Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Direction Cosines and Direction Ratios Solutions Exercise 6(b) will help students to clear their doubts quickly.

## Intermediate 1st Year Maths 1B Direction Cosines and Direction Ratios Solutions Exercise 6(b)

I.

Question 1.

Find the direction ratios of the line joining the points (3, 4, 0) and (4, 4, 4).

Solution:

A(3, 4, 0) and B(4, 4, 4) are the given points.

d.rs of AB are (x_{2} – x_{1}, y_{2} – y_{1}, z_{2} – z_{1})

(4-3, 4-4, 4-0) i.e.,(1, 0, 4)

Question 2.

The direction ratios of a line are (-6, 2, 3). Find its direction cosines.

Solution:

D.Rs of the line are -6, 2, 3

Dividing with \(\sqrt{36+4+9}\) = 7

Direction cosines of the line are – \(\frac{6}{7}, \frac{2}{7}, \frac{3}{7}\)

Question 3.

Find the cosine of the angle between the lines whose direction cosines are (\(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\)) and (\(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\), 0)

Solution:

cos θ = l_{1} + l_{2} + m_{1} m_{2} + n_{1} n_{2
}

Question 4.

Find the angle between the lines whose direction ratios are (1, 1, 2) (√3, -√3, 0).

Solution:

Question 5.

Show that the lines with direction cosines (\(\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}\)) and (\(\frac{4}{13}, \frac{12}{13}, \frac{3}{13}\)) are perpendicular to each other.

Solution:

If two lines, are perpendicular, then

l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2} = 0

l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2}

\(\frac{12}{13} \cdot \frac{4}{13}-\frac{3}{13} \cdot \frac{12}{13}-\frac{4}{13} \cdot \frac{3}{13}\)

∴ The given lines are perpendicular.

Question 6.

O is the origin, P(2, 3, 4) and Q(1, k, 1) are points such that \(\overline{\mathrm{O P}}\) ⊥ \(\overline{\mathrm{O Q}}\). Find k.

Solution:

d.rs of OP and 2, 3,4

d.rs of OQ are 1, k, 1

OP and OQ are perpendicular

⇒ a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0

2 + 3k + 4 = 0

3k = -6

k = -2.

II.

Question 1.

If the direction ratios of a line are (3, 4, 0), find its direction cosines are also the angles made with the co-ordinate axes.

Solution:

Direction ratios of the line are (3, 4, 0)

Dividing with \(\sqrt{9+16+0}\) = 5

D.C’s of the line are (\(\frac{3}{5}, \frac{4}{5}\), 0)

If α_{c}, βc² are the angles made By the line with the co-ordinate axes, then

cos α = \(\frac{3}{5}\) cos β = \(\frac{4}{5}\) cos γ = 0

α = cos^{-1}\(\frac{3}{5}\), β = cos^{-1}\(\frac{4}{5}\), γ = \(\frac{\pi}{2}\)

Angles made with co-ordinate axes are

cos^{-1}\(\frac{3}{5}\), cos^{-1}\(\frac{4}{5}\), γ = \(\frac{\pi}{2}\)

Question 2.

Show that the line through the points (1, -1,2) (3, 4, -2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).

Solution:

A(1, -1 2) B(3, 4, -2) C(0,3,2) and D(3, 5,6) are the given points

d.rs of AB are 3 – 1, 4 + 1, -2 -2 i.e., 2, 5, -4

d.rs of CD are 3 – 0, 5 – 3, 6 – 2 i.e., 3, 2 4

a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 2.3 + 5.2 – 4.4

= 6 + 10 – 16 = 0

AB and CD are perpendicular.

Question 3.

Find the angle between \(\overline{\mathrm{D C}}\) and \(\overline{\mathrm{A B}}\) where A = (3, 4, 5), B =(4, 6, 3) C = (-1, 2, 4) are D(l, 0, 5).

Solution:

A(3, 4, 5), B(4, 6, 3), C(-1, 2, 4), D(1, 0, 5) are the given points.

d.rs of AB are 4 – 3, 6 – 4, 3 – 5 i.e., 1, 2, -2

d.rs of CD are 1 + 1, 0 – 2, 5 – 4 i.e., 2, -2, 1

Question 4.

Find the direction cosines of a line which is perpendicular to the lines, whose direction ratios are (1, -1, 2) and (2, 1, -1).

Solution:

Let l, m, n be the d.cs of the required line it is perpendicular to the lines with

d.rs (1, -1, 2) and (2, 1, -1)

l – m + 2n = 0

2l + m – n = 0

d.rs of the line are – 1, 5, 3

Dividing with \(\sqrt{1+25+9} =\sqrt{35}\)

d.cs of the required line are ’

– \(\frac{1}{\sqrt{35}}, \frac{5}{\sqrt{35}}, \frac{3}{\sqrt{35}}\)

Question 5.

Show that the points (2, 3, -4), (1, -2, 3) and (3, 8, -11) are collinear.

Solution:

A(2, 3, -4),B( 1, -2, 3) and C(3, 8, -11) are the given points.

∴ A, B, C are collinear.

Question 6.

Show that the points (4, 7, 8), (2, 3, 4), (-1, -2,1), (1,2,5) are vertices of apare- llelogram.

Solution:

A(4, 7, 8), B(2, 3, 4), C(-1, -2, 1) and D (1, 2, 5) . are the given points.

∴ AB = CD and BC = DA

∴ A, B, C, D are the vertical of parallelogram.

III.

Question 1.

Show that the lines whose direction co – sines are given by l + m + n = 0, 2mn + 3nZ – 5Zm = 0 are perpendicular to each other.

Solution:

Given l + m + n = 0 ………… (1)

2mn + 3nl – 5lm = 0 …………… (2)

From (1), l = -(m + n)

Substituting in (2)

2mn – 3n(m + n) + 5m(m + n) = 0

2mn – 3mn – 3n2 + 5m2 + 5mn = 0

5m² + 4mn – 3n² = 0

\(\frac{m_{1} m_{2}}{n_{1} n_{2}}=-\frac{3}{5} \Rightarrow \frac{m_{1} m_{2}}{-3}=\frac{n_{1} n_{2}}{5}\) ………. (1)

From (1), n = -(l + m).

Substituting is (2)

-2m (l + m) – 3l(l + m) – 5lm = 0

– 2lm – 2m2 – 3l² – 3lm – 5lm = 0

3l² + 10lm + 2m² = 0

\(\frac{l_{1} l_{2}}{m_{1} m_{2}}=\frac{2}{3} \Rightarrow \frac{l_{1} l_{2}}{2}=\frac{m_{1} m_{2}}{3}\) ……… (2)

From (1) and (2) we get

\(\frac{I_{1} I_{2}}{2}=\frac{m_{1} m_{2}}{3}=\frac{n_{1} n_{2}}{-5}=1\)

l_{1}l_{2} = 2k, m_{1}m_{2} = 3k, n_{1} n_{2} = -5k

∴ l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2} = 2k + 3k – 5k = 0

The two lines are perpendicular.

Question 2.

Find the angle between the lines whose direction cosines satisfy the equations l + m + n = 0, l² + m² – n² = 0.

Solution:

Given l + m + n = 0

l² + m² – n² = 0

From (1), l = – (m + n)

Substituting in (2)

(m + n)² + m² – n² = 0

m² + n² + 2mn + m² – n² = 0

2m² + 2mn = 0

2m(m + n) = 0

∴ m = 0 or m + n = 0

Case (i) : m = 0, Substituting in (1) l + n = 0

l = -n ⇒ \(\frac{l}{1}\) = \(\frac{n}{-1}\)

D.Rs of the first line l_{1} are (1, 0, – 1)

Case (ii) :m + n = 0 ⇒ m = -n ⇒ \(\frac{m}{1}\) = \(\frac{n}{-1}\)

Substituting in (1) l = 0

D.Rs of the second line l_{2} are (0, 1 – 1)

Suppose θ is the angle between the two lines

Question 3.

If a ray makes angles α, α, γ and δ with the four diagonals of a cube find cos² α + cos² β + cos² γ + cos² δ.

Solution:

Let each side of the cube be of length a. Let one of the vertices of the cube be the origin O and the co-ordinate axes be along the three edges \(\overline{\mathrm{O A}}\), \(\overline{\mathrm{O B}}\) and \(\overline{\mathrm{O C}}\) passing through the origin. The co-ordinates of the vertices of the cube with respect to the frame of reference OABC are as shown in figure. The diagonals of he cubeare \(\overline{\mathrm{O P}}\) \(\overline{\mathrm{C D}}\), \(\overline{\mathrm{A E}}\) and \(\overline{\mathrm{B F}}\) .(a, a, a), (a, a, -a), (-a, a, a) and (a, -a, a) are direction . ratios of these diagonals respectively.

Let the direction cosines of the given ray be (l, m, n). If this fay makes the angles α, β, γ and δ with the four diagonals of the cube, then

cos² α + cos² β + cos² γ + cos² δ

\(\frac{1}{3}\){|l + m + n|² +|l + m – n|² + |-l + m + n|² + |l – m + n|²}

\(\frac{1}{3}\)[(l + m + n)² + (l + m – n)² + (-l + m + n)² + (l – m + n)²]

\(\frac{1}{3}\)[4(l² + m² + n²)] = \(\frac{4}{3}\) (since l² + m² + n² = 1)

Question 4.

If (l_{1}, m_{1}, n_{1}), (l_{2}, m_{2}, n_{2}) are d.c.s of two intersecting lines, show that d.c.s of two lines, bisecting the angles between them are proportional to l_{1} ± l_{2}, m_{1} ± m_{2} n_{1} ± n_{2}.

Solution:

OA and OB are the given lines

A and B are points at unit distances from O

Co-ordinates of A are (l_{1}, m_{1}, n_{1})

Co-ordinates of B are (l_{2}, m_{2}, n_{2})

P is the mid point of AB

Co-ordinates of P are \(\left(\frac{l_{1}+l_{2}}{2}, \frac{m_{1}+m_{2}}{2}, \frac{n_{1}+n_{2}}{2}\right)\)

∴ OP is the bisector of ∠AOB

D.Rs of OP are l_{1} + l_{2}, m_{1}+ m_{2}, n_{1}+ n_{2}

Suppose B’ is a point are OB such that OB’ OB = 1

Co-ordinates of B’ are (-l_{2}, -m_{2}, -n_{2})

Q is the mid point of AB’

Co-ordinates of Q are \(\left(\frac{l_{1}-l_{2}}{2}, \frac{m_{1}-m_{2}}{2}, \frac{n_{1}-n_{2}}{2}\right)\)

OQ is the other bisector of ∠AOB

D.Rs of OQ are l_{1} – l_{1}, m_{1} – m_{2}, n_{1} – n_{2}

Question 5.

A (-1, 2, -3), B(5, 0, -6), C(0, 4, -1) are three points. Show that the direction cosines of the bisectors of ∠BAC are propotional to (25, 8, 5) and (-11, 20, 23).

Solution:

A (-1, 2, -3), B (5, 0, -6) and C (0, 4, -1) are the given points.

D.Rs of AB are 5 +1, 0 -2, -6 +3 ie., 6, -2, -3

D.Rs of AB are \(\frac{6}{7}, \frac{-2}{7}, \frac{-3}{7}\)

D.Rs of AC are 0 + 1, 4 – 2, -1 + 3 i.e., 1, 2, 2

D.Rs of AC are \(\frac{1}{3}, \frac{2}{3}, \frac{2}{3}\)

D.Rs of one of the bisectors are

D.Rs of one of the bisectos are (25, 8, 5)

D.Rs of the other bisectors are

D.Rs of the second bisector are (-11, 20, 23)

Question 6.

If (6,10,10), (1, 0, -5), (6, -10, 0) are vertices of a triangle, find the direction ratios of its sides. Determine whether it is right angled or isosceles.

Solution:

A (6, 10, 10), B (1, 0, -5), C (6, -10, 0) are the vertices of ∆ABC

D.Rs of AB are 5, 10, 15 i.e., 1, 2, 3

D.Rs of BC are -5, 10, -5 i.e., 1, -2, 1

D.Rs of AC are 0,20,10, i.e., 0, 2, 1

∴ The given triangle is right angle a.

Question 7.

The vertices of a triangle are A (1, 4, 2), B (-2, 1, 2) C (2, 3, -4). Find ∠A, ∠B, ∠C.

Solution:

A (1, 4, 2), B (-2, 1, 2), C (2, 3, -4) and the vertices of OABC.

D.Rs of AB are 3, 3, 0 i.e., 1, 1, 0

D.Rs of BC are -4, -2, 6 i.e., 2, 1, -3

D.Rs of AC are -1, 1, 6.

Question 8.

Find the angle between the lines whose direction cosines are given by the equation 3l + m + 5n = 0 and 6mn – 2nl + 5lm = 0

Solution:

Given 3l + m + 5n = 0 ……….. (1)

6mn – 2nl + 5lm = 0 …………. (2)

From (1), m = – (3l + 5n)

Substituting in (2)

-6n(3l + 5n) -2nl – 5l(3l + 5n) = 0

-18ln – 30n² – 2nl – 15l² – 25ln = 0

-15l² – 45ln – 30n² = 0

l² + 3ln + 2n² = 0

(l + 2n) (l + n) — 0

l + 2n = 0 or l + n = 0

Case (i) : l_{1} + n_{1} = 0 ⇒ n_{1} = – l_{1}; ⇒ \(\frac{l_{1}}{1}=\frac{n_{1}}{-1}\)

But m_{1} = -(3l_{1} + 5n_{1}) = -(-3n_{1} + 5n_{1}) = -2n_{1}

∴ \(\frac{m_{1}}{+2}=\frac{n_{1}}{-1}\)

∴ \(\frac{l_{1}}{1}=\frac{m_{1}}{2}=\frac{n_{1}}{-1}\)

D.Rs of the first line lx are (1,2, -1)

Case (ii): l_{2} + 2n_{2} = 0

l_{2} = —2n_{2} ⇒ \(\frac{l_{2}}{-2}=\frac{n_{2}}{1}\)

m_{2} = -(3l_{2} + 5n_{2}) = – (-6n_{2} + 5n_{2}) = n_{2}

\(\frac{m_{2}}{1}=\frac{n_{2}}{1}\)

∴ \(\frac{l_{2}}{-2}=\frac{m_{2}}{1}=\frac{n_{2}}{1}\)

D.Rs of the second line Z2 are (-2, 1, 1)

Suppose ‘θ’ is the angle between the lines l_{1} and l_{2
}

Question 9.

If a variable line in two adjecent positions has direction cosines (l, m, n) and (l + δl, m + δm, n + δn), show that the small angle δθ between two position. Is given by (δθ)² = (δl)² + (δm)² + (δn)².

Solution:

Since (l, m, n) and (l + δl, m + δm, n + δn) are direction cosines, we have

l² + m² + n² = 1 …………. (1)

(l + δl)² + (m + δm)² + (n + δn)² — 1 ………. (2)

(2) – (1) gives (l + δl)² + (m + δm)² + (n + 8n)² – (l² + m² + n²) = 0

2(l.δl + m.δm + n.δn) = -((δl)² + (δm)² + (δn)²) …………. (3)

cos θ . δθ = l(l + δl) + m(m + δm) + n(n + δn)

= (l² + m² + n²) + (l.δl + m.δm + n.δn)

= 1 – \(\frac{1}{2}\) [(δl)² + (δm)² + (δn)²]

(δl)² + (δm)² + (δn)² = 1 = ² (1 – cos θ.δ θ)

δθ being small, sin \(\frac{\delta \theta}{2}\) = \(\frac{\delta \theta}{2}\)

∴ sin²θ \(\frac{\delta \theta}{2}\) = 4(\(\frac{\delta \theta}{2}\))² = (δθ)²

∴ (δθ)² = (δl)² + (δm)² + (δn)².