’15)

Solution:

Question 22.
Show that \(\left|\begin{array}{lll}
1 & a^{2} & a^{3} \\
1 & b^{2} & b^{3} \\
1 & c^{2} & c^{3}
\end{array}\right|\) = (a – b)(b – c)(c – a)(ab + bc + ca)
Solution:


= -(a – b)(b – c)(c – a) [(c² + ca + a²) – (b + c + a) (c + a)]
= -(a – b)(b – c)(c – a)[c² + ca + a² – b(c + a) — (c + a)²]
= -(a – b)(b – c)(c – a)[c² + ca + a² – bc – ab – c² – 2ca – a²]
= -(a – b) (b – c) (c – a) [-ab – bc – ca]
= (a – b)(b – c)(c – a) (ab + bc + ca)
∴ \(\left|\begin{array}{lll}
1 & a^{2} & a^{3} \\
1 & b^{2} & b^{3} \\
1 & c^{2} & c^{3}
\end{array}\right|\) = (a – b)(b – c)(c – a)(ab + bc + ca)

Question 23.
If ω is complex (non real) cube root of 1 then show that \(\left|\begin{array}{ccc}
1 & \omega & \omega^{2} \\
\omega & \omega^{2} & 1 \\
\omega^{2} & 1 & \omega
\end{array}\right|\) = 0 (Mar. ’14, ’11)
Solution:

Question 24.
Show that

= (a + b + c)³. (May. ’11)
Solution:

Question 25.
Show that the determinant of a skew-symmetric matrix of order three is always zero.
Solution:
Let us consider a skew-symmetric matrix of order 3 × 3, say

Question 26.
Find the value of x if

(T.S) (Mar. ’15
Mar. ’06)
Solution:

⇒ (x – 2)(30 – 24) – (2x – 3)(10 – 6) + (3x – 4)(4 – 3) = 0
⇒ 6x – 12 – 8x + 12 + 3x – 4 = 0
x – 4 = 0 ∴ x = 4
Question 27.
Find the adjoint and the inverse of the matrix A = \(\left[\begin{array}{cc}
1 & 2 \\
3 & -5
\end{array}\right]\).
Solution:
|A| = \(\left|\begin{array}{cc}
1 & 2 \\
3 & -5
\end{array}\right|\) = -5 – 6 = -11 ≠ 0
Hence A is invertible.
The cofactor matrix of A = \(\left[\begin{array}{cc}
-5 & -3 \\
-2 & 1
\end{array}\right]\)

Question 28.
Find the adjoint and the inverse of the matrix A = \(\left[\begin{array}{lll}
1 & 3 & 3 \\
1 & 4 & 3 \\
1 & 3 & 4
\end{array}\right]\)
Solution:
|A| = \(\left[\begin{array}{lll}
1 & 3 & 3 \\
1 & 4 & 3 \\
1 & 3 & 4
\end{array}\right]\)
= 1(16 – 9) – 3(4 – 3) + 3(3 – 4)
= 7 – 3 – 3
= 1 ≠ 0
∴ A is invertible.
The factor of A is B = \(\left[\begin{array}{ccc}
7 & -1 & -1 \\
-3 & 1 & 0 \\
-3 & 0 & 1
\end{array}\right]\)

Question 29.
Show that A = \(\left[\begin{array}{lll}
1 & 2 & 1 \\
3 & 2 & 3 \\
1 & 1 & 2
\end{array}\right]\) is non singular and find A^-1
Solution:
Let A = \(\left[\begin{array}{lll}
1 & 2 & 1 \\
3 & 2 & 3 \\
1 & 1 & 2
\end{array}\right]\)
det A = 1(4 – 3) – 2(6 – 3) + 1(3 – 2)
= 1 – 6 + 1 – 4
The cofactors of elements of A are
A₁₁ = +(4 – 3) = 1,
A₁₂ = -(6 – 3) = -3,
A₁₃ = +(3 – 2) = 1,
A₂₁ = -(4 – 1) = -3,
A₂₂ = +(2 -1) = 1,
A₂₃ = -(1 – 2) = 1,
A₃₁ = + (6 – 2) = 4,
A₃₂ = -(3 – 3) = 0,
A₃₃ = +(2 – 6) = -4

Question 30.
Find the rank of A = \(\left[\begin{array}{lll}
0 & 1 & 2 \\
1 & 2 & 3 \\
3 & 2 & 1
\end{array}\right]\) using elementary transformations.
Solution:

The last matrix is singular and \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) is a non singular submatrix of it. Hence its rank is 2.
∴ Rank (A) = 2.

Question 31.
Find the rank A = \(\left[\begin{array}{cccc}
1 & 2 & 0 & -1 \\
3 & 4 & 1 & 2 \\
-2 & 3 & 2 & 5
\end{array}\right]\) using elementary transformations.
Solution:

Question 32.
Apply the test of rank to examine whether the following equations are consistent.
2x – y + 3z = 8, —x + 2y + z = 4, 3x + y – 4z = 0 and, ¡f consistent, find the complete solution.
Solution:
The augmented matrix is


Hence rank (A) = rank [AD] = 3
The system has a unique solution.
We write the equivalent system of the equations from (F) : i.e.,
-x + 2y + z = 4
3y + 5z = 16
-38z = -76
∴ z = 2, y = 2, x = 2 is the solution.

Question 33.
Show that the following system of equations is consistent and solve it completely x + y + z = 3, 2x + 2y – z = 3, x + y – z = 1.
Solution:
Let A = \(\left[\begin{array}{ccc}
1 & 1 & 1 \\
2 & 2 & -1 \\
1 & 1 & -1
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) and D = \(\left[\begin{array}{l}
3 \\
3 \\
1
\end{array}\right]\)
Then the given equations can be written as
AX = D
Augmented Matrix
[AD] = \(\left[\begin{array}{cccc}
1 & 1 & 1 & 3 \\
2 & 2 & -1 & 3 \\
1 & 1 & -1 & 1
\end{array}\right]\)
Applying R₂ → R₂ – 2R₁, R₃ → R₃ – R₁,
~ \(\left[\begin{array}{cccc}
1 & 1 & 1 & 3 \\
0 & 0 & -3 & -3 \\
0 & 0 & -2 & -2
\end{array}\right]\)
Applying R₃ → 3R₃ – 2R₂
~ \(\left[\begin{array}{cccc}
1 & 1 & 1 & 3 \\
0 & 0 & -3 & -3 \\
0 & 0 & 0 & 0
\end{array}\right]\)
Clearly all the submatrices of order 3 × 3 of the above matrix are singular.
Hence rank of A ≠ 3, and rank of [AD] ≠ 3
Now consider the submatrix \(\left[\begin{array}{cc}
1 & 1 \\
0 & -3
\end{array}\right]\) of both
A and AD is non-singular.
Hence Rank (A) = 2 = Rank [AD]
∴ The system of equations is consistent and has infinitely many solutions.
From the above [AD] matrix
x + y + z = 3
-3z = -3 ⇒ z = 1
and x + y = 2
Hence x = k, y = 2 – k, z = 1, k ∈ R is a solution set.

Question 34.
Solve the following simultaneous linear equations by using ‘Cramers rule.
3x + 4y + 5z = 18
2x – y + 8z = 13
5x – 2y + 7z = 20
Solution:
Let A = \(\left[\begin{array}{ccc}
3 & 4 & 5 \\
2 & -1 & 8 \\
5 & -2 & 7
\end{array}\right]\) ; X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\), D = \(\left[\begin{array}{l}
18 \\
13 \\
20
\end{array}\right]\)
i.e., AX = D
Δ = det A = \(\left|\begin{array}{ccc}
3 & 4 & 5 \\
2 & -1 & 8 \\
5 & -2 & 7
\end{array}\right|\)
= 3(-7 + 16) – 4(14 – 40) + 5(-4 + 5)
= 27 + 104 + 5
= 136 ≠ 0

Hence by Cramer’s rule
x = \(\frac{\Delta_{1}}{\Delta}\) = \(\frac{408}{136}\) = 3
y = \(\frac{\Delta_{2}}{\Delta}\) = \(\frac{136}{136}\) = 1
z = \(\frac{\Delta_{3}}{\Delta}\) = \(\frac{136}{136}\) = 1
∴ The solution of the given system of equation is x = 3, y = z = 1.

Question 35.
Solve 3x + 4y + 5z = 18; 2x – y + 8z = 13 and 5x – 2y + 7z = 20 by using ‘Matrix inversion method’. (A.P) (Mar. ‘15, ’08)
Solution:
Let A = \(\left[\begin{array}{ccc}
3 & 4 & 5 \\
2 & -1 & 8 \\
5 & -2 & 7
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\), B = \(\left[\begin{array}{l}
18 \\
13 \\
20
\end{array}\right]\)
Given equations can be written as AX = B
By matrix inversion method X = A^-1B is the solution.
det A = 3(-7 + 16) – 4(14 – 40) + 5(-4 + 5)
= 27 + 104 + 5
= 136
The cofactors of elements of A are
A₁₁ = +(-7 + 16) = 9,
A₁₂ = -(-14 – 40) = 26,
A₁₃ = +(-4 + 5) = 1,
A₂₁ = -(28 + 10) = -38,
A₂₂ = +(21 – 25) = -4,
A₂₃ = -(-6 – 20) = 26,
A₃₁ = +(32 + 5) = 37,
A₃₂ = -(24 – 10) = -14,
A₃₃ = (-3 – 8) = -11

∴ x = 3, y = 1, z = 1 is the solution.

Question 36.
Solve the following equations by Gauss-Jordan method.
3x + 4y + 5z = 18,
2x – y + 8z = 13,
5x – 2y + 7z = 20.
Solution:
The augmented matrix is

Hence the solution is x = 3, y = 1, z = 1

Question 37.
Solve the following system of equations by Gauss-Jordan method,
x + y + z = 3,
2x + 2y – z = 3,
x + y – z = 1.
Solution:
The matrix equation is AX = D, where
A = \(\left[\begin{array}{ccc}
1 & 1 & 1 \\
2 & 2 & -1 \\
1 & 1 & -1
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\), D = \(\left[\begin{array}{l}
3 \\
3 \\
1
\end{array}\right]\)
The augmented matrix is
[AD] = \(\left[\begin{array}{cccc}
1 & 1 & 1 & 3 \\
2 & 2 & -1 & 3 \\
1 & 1 & -1 & 1
\end{array}\right]\)

Hence the following is the system of equations equivalent to the given system of equation: x + y + z = 3, -3z = -3
Hence z = 1, x + y = 2
∴ The solution set is
x = k, y = 2 – k, z = 1 where k ∈ R.

Question 38.
By using Gauss-Jordan method, show that the following system has no solution 2x + 4y – z = 0, x + 2y + 2z = 5,
3x + 6y – 7z = 2.
Solution:

Hence the given system of equations is equivalent to the following system of equations 2x + 4y – z = 0, 5z = 10 ⇒ z = 2 and 0(x) + 0(y) + 0(z) = 130.
Clearly no x, y, z satisfy the last equation.
∴ Given system of equations has no solution.

Question 39.
Find the non-trivial solutions, If any, for the following system of equations
2x + 5y + 6z = 0, x – 3y + 8z = 0, 3x + y -4z = 0.
Solution:
The coefficient matrix A = \(\left[\begin{array}{ccc}
2 & 5 & 6 \\
1 & -3 & -8 \\
3 & 1 & -4
\end{array}\right]\)
On interchanging R₁ and R₂, we get
A ~ \(\left[\begin{array}{ccc}
1 & -3 & 8 \\
2 & 5 & 6 \\
3 & 1 & -4
\end{array}\right]\)
R₂ → R₂ – 2R₁, R₃ → R₃ – 3R₁ we get

det A = 0 as R₂ and R₃ are identical. Clearly rank (A) = 2, as the sub-matrix
\(\left[\begin{array}{cc}
1 & -3 \\
0 & 1
\end{array}\right]\) is non-singular.
Hence the system has non-trivial solution. The following system of equations is equivalent to the given system of equations
x – 3y – 8z = 0
y + 2z = 0
On giving an arbitrary value k to z, we obtain the solution set is ‘
x = 2k, y = -2k, z = k, k ∈ R, for k ≠ 0.
We obtain non-trivial solutions.

Question 40.
Find whether the following system of linear homogeneous equations has a non-trivial solution. (T.S) (Mar. ’16)
x – y + z = 0,
x + 2y – z = 0,
2x + y + 3z = 0
Solution:
The coefficient matrix is \(\left[\begin{array}{ccc}
1 & -1 & 1 \\
1 & 2 & -1 \\
2 & 1 & 3
\end{array}\right]\)
Its determinant is 9 ≠ 0.
Hence the system has the trivial solution
x = y = z = 0 only.

Question 41.
Theorem : Matrix multiplication is associative. i.e., If conformability is assured for the matrices A, B and C, then
(AB)C = A(BC). (July ‘01, Inst. ’93, Oct.’83)
Solution:
Proof : Let A = (a_ij)m × n, B = (b_jk)n × p’
C = (C_kl)p × q

Question 42.
Theorem : Matrix multiplication is distri-butive over matrix addition i.e., If con-formability is assured for the matrices A, B and C, then. (Oct. ’99, inst. ’98)
i) A(B + C) = AB + AC
ii) (B + C)A = BA + CA
Proof:

Similarly we cna prove that
(B + C)A = BA + CA.

Question 43.
Theorem : If A is any matrix, then (A^T)^T = A. (Nov. ’80)
Solution:

Question 44.
Theorem: If A and B are two matrices of same type, then (A + B)^T = A^T + B^T. (July ‘01)
Proof:

Question 45.
Theorem: If A and B are two matrices for which conformability for multiplication is assured, then (AB)^T = B^TA^T.
Solution:
Proof:

Question 46.
Theorem : If A and B are two invertible matrices of same type then AB is also invertible and (AB)^-1 = B^-1A^-1.
Solution:
Proof:
A is invertible matrix ⇒ A^-1 exists and AA^-1 = A^-1A = I
B is an invertible matrix ⇒ B^-1 exists and
BB^-1 = B^-1B = I
Now (AB)(B^-1A^-1) = A(BB^-1) A^-1
= AIA^-1 = AA^-1 = 1
(B^-1A^-1)(AB) = B^-1(A^-1A)B = B^-1IB
= B^-1B = I
∴ (AB) (B^-1A^-1) = (B^-1A^-1) (AB) = I
∴ AB is invertible and (AB)^-1 = B^-1A^-1.
Question 47.
Theorem : If A is an invertible matrix then A’ is also invertible and (A^T)^-1 = (A^-1). (Nov. ’98)
Solution:
Proof : A is an invertible matrix ⇒ A^-1 exists
and AA^-1 = A^-1A = I
(AA^-1)’ = (A^-1A)’ = I’
⇒ (A^-1)’A’ = A’. (A^-1) = I .
⇒ By def. (A’)^-1 = (A^-1)’.

Question 48.
Theorem: If A \(\left[\begin{array}{lll}
a_{1} & b_{1} & c_{1} \\
a_{2} & b_{2} & c_{2} \\
a_{3} & b_{3} & c_{3}
\end{array}\right]\) is a non-singular matrix then A is invertible and
.
Proof:
Let A = \(\left[\begin{array}{lll}
a_{1} & b_{1} & c_{1} \\
a_{2} & b_{2} & c_{2} \\
a_{3} & b_{3} & c_{3}
\end{array}\right]\) be a non-singular matrix.
∴ det A ≠ 0

Question 49.
A certain bookshop has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are Rs. 80, Rs. 60 and Rs. 40 each respectively. Using matrix algebra, find the total value of the books in the shop.
Solution:
Number of books

Selling price (in rupees)

Total value of the books in the shop.

= [120 × 80 + 96 × 60 + 120 × 40]
= [9600 + 5760 + 4800]
= [20160] (in rupees)

Students get through Maths 1A Important Questions Inter 1st Year Maths 1A Mathematical Induction Important Questions which are most likely to be asked in the exam.

Intermediate 1st Year Maths 1A Mathematical Induction Important Questions

Question 1.
Use mathematical induction to prove the statement 1³ + 2³ + 3³ + … + n³ = \(\frac{n^{2}(n+1)^{2}}{4}\), ∀ n ∈ N
Solution:
Let p(n) be the statement:
1³ + 2³ + 3³ + … + n³ = \(\frac{n^{2}(n+1)^{2}}{4}\) and let S(n) be the sum on the L.H.S.
Since S(1) = 1³ = 1³ = \(\frac{1^{2}(1+1)^{2}}{4}\) = 1 = 1³
∴ The formula is true for n = 1.
Assume that the statement p(n) is true for n = k
(i.e.,) S(k) = 1³ + 2³ + 3³ + … + k³ = \(\frac{k^{2}(k+1)^{2}}{4}\)
We show that the formula is true for n = k + 1
(i.e.,) We show that S(k + 1) = \(\frac{(k+1)^{2}(k+2)^{2}}{4}\)
We observe that
S(k + 1) = 1³ + 2³ + 3³ + … + k³ + (k + 1)³
= S(k) + (k + 1)³
= \(\frac{k^{2}(k+1)^{2}}{4}\) + (k + 1)³

∴ The formula holdes for n = k + 1
∴ By the principle of mathematical induction p(n) is true for all n ∈ N
(i.e.,) the formula
1³ + 2³ + 3³ + … + k³ = \(\frac{n^{2}(n+1)^{2}}{4}\) is true for all n ∈ N.

Question 2.
Use mathematical induction to prove the statement
\(\sum_{k=1}^{n}(2 k-1)^{2}\) = \(\frac{n(2 n-1)(2 n+1)}{3}\) for all n ∈ N.
Solution:
Let p(n) be the statement:
1² + 3² + 5² + … + (2n – 1)² = \(\frac{n(2 n-1)(2 n+1)}{3}\) for all n ∈ N
Let S(n) be the sum on the L.H.S.
∴ s(1) = 1² = \(\frac{1(2-1)(2+1)}{3}\) = 1
∴ The formula is true for n = 1
Assume that the statement p(n) is true for n = k
(i.e.,) S(k) = 1² + 3² + 5² + … + (2k – 1)² = \(\frac{k(2 k-1)(2 k+1)}{3}\)
We show that the formula is true for n = k + 1 (i.e.,) we show that S(k + 1) = \(\frac{(k+1)(2 k+1)(2 k+3)}{3}\)
We observe that
S(k + 1) = 1² + 3² + 5² + ….. + (2k – 1)² + (2k + 1)²
= S(k) + (2k + 1)²

∴ The formula holds for n = k + 1
∴ By the principle of mathematical induction
p(n) is true ∀ n ∈ N.
i.e., \(\sum_{k=1}^{n}(2 k-1)^{2}\) = \(\frac{n(2 n-1)(2 n+1)}{3}\) for all n ∈ N.

Question 3.
Use mathematical induction to prove the statement 2 + 3.2 + 4.2² + … upto n terms = n.2ⁿ, ∀ n ∈ N. (May ’07)
Solution:
Let p(n) be the statement:
2 + 3.2 + 4.2² + … + (n + 1) . 2^n-1 = n.2ⁿ and let S(n) be the sum on the L.H.S.
∵ S(1) = 2 = (1) . 2¹
∵ The statement is true for n = 1
Assume that the statement p(n) is true for n = k.
(i.e.,) S(k) = 2 + 3.2 + 4.2² + … + (k + 1).2^k-1 = k . 2^k
We show that the formula is true for n = k + 1 (i.e.,) we show that S(k + 1) = (k + 1). 2^{k + 1}
We observe that
S(k + 1)= 2 + 3.2 + 4.2² + …(k + 1)2^{k + 1} + (k + 2) . 2^k
= S(k) + (k + 2) . 2^k
= k. 2^k + (k + 2) . 2^k
= (k + k + 2) 2^k
= 2(k + 1) . 2^k = 2^{k + 1} (k + 1)
∴ The formula holds for n = k + 1
∴ By the principle of mathematical induction, p(n) is true for all n ∈ N
(i.e.,) the formula
2 + 3.2 + 4.2² + … + (n + 1)2^{n – 1}
= n.2ⁿ ∀ n ∈ N (i.e.,) the formula

Question 4.
Show that ∀ n ∈ N, \(\frac{1}{1.4}\) + \(\frac{1}{4.7}\) + \(\frac{1}{7.10}\) + …….. upto n terms = \(\frac{n}{3 n+1}\) (Mar. ’05)
Solution:
1, 4, 7, … are in A.R, whose n^th term is
1 + (n – 1)3 = 3n – 2
4, 7, 10, … are in A.R, whose n^th term is
4 + (n – 1)3 = 3n + 1
∴ The n^th term of the given sum is
\(\frac{1}{(3 n-2)(3 n+1)}\)
Let p(n) be the statement:
\(\frac{1}{1.4}\) + \(\frac{1}{4.7}\) + \(\frac{1}{7.10}\) + ….. + \(\frac{1}{(3 n-2)(3 n+1)}\) = \(\frac{n}{3 n+1}\)
and S(n) be the sum on the L.H.S
Since S(1) = \(\frac{1}{1.4}\) = \(\frac{1}{3(1)+1}\) = \(\frac{1}{4}\)
∴ p(1) is true.
Assume the statement p(n) is true for n = k
(i.e)S(k) = \(\frac{1}{1.4}\) + \(\frac{1}{4.7}\) + \(\frac{1}{7.10}\) + …….. + \(\frac{1}{(3 k-2)(3 k+1)}\) = \(\frac{k}{3 k+1}\)
We show that the statement s p(n) is true for n = k + 1
(i.e) we show that S(k + 1) = \(\frac{k+1}{3 k+4}\)
We observe that

∴ The statement holds for n = k + 1.
∴ By the principle of mathematical induction, p(n) is true ∀ n ∈ N.
(ie.) \(\frac{1}{1.4}\) + \(\frac{1}{4.7}\) + \(\frac{1}{7.10}\) + ……. + \(\frac{1}{(3 n-2)(3 n+1)}\) = \(\frac{n}{3 n+1}\) for all n Σ N.

Question 5.
Use mathematical induction to prove that 2n – 3 ≤ 2^{2n – 2} for all n ≥ 5, n ∈ N.
Solution:
Let P(n) be the statement:
2n – 3 ≤ 2^{n – 2}, ∀ n ≥ 5, n ∈ N.
Here we note that the basis of induction is 5.
Since 2.5 – 3 ≤ 2^{5 – 2}, the statement is true for n = 5.
Assume the statement is true for n = k, k ≥ 5.
i.e., 2K – 3 ≤ 2^{k – 2}, for k ≥ 5.
We show that the statement is true for
n = k + 1, k ≥ 5
i.e., [2(k + 1) – 3] ≤ 2^{(k + 1) – 2}, for k ≥ 5.
We observe that [2(k + 1) – 3]
= (2k – 3) + 2
≤ 2, (By inductive hypothesis)
≤ 2^{k – 2} + 2^{k – 2} for k ≥ 5
= 2.2^{k – 2}
= 2^{(k + 1) – 2}
∴ The statement P(n) is true for
n = k + 1, k ≥ 5.
∴ By the principle of mathematical induction, the statement is true for all n ≥ 5, n ∈ N.

Question 6.
Use Mathematical induction to prove that (1 + x)ⁿ > 1 + nx for n ≥ 2, x > -1, x ≠ 0.
Solution:
Let the statement P (n) be : (1 + x)ⁿ > 1 + nx
Here we note that the basis of induction is 2 and that x ≠ 0, x > -1
⇒ 1 + x > 0.
Since (1 + x)² = t + 2x + x² > 1 + 2x, the statement is true for n = 2.
Assume that the statement is true for n = k, k ≥ 2.
i.e., (1 + x)^k > 1 + k x for k ≥ 2.
We show that the statement is true for n = k + 1,
i.e., (1 + x)^{k + 1} > + (k + 1)x.
We observe that (1 + x)^{k + 1}
= (1 + x)^k . (1 + x)
> (1 + kx) . (1 + x), (By inductive hypothesis)
= 1 + (k + 1)x + kx²
> 1 + (k + 1)x, (since kx² > 0)
∴ The statement is true for n = k + 1.
∴ By the principle of mathematical induction, the statement P(n) is true for all n ≥ 2.
i.e.,(1 + x)ⁿ > 1 + nx, ∀ n ≥ 2, x > -1, x ≠ 0.

Question 7.
If x and y are natural numbers and x ≠ y, using mathematical induction, show that xⁿ – yⁿ is divisible by x – y for all n ∈ N. (June ’04)
Solution:
Let p(n) be the statement:
xⁿ – yⁿ is divisible by (x – y)
Since x¹ – y¹ = x – y is divisible by (x – y)
∴ The statement is true for n = 1
Assume that the statement is true for n = k
(i.e) x^k – y^k is divisible by x – y
Then x^k – y^k = (x – y)p, for some integer p  (1)
We show that the statement is true for n = k + 1. (i.e) we show that x^{k + 1} – y^{k + 1} is divisible by x – y from (1), we have
x^k – y^k = (x – y) p
⇒ x^k = (x – y)p + y^k
∴ x^{k + 1} = (x – y)p x + y^k . x
⇒ x^{k + 1} – y^{k + 1} = (x – y) p x + y^k x – y^{k + 1}
= (x – y)p x + y^k(x – y)
= (x – y)[px + y^k],
(where px + y^k is an integer)
∴ x^{k + 1} – y^{k + 1} is divisible by (x – y)
∴ The statement p(n) is true for n = k + 1
∴ By mathematical induction, p (n) is true for all n ∈ n
(i.e) x^k – y^k is divisible by (x – y) for all n Σ N.

Question 8.
Using mathematical induction, show that x^m + y^m is divisible by x + y, if m is an odd natural number and x, y are natural numbers.
Solution:
Since m is an odd natural number,
Let m = 2n + 1 where n is a non-negative integer.
∴ Let p(n) be the statement:
x^{2n + 1} + y^{2n + 1} is divisible by (x + y)
Since x¹ + y¹ = x + y is divisible by x + y
∴ The statement is true for n = 0 and since x^{2.1 + 1} + y^{2.1 + 1} = x³ + y³ = (x + y)(x² – xy + y²) is divisible by x + y, the statement is true for n = 1.
Assume that the statement p(n) is true for n = k (i.e.) x^{2k + 1} + y^{2k + 1} is divisible by x + y. Then x^{2k + 1} + y^{2k + 1} = (x + y) p,
Where p is an integer ———— (1)
We show that the statement is true for
n = k + 1 (ie.) We show that
x^{2k + 3} + y^{2k + 3} is divisible by (x + y)
From (1)
x^{2k + 1} + y^{2k + 1} = (x + y)p
∴ x^{2k + 1} = (x + y) p – y^{2k + 1}
∴ x^{2k + 1} . x² = (x + y) p x² – y^{2k + 1} . x²
∴ x^{2k + 3} = (x + y) p x² – y^{2k + 1} . x²
∴ x^{2k + 3} + y^{2k + 3} = (x + y) p. (x²) – y^{2k + 1}.x² + y^{2k + 3}
= (x + y) p.x² – y^{2k + 1} (x² – y²)
= (x + y)p x² – y^{2k + 1} . (x + y)(x – y)
= (x + y)[p . x² – y^{2k + 1}(x – y)]
Here p x² – y^{2k + 1} (x – y) is an integer.
∴ x^{2k + 3} + y^{2k + 3} is divisible by (x + y)
∴ The statement is true for n = k + 1
∴ By the principle of mathematical induction, p(n) is true for all n
(i.e.,) x^{2n + 1} + y^{2n + 1} is divisible by (x + y), for all non-negative integers.
(i.e.,) x^m + y^m is divisible by (x + y), if m is an odd natural number.

Question 9.
Show that 49ⁿ +16n – 1 is divisible by 64 for all positive integers n. (May ’05)
Solution:
Let p(n) be the statement:
49ⁿ + 16n – 1 is divisible by 64
Since 49¹ + 16(1) – 1 = 64 is divisible by 64
The statement is true for n = 1
Assume that the statement p(n) is true for n = k
(i.e.,) 49^k + 16k – 1 is divisible by 64
Then (49^k + 16k – 1) = 64t, for some t ∈ N ——– (1)
we show that the statement is true for
n = k + 1.
(i.e.,) we show that
49^{k + 1} + 16 (k + 1) – 1 is divisible by 64
From (1), we have
49^k + 16k – 1 = 64t
∴ 49^k = 64t – 16k + 1
∴ 49^k . 49 = (64t – 16k + 1). 49
(64t – 16k + 1)49 + 16(k + 1) – 1
∴ 49^{k + 1} + 16(k + 1) – 1
= (64t – 16 K + 1)49 + 16(k + 1) – 1
∴ 49^{k + 1} + 16(k + 1) – 1 = 64(49t – 12k + 1)
here (49t — 12k + 1) is an integer
∴ 49^{k + 1} + 16(k + 1) – 1 is divisible by 64
∴ The statement is true for n = k + 1
∴ By the principle of mathematical induction,
∴ p(n) is true for all n ∈ N.
(i.e.,) 49ⁿ + 16n – 1 is divisible by 64, ∀ n ∈ N.

Question 10.
Use mathematical induction to prove that 2.4^{(2n + 1)} + 3^{(3n + 1)} is divisible by 11, ∀ n ∈ N.
Solution:
Let p(n) be the statement.
2.4^{(2n + 1)} + 3^{(3n + 1)} is divisible by 11
since 2.4^{(2.1 + 1)} + 3^{(3.1 + 1)} = 2.4³ + 3⁴
= 2(64) + 81
= 209 = 11 × 19 is divisible by 11
∴ The statement p(1) is true
Assume that the statement p(n) is true for n = k
(i.e.,) 24^{(2k + 1)} + 3^{(3k + 1)} is divisible by 11
Then 2.4^{(2k + 1)} + 3^{(3k + 1)} = (11)t, for some integer t ——– (1)
we show that the statement P(n) is true for n = K + 1
(i.e.,) we show that 2.4^{(2k + 3)} + 3^{(3k + 4)} is divisible by 11.
From (1), we have
2.4^{(2k + 1)} + 3^{(3k + 1)} = 11t
∴ 2.4^{(2k + 1)} = 11t – 3^{(3k + 1)}
∴ 2.4^{(2k + 1)} . 4² = (11t – 3^{(3k + 1)})
2.4^{(2k + 3)} + 3^{(3k + 4)} = (11t – 3^{(3k + 1)}) 16 + 3^{(3k + 4)}
= (11t)(16) + 3^{(3k + 1)}[27 – 16]
= 11[16t + 3^{3k + 1}]
Here 16t + 3^{(3k + 1)} is an integer
∴ 2.4^{(2k + 3)} + 3^{(3k + 4)} is divisible by 11
∴ The statement p(n) is true for n = k + 1.
∴ By the principle of mathematical induction, p(n) is true for all n ∈ N.
(i.e.,) 2.4^{(2n + 1)} + 3^{(3n + 1)} is divisible by 11, ∀ n ∈ N.