Mahesh

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Matrices Solutions Exercise 3(d) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Matrices Solutions Exercise 3(d)

I.

Question 1.
Find the determinants of the following matrices.
(i) \(\left[\begin{array}{cc}
2 & 1 \\
1 & -5
\end{array}\right]\)
Solution:
det A = ad – bc
= 2(-5) – 1(1)
= -10 – 1
= -11

(ii) \(\left[\begin{array}{cc}
4 & 5 \\
-6 & 2
\end{array}\right]\)
Solution:
det A = 4(2) – (-6)(5)
= 8 + 30
= 38

(iii) \(\left[\begin{array}{cc}
\mathbf{i} & 0 \\
0 & -\mathbf{i}
\end{array}\right]\)
Solution:
det A = -i² – 0
= 1 – 0
= 1

(iv) \(\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right]\)
Solution:
det A = 0(0 – 1) – 1(0 – 1) + 1(1 – 0)
= 1 + 1
= 2

(v) \(\left[\begin{array}{ccc}
1 & 4 & 2 \\
2 & -1 & 4 \\
-3 & 7 & 6
\end{array}\right]\)
Solution:
det A = 1(-6 – 28) – 4(12 + 12) + 2(14 – 3)
= -34 – 96 + 22
= -108

(vi) \(\left[\begin{array}{ccc}
2 & -1 & 4 \\
4 & -3 & 1 \\
1 & 2 & 1
\end{array}\right]\)
Solution:
det A = 2(-3 – 2) + 1(4 – 1) + 4(8 + 3)
= -10 + 3 + 44
= 37

(vii) \(\left[\begin{array}{ccc}
1 & 2 & -3 \\
4 & -1 & 7 \\
2 & 4 & -6
\end{array}\right]\)
Solution:
det A = 0 since R₁ and R₃ are proportional.

(viii) \(\left[\begin{array}{lll}
a & h & g \\
\text { h } & b & f \\
g & f & c
\end{array}\right]\)
Solution:
det A = a(bc – f²) – h(ch – fg) + g(hf – bg)
= abc – af² – ch² + fgh + fgh – bg²
= abc + 2fgh – af² – bg² – ch²

(ix) \(\left[\begin{array}{lll}
\mathbf{a} & \mathbf{b} & \mathbf{c} \\
\mathbf{b} & \mathbf{c} & \mathbf{a} \\
\mathbf{c} & \mathbf{a} & \mathbf{b}
\end{array}\right]\)
Solution:
det A = a(bc – a²) – b(b² – ac) + c(ab – c²)
= abc – a³ – b³ + abc + abc – c³
= 3abc – a³ – b³ – c³

(x) \(\left[\begin{array}{ccc}
1^{2} & 2^{2} & 3^{2} \\
2^{2} & 3^{2} & 4^{2} \\
3^{2} & 4^{2} & 5^{2}
\end{array}\right]\)
Solution:
det A = \(\left|\begin{array}{ccc}
1 & 4 & 9 \\
4 & 9 & 16 \\
9 & 16 & 25
\end{array}\right|\)
= 1(225 – 256) – 4(100 – 144) – 9(64 – 81)
= -31 + 176 – 153
= -184 + 176
= -8

Question 2.
If A = \(\left[\begin{array}{ccc}
1 & 0 & 0 \\
2 & 3 & 4 \\
5 & -6 & x
\end{array}\right]\) and det A = 45 then find x.
Solution:
det A = 45
\(\left|\begin{array}{ccc}
1 & 0 & 0 \\
2 & 3 & 4 \\
5 & -6 & x
\end{array}\right|\) = 45
⇒ 3x + 24 = 45
⇒ 3x – 45 + 24 = 0
⇒ 3x – 21 = 0
⇒ x = 7

II.

Question 1.
Show that \(\left|\begin{array}{lll}
b c & b+c & 1 \\
c a & c+a & 1 \\
a b & a+b & 1
\end{array}\right|\) = (a – b)(b – c)(c – a)
Solution:

Question 2.
Show that \(\left|\begin{array}{ccc}
\mathbf{b}+\mathbf{c} & \mathbf{c}+\mathbf{a} & \mathbf{a}+\mathbf{b} \\
\mathbf{a}+\mathbf{b} & \mathbf{b}+\mathbf{c} & \mathbf{c}+\mathbf{a} \\
\mathbf{a} & \mathbf{b} & \mathbf{c}
\end{array}\right|\) = a³ + b³ + c³ – 3abc
Solution:

= (a + b + c) [(-ac + b²) – (-c² + ab) + (-bc + a²)]
= (a + b + c) (-ac + b² + c² – ab – bc + a²)
= (a + b + c) (a² + b² + c² – ab – bc – ca)
= a³ + b³ + c³ – 3abc

Question 3.
Show that \(\left|\begin{array}{ccc}
\mathbf{y}+\mathbf{z} & \mathbf{x} & \mathbf{x} \\
\mathbf{y} & \mathbf{z}+\mathbf{x} & \mathbf{y} \\
\mathbf{z} & \mathbf{z} & \mathbf{x}+\mathbf{y}
\end{array}\right|\) = 4xyz
Solution:
L.H.S = \(\left|\begin{array}{ccc}
\mathbf{y}+\mathbf{z} & \mathbf{x} & \mathbf{x} \\
\mathbf{y} & \mathbf{z}+\mathbf{x} & \mathbf{y} \\
\mathbf{z} & \mathbf{z} & \mathbf{x}+\mathbf{y}
\end{array}\right|\)
= (y + z) [(z + x) (x + y) – yz] – x[y(x + y) – yz] + x[yz – z(z + x)]
= (y + z) (zx + yz + x² + xy – yz) – x(xy + y² – yz) + x(yz – z² – zx)
= (y + z) (zx + x² + xy) – x(xy + y² – yz) + x(yz – z² – zx)
= xyz + x²y + xy² + xz² + x²z + xyz – x²y – xy² + xyz + xyz – xz² – x²z
= 4xyz
= R.H.S

Question 4.
If \(\left|\begin{array}{ccc}
a & a^{2} & 1+a^{3} \\
b & b^{2} & 1+b^{3} \\
c & c^{2} & 1+c^{3}
\end{array}\right|\) = 0 and \(\left|\begin{array}{ccc}
a & a^{2} & 1 \\
b & b^{2} & 1 \\
c & c^{2} & 1
\end{array}\right|\) ≠ 0 then show that abc = -1
Hint: If each element in a row (column) of a square matrix is the sum of two numbers, then its discriminant can be expressed as the sum of discriminants of two square matrices.
Solution:

Question 5.
Without expanding the determinant, prove that
(i) \(\left|\begin{array}{ccc}
a & a^{2} & b c \\
b & b^{2} & c a \\
c & c^{2} & a b
\end{array}\right|=\left|\begin{array}{ccc}
1 & a^{2} & a^{3} \\
1 & b^{2} & b^{3} \\
1 & c^{2} & c^{3}
\end{array}\right|\)
Solution:

(ii) \(\left|\begin{array}{ccc}
a x & b y & c z \\
x^{2} & y^{2} & z^{2} \\
1 & 1 & 1
\end{array}\right|=\left|\begin{array}{ccc}
a & b & c \\
x & y & z \\
y z & z x & x y
\end{array}\right|\)
Solution:

(iii) \(\left|\begin{array}{lll}
1 & b c & b+c \\
1 & c a & c+a \\
1 & a b & a+b
\end{array}\right|=\left|\begin{array}{ccc}
1 & a & a^{2} \\
1 & b & b^{2} \\
1 & c & c^{2}
\end{array}\right|\)
Solution:
L.H.S = \(\left|\begin{array}{ccc}
1 & b c & b+c \\
1 & c a & c+a \\
1 & a b & a+b
\end{array}\right|\)

= (b – a) (c – a) (c + a – b – a)
= (a – b) (b – c) (c – a)
∴ LHS = RHS

Question 6.
If ∆₁ = \(\left|\begin{array}{ccc}
a_{1}^{2}+b_{1}+c_{1} & a_{1} a_{2}+b_{2}+c_{2} & a_{1} a_{3}+b_{3}+c_{3} \\
b_{1} b_{2}+c_{1} & b_{2}^{2}+c_{2} & b_{2} b_{3}+c_{3} \\
c_{3} c_{1} & c_{3} c_{2} & c_{3}^{2}
\end{array}\right|\) and ∆₂ = \(\left|\begin{array}{lll}
a_{1} & b_{2} & c_{2} \\
a_{2} & b_{2} & c_{2} \\
a_{3} & b_{3} & c_{3}
\end{array}\right|\), then find the value of \(\frac{\Delta_{1}}{\Delta_{2}}\)
Solution:

Question 7.
If ∆₁ = \(\left|\begin{array}{ccc}
1 & \cos \alpha & \cos \beta \\
\cos \alpha & 1 & \cos \gamma \\
\cos \beta & \cos \alpha & 1
\end{array}\right|\), ∆₂ = \(\left|\begin{array}{ccc}
0 & \cos \alpha & \cos \beta \\
\cos \alpha & 0 & \cos \gamma \\
\cos \beta & \cos \gamma & 0
\end{array}\right|\) and ∆₁ = ∆₂, then show that cos²α + cos²β + cos²γ = 1
Solution:
∆₁ = \(\left|\begin{array}{ccc}
1 & \cos \alpha & \cos \beta \\
\cos \alpha & 1 & \cos \gamma \\
\cos \beta & \cos \alpha & 1
\end{array}\right|\)
= 1(1 – cos²γ) – cos α (cos α – cos β cos γ) + cos β (cos α cos γ – cos β)
= 1 – cos²γ – cos²α + cos α cos β cos γ + cos α cos β cos γ – cos²β
= 1 – cos²γ – cos²α – cos²β + 2 cos α cos β cos γ
∆₂ = \(\left|\begin{array}{ccc}
0 & \cos \alpha & \cos \beta \\
\cos \alpha & 0 & \cos \gamma \\
\cos \beta & \cos \gamma & 0
\end{array}\right|\)
= 0(0 – cos²γ) – cos α (0 – cos γ cos β) + cos β (cos α cos γ – 0)
= cos α cos β cos γ + cos α cos β cos γ
= 2 cos α cos β cos γ
Given ∆₁ = ∆₂
1 – cos²α – cos²β – cos²γ + 2 cos α cos β cos γ = 2 cos α cos β cos γ
1 – cos²α – cos²β – cos²γ = 0
1 = cos²α + cos²β + cos²γ

III.

Question 1.
Show that \(\left|\begin{array}{ccc}
\mathbf{a}+\mathbf{b}+2 \mathbf{c} & \mathbf{a} & \mathbf{b} \\
\mathbf{c} & \mathbf{b}+\mathbf{c}+\mathbf{2} \mathbf{a} & \mathbf{b} \\
\mathbf{c} & \mathbf{a} & \mathbf{c}+\mathbf{a}+\mathbf{2} \mathbf{b}
\end{array}\right|\) = 2(a + b + c)³
Solution:

Question 2.
Show that \(\left|\begin{array}{lll}
a & b & c \\
b & c & a \\
c & a & b
\end{array}\right|^{2}\) = \(\left|\begin{array}{ccc}
2 b c-a^{2} & c^{2} & b^{2} \\
c^{2} & 2 a c-b^{2} & a^{2} \\
b^{2} & a^{2} & 2 a b-c^{2}
\end{array}\right|\) = (a³ + b³ + c³ – 3abc)²
Solution:

Question 3.
Show that \(\left|\begin{array}{ccc}
a^{2}+2 a & 2 a+1 & 1 \\
2 a+1 & a+2 & 1 \\
3 & 3 & 1
\end{array}\right|\) = (a – 1)³
Solution:

Question 4.
Show that \(\left|\begin{array}{ccc}
a & b & c \\
a^{2} & b^{2} & c^{2} \\
a^{3} & b^{3} & c^{3}
\end{array}\right|\) = abc(a – b)(b – c)(c – a)
Solution:

Question 5.
Show that \(\left|\begin{array}{ccc}
-2 \mathbf{a} & \mathbf{a}+\mathbf{b} & \mathbf{c}+\mathbf{a} \\
\mathbf{a}+\mathbf{b} & -\mathbf{2} \mathbf{b} & \mathbf{b}+\mathbf{c} \\
\mathbf{c}+\mathbf{a} & \mathbf{c}+\mathbf{b} & -2 \mathbf{c}
\end{array}\right|\) = 4(a + b) (b + c) (c + a)
Solution:

∴ (c + a) is a factor for ∆
Similarly a + b, b + c are also factors ∆.
∵ ∆ is a third-degree expression in a, b, c.
∆ = k(a + b) (b + c) (c + a),
where k is a non-zero scalar.
Put a = 1, b = 1, c = 1, then
\(\left|\begin{array}{ccc}
-2 & 2 & 2 \\
2 & -2 & 2 \\
2 & 2 & -2
\end{array}\right|\) = k(1 + 1) (1 + 1) (1 + 1)
⇒ -2(4 – 4) – 2(-4 – 4) + 2(4 + 4) = 8k
⇒ 16 + 16 = 8k
⇒ k = 4
∴ ∆ = 4(a + b) (b + c) (c + a)
Hence \(\left|\begin{array}{ccc}
-2 \mathbf{a} & \mathbf{a}+\mathbf{b} & \mathbf{c}+\mathbf{a} \\
\mathbf{a}+\mathbf{b} & -\mathbf{2} \mathbf{b} & \mathbf{b}+\mathbf{c} \\
\mathbf{c}+\mathbf{a} & \mathbf{c}+\mathbf{b} & -2 \mathbf{c}
\end{array}\right|\) = 4(a + b) (b + c) (c + a)

Question 6.
Show that \(\left|\begin{array}{lll}
\mathbf{a}-\mathbf{b} & \mathbf{b}-\mathbf{c} & \mathbf{c}-\mathbf{a} \\
\mathbf{b}-\mathbf{c} & \mathbf{c}-\mathbf{a} & \mathbf{a}-\mathbf{b} \\
\mathbf{c}-\mathbf{a} & \mathbf{a}-\mathbf{b} & \mathbf{b}-\mathbf{c}
\end{array}\right|\)
Solution:
L.H.S = \(\left|\begin{array}{ccc}
0 & 0 & 0 \\
b-c & c-a & a-b \\
c-a & a-b & b-c
\end{array}\right|\) = 0
By R₁ → R₁ + (R₂ + R₃)

Question 7.
Show that \(\left|\begin{array}{ccc}
1 & a & a^{2}-b c \\
1 & b & b^{2}-c a \\
1 & c & c^{2}-a b
\end{array}\right|\) = 0
Solution:

Question 8.
Show that \(\left|\begin{array}{lll}
\mathbf{x} & \mathbf{a} & \mathbf{a} \\
\mathbf{a} & \mathbf{x} & \mathbf{a} \\
\mathbf{a} & \mathbf{a} & \mathbf{x}
\end{array}\right|\) = (x + 2a) (x – a)²
Solution:

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

Intermediate 1st Year Maths 1A Matrices Solutions Exercise 3(c)

I.

Question 1.
If A = \(\left[\begin{array}{ccc}
2 & 0 & 1 \\
-1 & 1 & 5
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
-1 & 1 & 0 \\
0 & 1 & -2
\end{array}\right]\), then find (AB’)’.
Solution:

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

Question 3.
If A = \(\left[\begin{array}{cc}
2 & -4 \\
-5 & 3
\end{array}\right]\), then find A + A’ and A.A’
Solution:

Question 4.
If A = \(\left[\begin{array}{ccc}
-1 & 2 & 3 \\
2 & 5 & 6 \\
3 & x & 7
\end{array}\right]\) is a symmetric matrix, then find x.
Hint: ‘A’ is a symmetric matrix ⇒ A^T = A
Solution:
A is a symmetric matrix
⇒ A’ = A

Equating 2nd row, 3rd column elements we get x = 6.

Question 5.
If A = \(\left[\begin{array}{ccc}
0 & 2 & 1 \\
-2 & 0 & -2 \\
-1 & x & 0
\end{array}\right]\) is a skew-symmetric matrix, find x.
Solution:
∵ A is a skew-symmetric matrix
⇒ A^T = -A

Equating second-row third column elements we get x = 2.

Question 6.
Is \(\left[\begin{array}{ccc}
0 & 1 & 4 \\
-1 & 0 & 7 \\
-4 & -7 & 0
\end{array}\right]\) symmetric or skewsymmetric?
Solution:

∴ A is a skew-symmetric matrix.

II.

Question 1.
If A = \(\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\), show that A.A’ = A’. A = I₂
Solution:

From (1), (2) we get A.A’ = A’. A = I₂

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

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

Question 4.
For any square matrix A, Show that AA’ is symmetric.
Solution:
A is a square matrix
(AA’)’ = (A’)’A’ = A.A’
∵ (AA’)’ = AA’
⇒ AA’ is a symmetric matrix.

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

Intermediate 1st Year Maths 1A Matrices Solutions Exercise 3(b)

I.

Question 1.
Find the following products wherever possible.
Hint: (1 × 3) by (3 × 1) = 1 × 1
(i) \(\left[\begin{array}{lll}
-1 & 4 & 2
\end{array}\right]\left[\begin{array}{l}
5 \\
1 \\
3
\end{array}\right]\)
(ii) \(\left[\begin{array}{ccc}
2 & 1 & 4 \\
6 & -2 & 3
\end{array}\right]\left[\begin{array}{l}
1 \\
2 \\
1
\end{array}\right]\)
(iii) \(\left[\begin{array}{cc}
3 & -2 \\
1 & 6
\end{array}\right]\left[\begin{array}{cc}
4 & -1 \\
2 & 5
\end{array}\right]\)
(iv) \(\left[\begin{array}{lll}
2 & 2 & 1 \\
1 & 0 & 2 \\
2 & 1 & 2
\end{array}\right]\left[\begin{array}{ccc}
-2 & -3 & 4 \\
2 & 2 & -3 \\
1 & 2 & -2
\end{array}\right]\)
(v) \(\left[\begin{array}{ccc}
3 & 4 & 9 \\
0 & -1 & 5 \\
2 & 6 & 12
\end{array}\right]\left[\begin{array}{ccc}
13 & -2 & 0 \\
0 & 4 & 1
\end{array}\right]\)
(vi) \(\left[\begin{array}{c}
1 \\
-2 \\
1
\end{array}\right]\left[\begin{array}{ccc}
2 & 1 & 4 \\
6 & -2 & 3
\end{array}\right]\)
(vii) \(\left[\begin{array}{cc}
1 & -1 \\
-1 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\)
(viii) \(\left[\begin{array}{ccc}
0 & c & -b \\
-c & 0 & a \\
b & -a & 0
\end{array}\right]\left[\begin{array}{ccc}
a^{2} & a b & a c \\
a b & b^{2} & b c \\
a c & b c & c^{2}
\end{array}\right]\)
Solution:

(v) \(\left[\begin{array}{ccc}
3 & 4 & 9 \\
0 & -1 & 5 \\
2 & 6 & 12
\end{array}\right]\left[\begin{array}{ccc}
13 & -2 & 0 \\
0 & 4 & 1
\end{array}\right]\)
First matrix is a 3 × 3 matrix and second matrix is 2 × 3 matrix.
No. of columns in the first matrix ≠ No. of rows in the second matrix.
∴ Matrix product is not possible.

(vi) \(\left[\begin{array}{c}
1 \\
-2 \\
1
\end{array}\right]\left[\begin{array}{ccc}
2 & 1 & 4 \\
6 & -2 & 3
\end{array}\right]\)
No. of columns in first matrix = 1
No. of rows in second matrix = 2
No. of columns in the first matrix ≠ No. of rows in the second matrix
Multiplication of matrices is not possible.

Question 2.
If A = \(\left[\begin{array}{ccc}
1 & -2 & 3 \\
-4 & 2 & 5
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
2 & 3 \\
4 & 5 \\
2 & 1
\end{array}\right]\), do AB and BA exist? If they exist, find them. Do A and B commute with respect to multiplication?
Solution:

AB ≠ BA
∴ A and B are not commutative with respect to the multiplication of matrices.

Question 3.
Find A² where A = \(\left[\begin{array}{cc}
4 & 2 \\
-1 & 1
\end{array}\right]\)
Solution:

Question 4.
If A = \(\left[\begin{array}{ll}
i & 0 \\
0 & i
\end{array}\right]\), find A².
Solution:

Question 5.
If A = \(\left[\begin{array}{cc}
i & 0 \\
0 & -i
\end{array}\right]\), B = \(\left[\begin{array}{cc}
0 & -1 \\
1 & 0
\end{array}\right]\) and C = \(\left[\begin{array}{ll}
0 & \mathbf{i} \\
\mathbf{i} & \mathbf{0}
\end{array}\right]\), and I is the unit matrix of order 2, then show that
(i) A² = B² = C² = -I
(ii) AB = -BA = -C
Solution:

Question 6.
If A = \(\left[\begin{array}{ll}
2 & 1 \\
1 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{lll}
3 & 2 & 0 \\
1 & 0 & 4
\end{array}\right]\), find AB. Find BA if it exists.
Solution:
Given A = \(\left[\begin{array}{ll}
2 & 1 \\
1 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{lll}
3 & 2 & 0 \\
1 & 0 & 4
\end{array}\right]\)

The order of AB is 2 × 3
BA does not exist since no. of columns in B ≠ No. of rows in A.

Question 7.
If A = \(\left[\begin{array}{cc}
2 & 4 \\
-1 & k
\end{array}\right]\) and A² = 0, then find the value of k.
Solution:

II.

Question 1.
If A = \(\left[\begin{array}{lll}
3 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 3
\end{array}\right]\) then find A⁴.
Solution:

Question 2.
If A = \(\left[\begin{array}{ccc}
1 & 1 & 3 \\
5 & 2 & 6 \\
-2 & -1 & -3
\end{array}\right]\) then find A³.
Solution:

Question 3.
If A = \(\left[\begin{array}{ccc}
1 & -2 & 1 \\
0 & 1 & -1 \\
3 & -1 & 1
\end{array}\right]\), then find A³ – 3A² – A – 3I, where I is unit matrix of order 3 × 3.
Solution:

Question 4.
If I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) and E = \(\left[\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right]\), show that (aI + bE)³ = a³I + 3a²bE, Where I is unit matrix of order 2.
Solution:

III.

Question 1.
If A = [a₁, a₂, a₃,], then for any integer n ≥ 1 show that Aⁿ = \(\left[\begin{array}{lll}
a_{1}, & a_{2}^{n}, & a_{3}^{n}
\end{array}\right]\)
Solution:
Given A = diag [a₁, a₂, a₃,] = \(\left[\begin{array}{ccc}
a_{1} & 0 & 0 \\
0 & a_{2} & 0 \\
0 & 0 & a_{3}
\end{array}\right]\)
Aⁿ = diag \(\left[\begin{array}{lll}
a_{1}^{n} & a_{2}^{n} & a_{3}^{n}
\end{array}\right]=\left[\begin{array}{ccc}
a_{1}^{n} & 0 & 0 \\
0 & a_{2}^{n} & 0 \\
0 & 0 & a_{3}^{n}
\end{array}\right]\)
This problem can be should by using Mathematical Induction
put n = 1
A¹ = \(\left[\begin{array}{ccc}
a_{1} & 0 & 0 \\
0 & a_{2} & 0 \\
0 & 0 & a_{3}
\end{array}\right]\)
∴ The result is true for n = 1
Assume the result is true for n = k
A^k = \(\left[\begin{array}{ccc}
a_{1}^{k} & 0 & 0 \\
0 & a_{2}^{k} & 0 \\
0 & 0 & a_{3}^{k}
\end{array}\right]\)
Consider

∴ The result is true for n = k + 1
Hence by the Principle of Mathematical Induction, the statement is true ∀ n ∈ N

Question 2.
If θ – φ = \(\frac{\pi}{2}\), then show that \(\left[\begin{array}{cc}
\cos ^{2} \theta & \cos \theta \sin \theta \\
\cos \theta \sin \theta & \sin ^{2} \dot{\theta}
\end{array}\right]\) \(\left[\begin{array}{cc}
\cos ^{2} \phi & \cos \phi \sin \phi \\
\cos \phi \sin \phi & \sin ^{2} \phi
\end{array}\right]\) = 0
Solution:

Question 3.
If A = \(\left[\begin{array}{rr}
3 & -4 \\
1 & -1
\end{array}\right]\) then show that Aⁿ = \(\left[\begin{array}{cc}
1+2 n & -4 n \\
n & 1-2 n
\end{array}\right]\), for any integer n ≥ 1 by using Mathematical Induction.
Solution:
We shall prove the result by Mathematical Induction.

∴ The given result is true for n = k + 1
By Mathematical Induction, the given result is true for all positive integral values of n.

Question 4.
Give examples of two square matrices A and B of the same order for which AB = 0 but BA ≠ 0.
Solution:

Question 5.
A Trust fund has to invest ₹ 30,000 in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide ₹ 30,000 among the two types of bonds if the trust fund must obtain an annual total interest of (a) ₹ 1800 (b) ₹ 2000
Solution:
Let the first bond be ‘x’ and the second bond be 30,000 – x respectively
The rate of interest is 0.05 and 0.07 respectively.
(a) \([x, 30,000-x]\left[\begin{array}{l}
0.05 \\
0.07
\end{array}\right] \quad=[1800]\)
[0.05x + 0.07(30,000 – x)] = 1800
\(\frac{5}{100} x+\frac{7}{100}(30,000-x)=1800\)
5x + 21,0000 – 7x = 1,80,000
-2x = 1,80,000 – 2,10,000 = -30,000
x = 15,000
∴ First bond = 15,000
Second bond = 30,000 – 15,000 = 15,000

(b) \(\left[\begin{array}{ll}
x & 30,000-x
\end{array}\right]\left[\begin{array}{l}
0.05 \\
0.07
\end{array}\right]=[2000]\)
[0.05x + 0.07(30,000 – x)] = [2000]
\(\frac{5 x}{100} \times \frac{7}{100}(30,000-x)=2000\)
5x + 2,10,000 – 7x = 2,00,000
-2x = 2,00,000 – 2,10,000
-2x = -10,000
x = 5,000
∴ First bond = 5000
Second bond = 30,000 – 5000 = 25,000

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

Intermediate 1st Year Maths 1A Matrices Solutions Exercise 3(a)

I.

Question 1.
Write the following as a single matrix.
(i) \(\left[\begin{array}{lll}
2 & 1 & 3
\end{array}\right]+\left[\begin{array}{lll}
1 & 0 & 0
\end{array}\right]\)
(ii) \(\left[\begin{array}{ccc}
3 & 9 & 0 \\
1 & 8 & -2
\end{array}\right]+\left[\begin{array}{ccc}
4 & 0 & 2 \\
7 & 1 & 4
\end{array}\right]\)
(iii) \(\left[\begin{array}{c}
0 \\
1 \\
-1
\end{array}\right]+\left[\begin{array}{l}
1 \\
1 \\
1
\end{array}\right]\)
(iv) \(\left[\begin{array}{cc}
-1 & 2 \\
2 & -2 \\
3 & 1
\end{array}\right]-\left[\begin{array}{cc}
0 & 1 \\
-1 & 0 \\
-2 & 1
\end{array}\right]\)
Solution:

Question 2.
If A = \(\left[\begin{array}{cc}
-1 & 3 \\
4 & 2
\end{array}\right]\), B = \(\left[\begin{array}{cc}
2 & 1 \\
3 & -5
\end{array}\right]\), X = \(\left[\begin{array}{ll}
x_{1} & x_{2} \\
x_{3} & x_{4}
\end{array}\right]\) and A + B = X, then find the values of x₁, x₂, x₃ and x₄.
Solution:
A + B = X

∴ x₁ = 1, x₂ = 4, x₃ = 7, x₄ = -3

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

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

Question 5.
If \(\left[\begin{array}{cc}
x-3 & 2 y-8 \\
z+2 & 6
\end{array}\right]=\left[\begin{array}{cc}
5 & 2 \\
-2 & a-4
\end{array}\right]\) then find the values of x, y, z and a.
Solution:
Given \(\left[\begin{array}{cc}
x-3 & 2 y-8 \\
z+2 & 6
\end{array}\right]=\left[\begin{array}{cc}
5 & 2 \\
-2 & a-4
\end{array}\right]\)
∴ x – 3 = 5 ⇒ x = 3 + 5 = 8
2y – 8 = 2 ⇒ 2y = 8 + 2 = 10 ⇒ y = 5
z + 2 = -2 ⇒ z = -2 – 2 = -4
a – 4 = 6 ⇒ a = 4 + 6 = 10

II.

Question 1.
If \(\left[\begin{array}{ccc}
x-1 & 2 & 5-y \\
0 & z-1 & 7 \\
1 & 0 & a-5
\end{array}\right]=\left[\begin{array}{ccc}
1 & 2 & 3 \\
0 & 4 & 7 \\
1 & 0 & 0
\end{array}\right]\) then find the values of x, y, z and a.
Solution:
Given \(\left[\begin{array}{ccc}
x-1 & 2 & 5-y \\
0 & z-1 & 7 \\
1 & 0 & a-5
\end{array}\right]=\left[\begin{array}{ccc}
1 & 2 & 3 \\
0 & 4 & 7 \\
1 & 0 & 0
\end{array}\right]\)
∴ x – 1 = 1 ⇒ x = 1 + 1 = 2
5 – y = 3 ⇒ y = 5 – 3 = 2
z – 1 = 4 ⇒ z = 4 + 1 = 5
a – 5 = 0 ⇒ a = 5

Question 2.
Find the trace of \(\left[\begin{array}{ccc}
1 & 3 & -5 \\
2 & -1 & 5 \\
1 & 0 & 1
\end{array}\right]\)
Solution:
Trace of \(\left[\begin{array}{ccc}
1 & 3 & -5 \\
2 & -1 & 5 \\
1 & 0 & 1
\end{array}\right]\) = Sum of the diagonal elements
= 1 – 1 + 1
= 1

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

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

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Mathematical Induction Solutions Exercise 2(a) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Mathematical Induction Solutions Exercise 2(a)

Using mathematical induction, prove each of the following statements for all n ∈ N.

Question 1.
1² + 2² + 3² + …… + n² = \(\frac{n(n+1)(2 n+1)}{6}\)
Solution:
Let p(n) be the given statement:
1² + 2² + 3² + ….. + n² = \(\frac{n(n+1)(2 n+1)}{6}\)
Since 1² = \(\frac{(1)(1+1)(2 \times 1+1)}{6}\)
⇒ 1 = 1 the formula is true for n = 1
i.e., p(1) is true.
Assume the statement p(n) is true for n = k
i.e., 1² + 2² + 3² + …… + 1k² = \(\frac{k(k+1)(2 k+1)}{6}\)
We show that the formula is true for n = k + 1
i.e., We show that p(k + 1) = \(\frac{(k+1)(k+2)(2 k+3)}{6}\)
(Where p(k) = 1² + 2² + 3² + … + k²)
We observe that
p(k + 1) = 1² + 2² + 3² + …… + (k)² + (k + 1)² = p(k) + (k + 1)²
Since p(k) = \(\frac{k(k+1)(2 k+1)}{6}\)
We have p(k + 1) = p(k) + (k + 1)²

∴ 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 1² + 2² + 3² + ……. + n² = \(\frac{n(n+1)(2 n+1)}{6}\) for all n ∈ N

Question 2.
2.3 + 3.4 + 4.5 + …… up to n terms = \(\frac{n\left(n^{2}+6 n+11\right)}{3}\)
Solution:
The nth term in the given series is (n + 1) (n + 2)
Let p(n) be the statement:
2.3 + 3.4 + 4.5 + …… + (n + 1) (n + 2) = \(\frac{n\left(n^{2}+6 n+11\right)}{3}\)
and let S(n) be the sum on the left-hand side.
Since S(1) = 2.3 = \(\frac{(1)(1+6+11)}{3}\) = 6
∴ The statement is true for n = 1
Assume that the statement p(n) is true for n = k
i.e., S(k) = 2.3 + 3.4 + …… + (k + 1) (k + 2) = \(\frac{k\left(k^{2}+6 k+11\right)}{3}\)
We show that the statement is true for n = k + 1
i.e., We show that S(k + 1) = \((k+1)\left[\frac{(k+1)^{2}+6(k+1)+11}{3}\right]\)
We observe that
S(k + 1) = 2.3 + 3.4 + 4.5 + + (k + 1) (k + 2) + (k + 2) (k + 3)
= S(k) + (k + 2) (k + 3)

∴ The statement holds for n = k + 1
∴ By the principle of mathematical induction,
p(n) is true for all n ∈ N
i.e., 2.3 + 3.4 + 4.5 + ……. + (n + 1) (n + 2) = \(\frac{n\left(n^{2}+6 n+11\right)}{3}\)

Question 3.
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1}\)
Solution:
Let p(n) be the statement:
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1}\)
and let S(n) be the sum on the L.H.S.
Since S(1) = \(\frac{1}{1.3}=\frac{1}{1(2+1)}=\frac{1}{1.3}\)
∴ P(1) is true.
Assume that the statement p(n) is true for n = k
i.e., S(k) = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 k-1)(2 k+1)}\) = \(\frac{k}{2 k+1}\)
We show that the statement p(n) is true for n = k + 1
i.e., we show that s(k + 1) = \(\frac{k+1}{2(k+1)+1}\)
We observe that

∴ The statement holds for n = k + 1
∴ By the principle of mathematical induction,
p(n) is true for all n ∈ N
i.e., \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1}\)

Question 4.
4³ + 8³ + 12³ + …… up to n terms = 16n²(n + 1)².
Solution:
4, 8, 12,….. are in A.P., whose nth term is (4n)
Let p(n) be the statement:
4³ + 8³ + 12³ + ………. + (4n)³ = 16n²(n + 1)²
and S(n) be the sum on the L.H.S.
S(1) = 4³ = 16(1²) (1 + 1)² = 16(4) = 64 = 4³
∴ p(1) is true
Assume that the statement p(n) is true for n = k
i.e., S(k) = 4³ + 8³ + (12)³ + …… + (4k)³ = 16k²(k + 1)²
We show that the statement is true for n = k + 1
i.e., We show that S(k + 1) = 16(k + 1)² (k + 2)²
We observe that
S(k + 1) = 4³ + 8³ + 12³ + …… + (4k)³ + [4(k + 1)]³
= S(k) + [4(k + 1)]³
= 16k² (k + 1)² + 4³ (k + 1)³
= 16(k + 1)² [k² + 4(k + 1)]
= 16(k + 1)² [k² + 4k + 4]
= 16(k + 1)² (k + 2)²
= 16(k + 1)² \((\overline{k+1}+1)^{2}\)
∴ The formula holds for n = k + 1
∴ By the principle of mathematical induction,
p(n) is true for all n ∈ N
(i.e.,) 4³ + 8³ + 12³ + …… + (4n)³ = 16n²(n + 1)²

Question 5.
a + (a + d) + (a + 2d) + ……. up to n terms = \(\frac{n}{2}\) [2a + (n – 1)d]
Solution:
Let p(n) be the statement:
a + (a + d) + (a + 2d) + …… + [a + (n – 1)d] = \(\frac{n}{2}\) [2a + (n – 1)d]
and let the sum on the L.H.S. is denoted by S(n)
Since S(1) = a = \(\frac{1}{2}\) [2a + (1 – 1)d] = a
∴ p(1) is true.
Assume that the statement is true for n = k
(i.e.,) S(k) = a + (a + d) + (a + 2d) + ……. + [a + (k – 1)d] = \(\frac{k}{2}\) [2a + (k – 1 )d]
We show that the statement is true for n = k + 1
(i.e.,) we show that S(k + 1) = \(\left(\frac{k+1}{2}\right)[2 a+k d]\)
We observe that
S(k + 1) = a + (a + d) + (a + 2d) + …… + [a + (k – 1)d] + (a + kd)
= S(k) + (a + kd)
= \(\frac{k}{2}\) [2a + (k – 1)d] + (a + kd)
= \(\frac{k[2 a+(k-1) d]+2(a+k d)}{2}\)
= \(\frac{1}{2}\) [2ak + k(k – 1)d + 2a + 2kd]
= \(\frac{1}{2}\) [2a(k + 1) + k(k – 1 + 2)d]
= \(\frac{1}{2}\) (k + 1)(2a + kd)
∴ The statement holds for n = k + 1
∴ By the principle of mathematical inductions,
p(n) is true for all n ∈ N
(i.e.,) a + (a + d) + (a + 2d) + …… + [a + (n – 1)d] = \(\frac{n}{2}\) [2a + (n – 1)d]

Question 6.
a + ar + ar² + ……… up to n terms = \(\frac{a\left(r^{n}-1\right)}{r-1}\); r ≠ 1
Solution:
Let p(n) be the statement:
a + ar + a.r² + …… + a. r^n-1 = \(\frac{a\left(r^{n}-1\right)}{r-1}\), r ≠ 1
and let S(n) be the sum on the L.H.S
Since S(1) = a = \(\frac{a\left(r^{1}-1\right)}{r-1}\) = a
∴ p(1) is true
Assume that the statement is true for n = k
(i.e) S(k) = a + ar + ar² + ……… + a . r^k-1 = \(\frac{a\left(r^{k}-1\right)}{r-1}\)
We show that the statement is true for n = k + 1
(i.e) S(k + 1) = \(\frac{a\left(r^{k+1}-1\right)}{r-1}\)
Now S(k + 1) = a + ar + ar² + ……. + a r^k-1 + ar^k
= S(k) + a . r^k

∴ The statement holds for n = k + 1
∴ By the principle of mathematical induction,
p(n) is true for all n ∈ N
(i.e) a + ar + ar² + ……. + a.r^n-1 = \(\frac{a\left(r^{n}-1\right)}{r-1}\), r ≠ 1

Question 7.
2 + 7 + 12 + ……. + (5n – 3) = \(\frac{n(5 n-1)}{2}\)
Solution:
Let p(n) be the statement:
2 + 7 + 12 + ……. + (5n – 3) = \(\frac{n(5 n-1)}{2}\)
and let S(n) be the sum on the L.H.S
Since S(1) = 2 = \(\frac{1(5 \times 1-1)}{2}=\frac{4}{2}\) = 2
∴ p(1) is true
Assume that the statement is true for n = k
(i.e) S(k) = 2 + 7 + 12 + …….. + (5k – 3) = \(\frac{k(5 k-1)}{2}\)
We have to show that S(k + 1) = \(\frac{(k+1)(5 k+4)}{2}\)
We observe that S(k + 1) = 2 + 7 + 12 + ……. + (5k – 3) + (5k + 2)
= S(k) + (5k + 2)
= \(\frac{k(5 k-1)}{2}\) + (5k + 2)
= \(\frac{5 k^{2}-k+2(5 k+2)}{2}\)
= \(\frac{1}{2}\) [5k² + 9k + 4]
= \(\frac{1}{2}\) (k + 1) (5k + 4)
= \(\frac{1}{2}\) (k + 1) [5(k + 1) – 1]
∴ p(k + 1) is true
∴ By the principle of mathematical induction,
p(n) is true for all n ∈ N.
(i.e.,) 2 + 7 + 12 + …… + (5n – 3) = \(\frac{n(5 n-1)}{2}\)

Question 8.
\(\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \ldots \ldots\left(1+\frac{2 n+1}{n^{2}}\right)\) = (n + 1)²
Solution:
Let p(n) be the statement:
\(\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \ldots \ldots\left(1+\frac{2 n+1}{n^{2}}\right)\) = (n + 1)²
and let S(n) be the product on the LHS
since S(1) = 1 + 3 = 4 = (1 + 1)² = 4
∴ P(a) 4 time for n = 1
Assume that p(n) is true for n = k

= (k + 1)² + 2k + 3
= k² + 2k + 1 + 2k + 3
= k² + 4k + 4
= (k + 2)²
= (k + 1 + 1)²
∴ P(n) is true for n = k + 1
By the principle of Mathematical Induction,
p(n) is true & n ∈ N

Question 9.
(2n + 7) < (n + 3)²
Solution:
Let p (n) be the statement
When n = 1, 9 < 16
∴ p(n) is true for n = 1
Assume p (n) is true for n = k
(2k + 7) < (k + 3)²
We show that p(n) is true for n = k + 1
2(k + 1) + 7 = 2k + 7 + 2
< (k + 3)² + 2
< k² + 6k + 9 + 2 + 2k + 5 – 2k – 5
< (k + 4)² – (2k + 5)
< (k + 4)²
< (k + 1 + 3)²
∴ p(n) is true for n = k + 1
By the principle of Mathematical Induction
p(n) is true ∀ n ∈ N

Question 10.
1² + 2² + …… + n² > \(\frac{n^{3}}{3}\)
Solution:
Let P(n) by the statement
when n = 1, 1 > \(\frac{1}{3}\)
∴ p(n) is true for n = 1
Assume p (n) is true for n = k
1² + 2² + …… + k² > \(\frac{k^{3}}{3}\)
We show that p(n) is true for n = k + 1

∴ p(n) is true for n = k + 1
By the principle of Mathematical Induction,
p(n) is true ∀ n ∈ N

Question 11.
4ⁿ – 3n – 1 is divisible by 9.
Solution:
Let p(n) be the statement:
4ⁿ – 3n – 1 is divisible by 9
Since 4¹ – 3(1) – 1 = 0 is divisible by 9.
The statement is true for n = 1
Assume that p(n) is true for n = k
(i.e) 4^k – 3k – 1 is divisible by 9
Then 4^k – 3k – 1 = 9t, for some t ∈ N ……..(1)
Show that the statement p(n) is true for n = k + 1
(i.e.,) we show that S(k + 1) = 4^k+1 – 3(k+1) – 1 is divisible by 9
From (1), we have
4^k = 9t + 3k + 1
∴ S(k + 1) = 4 . 4^k – 3(k + 1) – 1
= 4(9t + 3k + 1) – 3k – 3 – 1
= 4(9t) + 9k
= 9[4t + k]
Hence s(k + 1) is divisible by 9
Since 4t + k is an integer
∴ 4^k+1 – 3(k+1) – 1 is divisible by 9
∴ The statement is true for n = k + 1
∴ By the principle of mathematical induction,
p(n) is true for all n ∈ k
(i.e.,) 4ⁿ – 3n – 1 is divisible by 9

Question 12.
3 . 5²ⁿ⁺¹ + 2³ⁿ⁺¹ is divisible by 17.
Solution:
Let p(n) be the statement:
3. 5²ⁿ⁺¹ + 2³ⁿ⁺¹ is divisible by 17
Since 3 . 5²⁽¹⁾⁺¹ + 2³⁽¹⁾⁺¹
= 3 . 5³ + 2⁴
= 3(125) + 16
= 375 + 16
= 391
= 17(23) is divisible by 17
∴ The statement is true for n = 1
Assume that the statement is true for n = k
(i.e) 3 . 5^2k+1 + 2^3k+1 is divisible by 17
Then 3 . 5^2k+1 + 2^3k+1 = 17t, for some t ∈ N ……..(1)
Show that the statement p(n) is true for n = k + 1
(i.e.,) We have to show that
\(\text { 3. } 5^{2(k+1)+1}+2^{3(k+1)+1}\) is divisible by 17
From (1) we have

Here 25t + 2^3k+1 is an integer
∴ \(\text { 3. } 5^{2(k+1)+1}+2^{3(k+1)+1}\) is divisible by 17
∴ The statement is true for n = k + 1
∴ By the principle of mathematical induction,
p(n) is true for all n ∈ N
(i.e.,) 3 . 5²ⁿ⁺¹ + 2³ⁿ⁺¹ is divisible by 17.

Question 13.
1.2.3 + 2.3.4 + 3.4.5 + ……. upto n terms = \(\frac{n(n+1)(n+2)(n+3)}{4}\)
Solution:
The nth term of the given series is (n) (n + 1) (n + 2)
Let p(n) be the statement:
1.2.3 + 2.3.4 + 3.4.5 +……. + (n) (n+1) (n+2) = \(\frac{n(n+1)(n+2)(n+3)}{4}\)
and S(n) be the sum on the L.H.S.
∵ S(1) = 1.2.3 = \(\frac{(1)(1+1)(1+2)(1+3)}{4}\) = 1.2.3
∴ p(1) is true
Assume that the statement p(n) is true for n = k
(i.e) S(k) = 1.2.3 + 2.3.4 + 3.4.5 + ……. + k(k + 1) (k + 2) = \(\frac{k(k+1)(k+2)(k+3)}{4}\)
We show that the statement is true for n = k + 1
(i.e) We show that S(k + 1) = \(\frac{(k+1)(k+2)(k+3)(k+4)}{4}\)
We observe that
S(k + 1) = 1.2.3 + 2.3.4 + …… + k(k + 1) (k + 2) + (k + 1) (k + 2) (k + 3)
= S(k) + (k + 1) (k + 2) (k + 3)
= \(\frac{k(k+1)(k+2)(k+3)}{4}\) + (k + 1)(k + 2)(k + 3)
= (k + 1)(k + 2)(k + 3) \(\left(\frac{k}{4}+1\right)\)
= \(\frac{(k+1)(k+2)(k+3)(k+4)}{4}\)
∴ The statement holds for n = k + 1
∴ By the principle of mathematical induction,
p(n) is true for all n ∈ N
(i.e.,) 1.2.3 + 2.3.4 + 3.4.5 + ……. + (n)(n + 1)(n + 2) = \(\frac{n(n+1)(n+2)(n+3)}{4}\)

Question 14.
\(\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}\) + …. up to n terms = \(\frac{n}{24}\) [2n² + 9n + 13]
Solution:
The nth term of the given series is \(\frac{1^{3}+2^{3}+3^{3}+\ldots .+n^{3}}{1+3+5+\ldots+(2 n-1)}\)
Let p(n) be the statement :

and let S(n) be the sum on the L.H.S.
∵ S(1) = \(\frac{1^{3}}{1}=\frac{1}{24}(2+9+13)=1=\frac{1^{3}}{1}\)
∴ p(1) is true
Assume that p(k) is true
(i.e.,) S(k) = \(\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\ldots+\frac{1^{3}+2^{3}+\ldots \pm k^{3}}{1+3+\ldots+(2 k-1)}\) = \(\frac{k}{24}\) [2k² + 9k + 13]
We show that p(k + 1) is true
(i.e,) we show that


∴ The statement holds for n = k + 1
∴ By the principle of mathematical induction,
p(n) is true for all n
(i.e.,) \(\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\ldots+\frac{1^{3}+2^{3}+\ldots \ldots+n^{3}}{1+3+\ldots+(2 n-1)}\) = \(\frac{n}{24}\) [2n² + 9n + 13]

Question 15.
1² + (1² + 2²) + (1² + 2² + 3²) + ……. up to n terms = \(\frac{n(n+1)^{2}(n+2)}{12}\)
Solution:
The nth term of the given series is (1² + 2² + 3² + …… + n²)
Let p(n) be the statement:
1² + (1² + 2²) + (1² + 2² + 3²) + ……. + (1² + 2² + …… + n²) = \(\frac{\mathrm{n}(\mathrm{n}+1)^{2}(\mathrm{n}+2)}{12}\)
and the sum on the LH.S. is denoted by S(n).
Since S(1) = 1² = \(\frac{1(1+1)^{2}(1+2)}{12}\) = 1 = 1²
∴ p(1) is true.
Assume that the statement is true for n = k
(i.e.,) S(k) = 1² + (1² + 2²) + ……. + (1² + 2² + ……. + k²)
= \(\frac{k(k+1)^{2}(k+2)}{12}\)
We show that S(k + 1) = \(\frac{(k+1)(k+2)^{2}(k+3)}{12}\)
We observe that


∴ The statement holds for n = k + 1.
∴ By the principle of mathematical induction,
p(n) is true for all n ∈ N.
(i.e.,) 1² + (1² + 2²) + (1² + 2² + 3²) + …….. (1² + 2² + ………. + n²) = \(\frac{n(n+1)^{2}(n+2)}{12}\)

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Exercise 8(a) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Inverse Trigonometric Functions Solutions Exercise 8(a)

I.

Question 1.
Evaluate the following.
(i) \(\sin ^{-1}\left(\frac{-\sqrt{3}}{2}\right)\)
Solution:

(ii) \(\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)\)
Solution:
\(\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\cos ^{-1}\left(\cos \frac{\pi}{4}\right)=\frac{\pi}{4}\)

(iii) sec^-1(-√2)
Solution:

(iv) cot^-1(-√3)
Solution:

(v) \(\sin \left(\frac{\pi}{3}-\sin ^{-1}\left(\frac{-1}{2}\right)\right)\)
Solution:

(vi) \(\sin ^{-1}\left(\sin \frac{5 \pi}{6}\right)\)
Solution:

(vii) \(\cos ^{-1}\left(\cos \frac{5 \pi}{4}\right)\)
Solution:

Question 2.
Find the values of
(i) \(\sin \left(\cos ^{-1} \frac{3}{5}\right)\)
Solution:
\(\sin \left(\cos ^{-1} \frac{3}{5}\right)=\sin \left(\sin ^{-1} \frac{4}{5}\right)=\frac{4}{5}\)

(ii) \(\tan \left({cosec}^{-1} \frac{65}{63}\right)\)
Solution:
\(\tan \left({cosec}^{-1} \frac{65}{63}\right)=\tan \left(\tan ^{-1} \frac{63}{16}\right)\) = \(\frac{63}{16}\)

(iii) \(\sin \left(2 \sin ^{-1} \frac{4}{5}\right)\)
Solution:

(iv) \(\sin ^{-1}\left(\sin \frac{33 \pi}{7}\right)\)
Solution:

(v) \(\cos ^{-1}\left(\cos \frac{17 \pi}{6}\right)\)
Solution:

Question 3.
Simplify each of the following.
(i) \(\tan ^{-1}\left[\frac{\sin x}{1+\cos x}\right]\)
Solution:

(ii) tan^-1(sec x + tan x)
Solution:

(iii) \(\tan ^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}}\)
Solution:

(iv) sin^-1(2 cos²θ – 1) + cos-1(1 – 2 sin²θ)
Solution:
sin^-1(cos 2θ) + cos^-1(cos 2θ)
= sin^-1[sin (90° – 2θ)] + cos^-1(cos 2θ)
= 90° – 2θ + 2θ
= 90°

(v) \(\tan ^{-1}\left(x+\sqrt{1+x^{2}}\right)\); x ∈ R
Solution:

II.

Question 1.
Prove that
(i) \(\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{8}{17}=\cos ^{-1} \frac{36}{85}\)
Solution:

(ii) \(\sin ^{-1} \frac{3}{5}+\cos ^{-1} \frac{12}{13}=\cos ^{-1} \frac{33}{65}\)
Solution:

(iii) \(\tan \left[\cot ^{-1} 9+{cosec}^{-1} \frac{\sqrt{41}}{4}\right]=1\)
Solution:

(iv) \(\cos ^{-1} \frac{4}{5}+\sin ^{-1} \frac{3}{\sqrt{34}}=\tan ^{-1} \frac{27}{11}\)
Solution:

Question 2.
Find the values of
(i) \(\sin \left(\cos ^{-1} \frac{3}{5}+\cos ^{-1} \frac{12}{13}\right)\)
Solution:

(ii) \(\tan \left(\sin ^{-1} \frac{3}{5}+\cos ^{-1} \frac{5}{\sqrt{34}}\right)\)
Solution:

(iii) \(\cos \left(\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}\right)\)
Solution:
Let \(\sin ^{-1} \frac{3}{5}\) = α and \(\sin ^{-1} \frac{5}{13}\) = β

Question 3.
Prove that
(i) \(\cos \left[2 \tan ^{-1} \frac{1}{7}\right]=\sin \left[2 \tan ^{-1} \frac{3}{4}\right]\)
Solution:

(ii) \(\tan \left[2 \tan ^{-1}\left(\frac{\sqrt{5}-1}{2}\right)\right]=2\)
Solution:

(iii) \(\cos \left\{2\left[\tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{2}{9}\right)\right]\right\}=\frac{3}{5}\)
Solution:

Question 4.
Prove that
(i) \(\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{13}-\tan ^{-1} \frac{2}{9}=0\)
Solution:

(ii) \(\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}=\frac{\pi}{4}\)
Solution:

(iii) \(\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{3}{5}-\tan ^{-1} \frac{8}{19}=\frac{\pi}{4}\)
Solution:

(iv) \(\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{8}\) = \(\cot ^{-1} \frac{201}{43}+\cot ^{-1} 18\)
Solution:

Question 5.
Show that
(i) sec² (tan^-1 2) + cosec² (cot^-1 2) = 10
Solution:
Let a = tan^-1 2 ⇒ tan α = 2
sec² α = 1 + tan^-1 α = 1 + 4 = 5
Let β = cot^-1 2 ⇒ cot β = 2
cosec² β = 1 + cot² β = 1 + 4 = 5
LHS = sec² (tan^-1 2) + cosec² (cot^-1 2)
= 5 + 5
= 10
= RHS

(ii) Find the value of \(\left(\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right)\)
Solution:

(iii) If sin^-1 x – cos^-1 x = \(\frac{\pi}{6}\) then find x.
Solution:

III.

Question 1.
Prove that
(i) \(2 \sin ^{-1} \frac{3}{5}-\cos ^{-1} \frac{5}{13}=\cos ^{-1} \frac{323}{325}\)
Solution:

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

(iii) \(4 \tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{99}-\tan ^{-1} \frac{1}{70}=\frac{\pi}{4}\)
Solution:

Question 2.
(i) If α = \({tan}^{-1}\left(\frac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}\right)\) then prove that x² = sin 2α.
Solution:

(ii) Prove that tan\(\left\{2-\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)\right\}=\mathbf{x}\)
Solution:

(iii) Prove that \(\sin \left[\cot ^{-1} \frac{2 x}{1-x^{2}}+\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\right]\) = 1
Solution:

(iv) Prove that \(\left\{\frac{\pi}{4}+\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)\right\}\)
Solution:

Question 3.
(i) If cos^-1 p + cos^-1 q + cos^-1 r = π, then prove that p² + q² + r² + 2pqr = 1
Solution:
Let cos^-1 p = A, cos^-1 q = B and cos^-1 r = C
then A + B + C = π ………(1)
and p = cos A, q = cos B and r cos C
Now p² + q² + r² = cos² A + cos² B + cos² C
= cos² A + (1 – sin² B + cos² C)
= 1 + (cos² A – sin² B) + cos² C
= 1 + cos (A + B) . cos (A – B) + cos² C
= 1 + cos (π – C) cos (A – B) + cos² C (By (1))
= 1 – cos C cos (A – B) + cos² C
= 1 – cos C [cos (A – B) – cos C]
= 1 – cos C [cos (A – B) – cos(180° – \(\overline{A+B}\)]
= 1 – cos C [cos (A – B) + cos (A + B)]
= 1 – cos C [2 cos A cos B]
= 1 – 2 pqr
∴ p² + q² + r² + 2pqr = 1

(ii) If \(\sin ^{-1}\left(\frac{2 p}{1+p^{2}}\right)-\cos ^{-1}\left(\frac{1-q^{2}}{1+q^{2}}\right)\) = \(\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)\), then prove that x = \(\frac{p-q}{1+p q}\)
Solution:

(iii) If a, b, c are distinct non-zero real numbers having the same sign, prove that \(\cot ^{-1}\left(\frac{a b+1}{a-b}\right)+\cot ^{-1}\left(\frac{b c+1}{b-c}\right)\) + \(\cot ^{-1}\left(\frac{c a+1}{c-a}\right)\) = π or 2π
Solution:
Since (a – b) + (b – c) + (c – a) = 0.
(a – b), (b – c), (c – a) all cannot have the same sign.
Now two cases arise, namely, either two of these numbers are positive and one negative (or) two of these numbers are negative and one is positive.
Case (i): Without loss of generality, we assume that (a – b), (b – c) are both positive and (c – a) is negative

Case (ii): Without loss of generality, we assume that (a – b) and (b – c) are both negative and (c – a) is positive.

(iv) If sin^-1 (x) + sin^-1 (y) + sin^-1 (z) = π, then prove that \(x \sqrt{1-x^{2}}+y \sqrt{1-y^{2}}+z \sqrt{1-z^{2}}=2 x y z\)
Solution:
Let sin^-1 (x) = A, sin^-1 (y) = B and sin^-1 (z) = C
Then A + B + C = π …………(1)
and x = sin A, y = sin B and z = sin C
Now LHS = \(x \sqrt{1-x^{2}}+y \sqrt{1-y^{2}}+z \sqrt{1-z^{2}}\)
= sin A \(\sqrt{1-\sin ^{2} A}\) + sin B \(\sqrt{1-\sin ^{2} B}\) + sin C \(\sqrt{1-\sin ^{2} C}\)
= sin A cos A + sin B cos B + sin C cos C
= \(\frac{1}{2}\) [sin 2A + sin 2B + sin 2C]
= \(\frac{1}{2}\) [2 . sin (A + B) cos (A – B) + sin 2C]
= \(\frac{1}{2}\) [2 sin (π – c). cos (A – B) + sin 2C]
= \(\frac{1}{2}\) [2 sin C cos (A – B) + 2 sin C cos C]
= \(\frac{1}{2}\) (2 sin C) [cos (A – B) + cos C]
= sin C [cos (A – B) + cos (180° – \(\overline{A+B}\)]
= sin C [cos (A – B) – cos (A + B)]
= sin C [2 sin A sin B]
= 2 xyz
∴ \(x \sqrt{1-x^{2}}+y \sqrt{1-y^{2}}+z \sqrt{1-z^{2}}=2 x y z\)

(v) (a) If tan^-1 x + tan^-1 y + tan^-1 z = π, then prove that x + y + z = xyz.
Solution:
Let A = tan^-1 x, B = tan^-1 y, C = tan^-1 z
tan A = x, tan B = y, tan C = z
Given A + B + C = π ……….(1)
A + B = π – C
tan (A + B) = tan (π – C)
\(\frac{\tan A+\tan B}{1-\tan A \tan B}\) = -tan C
\(\frac{x+y}{1-x y}\) = -z
x + y = -z + xyz
∴ x + y + z = xyz

(b) If tan^-1 x + tan^-1 y + tan^-1 z = \(\frac{\pi}{2}\), then prove that xy + yz + zx = 1.
Solution:
Let tan^-1 x = A, tan^-1 y = B and tan^-1 z = C
Then A + B + C = \(\frac{\pi}{2}\) …….(1)
and x = tan A, y = tan B and z = tan C
∵ A + B + C = \(\frac{\pi}{2}\)
A + B = \(\frac{\pi}{2}\) – C
⇒ tan (A + B) = tan(\(\frac{\pi}{2}\) – C)
⇒ \(\frac{\tan A+\tan B}{1-\tan A \cdot \tan B}\) = cot C
⇒ \(\frac{x+y}{1-x y}=\frac{1}{z}\)
⇒ zx + yz = 1 – xy (or) xy + yz + zx = 1

Question 4.
Solve the following equations for x:
(i) \({Tan}^{-1}\left(\frac{x-1}{x-2}\right)+{Tan}^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}\)
Solution:

(ii) \(\tan ^{-1}\left(\frac{1}{2 x+1}\right)+\tan ^{-1}\left(\frac{1}{4 x+1}\right)\) = \(\tan ^{-1} \frac{2}{x^{2}}\)
Solution:
Given that

⇒ x²(3x + 1) = 2x(4x + 3)
⇒ x [x(3x + 1) – 2(4x + 3)] = 0
⇒ x = 0 (or) 3x² – 7x – 6 = 0
⇒ x = 0 (or) 3x² – 9x + 2x – 6 = 0
⇒ x = 0 (or) 3x(x – 3) + 2(x – 3) = 0
⇒ x = 0 (or) (3x + 2) (x – 3) = 0
⇒ x = 0 (or) 3 (or) \(\frac{-2}{3}\)

(iii) \(3 \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)-4 \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\) + \(2 \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{\pi}{3}\)
Solution:

(iv) sin^-1(1 – x) – 2 sin^-1 x = \(\frac{\pi}{2}\)
Solution:
Given sin^-1(1 – x) – 2 sin^-1x = \(\frac{\pi}{2}\)
Let sin^-1(1 – x) = α and sin^-1(x) = β
Then sin α = 1 – x and sin β = x
cos α = \(\sqrt{1-(1-x)^{2}}\) and cos β = \(\sqrt{1-x^{2}}\)
Now sin^-1(1 – x) – 2 sin^-1(x) = \(\frac{\pi}{2}\)
⇒ α – 2β = \(\frac{\pi}{2}\)
⇒ α = \(\frac{\pi}{2}\) + 2β
⇒ sin α = sin (\(\frac{\pi}{2}\) + 2β)
⇒ sin α = cos 2β
⇒ 1 – x = 1 – 2 sin2β
⇒ 1 – x = 1 – 2x²
⇒ 2x² – x = 0
⇒ x(2x – 1) = 0
⇒ x = 0 (or) x = \(\frac{1}{2}\)
But when x = \(\frac{1}{2}\)

Hence x = 0 is the only solution for the given equation.

Question 5.
Solve the following equations.
(i) \(\cot ^{-1}\left(\frac{1+x}{1-x}\right)=\frac{1}{2} \cot ^{-1}\left(\frac{1}{x}\right)\), x > 0 and x ≠ 1
Solution:

(ii) \(\tan \left[\cos ^{-1} \frac{1}{x}\right]=\sin \left[\cot ^{-1} \frac{1}{2}\right]\); x ≠ 0
Solution:
Let \(\cos ^{-1}\left(\frac{1}{x}\right)=\alpha, \cot ^{-1} \frac{1}{2}=\beta\)

(iii) \(\cos ^{-1} x+\sin ^{-1} \frac{x}{2}=\frac{\pi}{6}\)
Solution:

(iv) \(\cos ^{-1}(\sqrt{3} \cdot x)+\cos ^{-1} x=\frac{\pi}{2}\)
Solution:

(v) \(\sin \left[\sin ^{-1}\left(\frac{1}{5}\right)+\cos ^{-1} x\right]=1\)
Solution:

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Exercise 6(d) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Exercise 6(d)

I.

Question 1.
Simplify
(i) \(\frac{\sin 2 \theta}{1+\cos 2 \theta}\)
Solution:
\(\frac{\sin 2 \theta}{1+\cos 2 \theta}\)
= \(\frac{2 \sin \theta \cos \theta}{2 \cos ^{2} \theta}\)
= tan θ

(ii) \(\frac{3 \cos \theta+\cos 3 \theta}{3 \sin \theta-\sin 3 \theta}\)
Solution:
\(\frac{3 \cos \theta+\cos 3 \theta}{3 \sin \theta-\sin 3 \theta}\)
= \(\frac{3 \cos \theta+4 \cos ^{3} \theta-3 \cos \theta}{3 \sin \theta-\left(3 \sin \theta-4 \sin ^{3} \theta\right)}\)
= \(\frac{4 \cos ^{3} \theta}{4 \sin ^{3} \theta}\)
= cot³θ

Question 2.
Evaluate the following.
(i) 6 sin 20° – 8 sin3 20°
Solution:
6 sin 20° – 8 sin³20°
= 2(3 sin 20° – 4 sin³20°)
= 2 sin (3 × 20)
= 2 sin 60°
= 2 \(\left[\frac{\sqrt{3}}{2}\right]\)
= √3

(ii) cos²72° – sin²54°
Solution:

(iii) sin²42° – sin²12°
Solution:
sin (42 + 12) sin (42 – 12)
= sin 54° . sin 30°
= \(\left[\frac{\sqrt{5}+1}{4}\right] \frac{1}{2}\)
= \(\frac{\sqrt{5}+1}{8}\)

Question 3.
(i) Express \(\frac{\sin 4 \theta}{\sin \theta}\) interms of cos³θ and cos θ.
Solution:
consider sin 4θ = sin(3θ + θ)
= sin 3θ cos θ + cos 3θ sin θ
= (3 sin θ – 4 sin³θ) cos θ + (4 cos³θ – 3 cos θ) sin θ
= 3 sin θ cos θ – 4 sin³θ cos θ + 4 cos³θ sin θ – 3 cos θ sin θ
= 4 cos³θ sin θ – 4 sin³θ cos θ
= sin θ (4 cos³θ – 4 sin²θ cos θ)
\(\frac{\sin 4 \theta}{\sin \theta}=\frac{\sin \theta\left(4 \cos ^{3} \theta-4 \sin ^{2} \theta \cos \theta\right)}{\sin \theta}\)
= 4 cos³θ – 4(1 – cos²θ) cos θ
= 4 cos³θ – 4 cos θ + 4 cos³θ
= 8 cos³θ – 4 cos θ

(ii) Express cos⁶A + sin⁶A in terms of sin 2A.
Solution:
cos⁶A + sin⁶A
= (cos²A)³ + (sin²A)³
= (cos²A + sin²A)³ – 3 cos²A sin²A (cos²A + sin²A)
= 1 – 3 cos²A sin²A
= 1 – \(\frac{3}{4}\) (4 cos²A sin²A)
= 1 – \(\frac{3}{4}\) (sin²2A)

(iii) Express \(\frac{1-\cos \theta+\sin \theta}{1+\cos \theta+\sin \theta}\) in terms of tan θ/2
Solution:

Question 4.
(i) If sin α = \(\frac{3}{5}\), where \(\frac{\pi}{2}\) < α < π, evaluate cos 3α and tan 2α.
Solution:

(ii) If cos A = \(\frac{7}{25}\) and \(\frac{3 \pi}{2}\) < A < 2π, then find the value of cot A/2.
Solution:

(iii) If 0 < θ < \(\frac{\pi}{8}\), show that \(\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 4 \theta)}}}=2 \cos (\theta / 2)\)
Solution:

Question 5.
Find the extreme values of
(i) cos 2x + cos2x
Solution:
cos 2x + cos²x
= 2 cos²x – 1 + cos²x
= 3 cos²x – 1
-1 ≤ cos x ≤ 1
0 ≤ cos²x ≤ 1
0 ≤ 3 cos²x ≤ 3
-1 ≤ 3 cos²x – 1 ≤ 2
maximum value = 2
minimum value = -1

(ii) 3 sin²x + 5 cos²x
Solution:
3 sin²x + 5 cos²x
= 3(1 – cos²x) + 5 cos²x
= 3 – 3 cos²x + 5 cos²x
= 3 + 2 cos²x
-1 ≤ cos x ≤ 1
0 ≤ cos²x ≤ 1
0 ≤ 2 cos²x ≤ 2
3 ≤ 3 + 2 cos²x ≤ 5
maximum value = 5
minimum value = 3

Question 6.
If a ≤ cos θ + 3√2 sin[θ + \(\frac{\pi}{4}\)] + 6 ≤ b, then find the largest value of a and smallest value of b.
Solution:
a ≤ cos θ + 3√2 sin[θ + \(\frac{\pi}{4}\)] + 6 ≤ b
consider cos θ + 3√2 sin[θ + \(\frac{\pi}{4}\)] + 6
= cos θ + 3√2[sin θ cos \(\frac{\pi}{4}\) + cos θ sin \(\frac{\pi}{4}\)] + 6
= cos θ + 3√2 sin θ \(\frac{1}{\sqrt{2}}\) + 3√2 cos θ \(\frac{1}{\sqrt{2}}\) + 6
= cos θ + 3 sin θ + 3 cos θ + 6
= 4 cos θ + 3 sin θ + 6
∴ a = 4, b = 3, c = 6
maximum value = c + \(\sqrt{a^{2}+b^{2}}\)
= 6 + \(\sqrt{16+9}\)
= 6 + 5
= 11
minimum value = c – \(\sqrt{a^{2}+b^{2}}\)
= 6 – 5
= 1

Question 7.
Find the periods for the following functions.
(i) cos⁴x
Solution:

(ii) \(2 \sin \left[\frac{\pi x}{4}\right]+3 \cos \left[\frac{\pi x}{3}\right]\)
Solution:

(iii) sin²x + 2 cos²x
Solution:
Let f(x) = sin²x + 2 cos²x
= 1 – cos²x + 2 cos²x
= 1 + cos²x
= 1 + \(\frac{1+\cos 2 x}{2}\)
∴ period of cos 2x = \(\frac{2 \pi}{2}\) = π
∴ period of f(x) = π

(iv) \(2 \sin \left[\frac{\pi}{4}+x\right] \cos x\)
Solution:

(v) \(\frac{5 \sin x+3 \cos x}{4 \sin 2 x+5 \cos x}\)
Solution:
Let f(x) = \(\frac{5 \sin x+3 \cos x}{4 \sin 2 x+5 \cos x}\)
period of sin x = 2π
period of cos x = 2π
period of sin 2x = \(\frac{2 \pi}{2}\) = π
period of cos x = 2π
L.C.M. of (2π, 2π, π, 2π) = 2π
∴ period of f(x) = 2π

II.

Question 1.
(i) If 0 < A < \(\frac{\pi}{4}\) and cos A = \(\frac{4}{5}\), find the values of sin 2A and cos 2A.
Solution:

(ii) For what values of A in the first quadrant, the expression \(\frac{\cot ^{3} A-3 \cot A}{3 \cot ^{2} A-1}\) is positive?
Solution:

(iii) Prove that \(\frac{\cos 3 A+\sin 3 A}{\cos A-\sin A}=1+2 \sin A\)
Solution:

Question 2.
(i) Prove that \(\cot \left[\frac{\pi}{4}-\theta\right]=\frac{\cos 2 \theta}{1-\sin 2 \theta}\) and hence find the value of cot 15°.
Solution:

(ii) If θ lies in third quadrant and sin θ = \(\frac{-4}{5}\), find the values of cosec (θ/2) and tan (θ/2).
Solution:

(iii) If 450° < θ < 540° and sin θ = \(\frac{12}{13}\), then calculate sin (θ/2) and cos (θ/2)
Solution:

(iv) Prove that \(\frac{1}{\cos 290^{\circ}}+\frac{1}{\sqrt{3} \sin 250^{\circ}}=\frac{4}{\sqrt{3}}\)
Solution:

Question 3.
Prove that
(i) \(\frac{\sin 2 A}{(1-\cos 2 A)} \cdot \frac{(1-\cos A)}{\cos A}=\tan \frac{A}{2}\)
Solution:

(ii) \(\frac{\sin 2 x}{(\sec x+1)} \cdot \frac{\sec 2 x}{(\sec 2 x+1)}\) = \(\tan \frac{x}{2}\)
Solution:

(iii) \(\frac{\left(\cos ^{3} \theta-\cos 3 \theta\right)}{\cos \theta}+\frac{\left(\sin ^{3} \theta+\sin 3 \theta\right)}{\sin \theta}=\mathbf{3}\)
Solution:

Question 4.
(i) Show that cos A = \(\frac{\cos 3 A}{(2 \cos 2 A-1)}\). Hence find the value of cos 15°.
Solution:

(ii) Show that sin A = \(\frac{\sin 3 A}{1+2 \cos 2 A}\). Hence find the value of sin 15°.
Solution:

(iii) Prove that tan α = \(\frac{\sin 2 \alpha}{1+\cos 2 \alpha}\) and hence deduce the values tan 15° and tan 22\(\frac{1^{\circ}}{2}\).
Solution:

Question 5.
Prove that
(i) \(\frac{1}{\sin 10^{\circ}}-\frac{\sqrt{3}}{\cos 10^{\circ}}=4\)
Solution:

(ii) √3 cosec 20° – sec 20° = 4
Solution:

(iii) tan 9° – tan 27° – cot 27° + cot 9° = 4
Solution:

(iv) If \(\frac{\sin \alpha}{a}=\frac{\cos \alpha}{b}\), then prove that a sin 2α + b cos 2α = b
Solution:
Given \(\frac{\sin \alpha}{a}=\frac{\cos \alpha}{b}\)
b sin α = a cos α
L.H.S. = a sin 2α + b cos 2α
= a 2 sin α cos α + b (1 – 2 sin²α)
= 2 sin α (a cos α) + b – 2b sin²α
= 2 sin α (b sin α) + b – 2b sin²α
= 2b sin²α + b – 2b sin²α
= b

Question 6.
(i) In a ∆ABC; if \(\tan \frac{A}{2}=\frac{5}{6}\) and tan\(\tan \frac{B}{2}=\frac{20}{37}\), then show that \(\tan \frac{C}{2}=\frac{2}{5}\)
Solution:

(ii) If cos θ = \(\frac{5}{13}\) and 270° < θ < 360°, evaluate sin (θ/2) and cos (θ/2).
Solution:
cos θ = \(\frac{5}{13}\)
given 270 < θ < 360°
⇒ 135 < \(\frac{\theta}{2}\) < 180°
∴ θ lies in the fourth quadrant
θ/2 lies is second quadrant

(iii) If 180° < θ < 270° and sin θ = \(\frac{-4}{5}\) calculate sin[θ/2] cos[θ/2]
Solution:
Given sin θ = \(\frac{-4}{5}\)
⇒ cos θ = \(\frac{-3}{5}\)
given 180 < θ < 270
∴ θ in the III quadrant
Now 90 < \(\frac{\theta}{2}\) < 135
∴ \(\frac{\theta}{2}\) is in II quadrant

Question 7.
(i) Prove that \(\cos ^{2} \frac{\pi}{8}+\cos ^{2} \frac{3 \pi}{8}+\cos ^{2} \frac{5 \pi}{8}+\cos ^{2} \frac{7 \pi}{8}\) = 2
Solution:

(ii) Show that \(\cos ^{4}\left(\frac{\pi}{8}\right)+\cos ^{4}\left(\frac{3 \pi}{8}\right)+\cos ^{4}\left(\frac{5 \pi}{8}\right)+\cos ^{4}\left(\frac{7 \pi}{8}\right)\) = \(\frac{3}{2}\)
Solution:

III.

Question 1.
(i) If tan x + tan(x + \(\frac{\pi}{3}\)) + tan(x + \(\frac{2 \pi}{3}\)) = 3, show that tan 3x = 1
Solution:

(ii) Prove that \(\sin \frac{\pi}{5} \sin \frac{2 \pi}{5} \cdot \sin \frac{3 \pi}{5} \cdot \sin \frac{4 \pi}{5}\) = \(\frac{5}{16}\)
Solution:

(iii) Show that \(\cos ^{2}\left(\frac{\pi}{10}\right)+\cos ^{2}\left(\frac{2 \pi}{5}\right)+\cos ^{2}\left(\frac{3 \pi}{5}\right)\) + \(\cos ^{2}\left(\frac{9 \pi}{10}\right)\) = 2
Solution:

Question 2.
(i) \(\frac{1-\sec 8 \alpha}{1-\sec 4 \alpha}=\frac{\tan 8 \alpha}{\tan 2 \alpha}\)
Solution:

(ii) \(\left[1+\cos \frac{\pi}{10}\right]\left[1+\cos \frac{3 \pi}{10}\right]\left[1+\cos \frac{7 \pi}{10}\right]\) \(\left[1+\cos \frac{9 \pi}{10}\right]=\frac{1}{16}\)
Solution:

Question 3.
(i) Prove that \(\cos \frac{2 \pi}{7} \cdot \cos \frac{4 \pi}{7} \cdot \cos \frac{8 \pi}{7}=\frac{1}{8}\)
Solution:

(ii) \(\cos \frac{\pi}{11} \cdot \cos \frac{2 \pi}{11} \cdot \cos \frac{3 \pi}{11} \cdot \cos \frac{4 \pi}{11}\) \(\cos \frac{5 \pi}{11}=\frac{1}{32}\)
Solution:

Question 4.
(i) If cos α = \(\frac{3}{5}\) and cos β = \(\frac{5}{13}\) and α, β are acute angles, then prove that
(a) \(\sin ^{2}\left[\frac{\alpha-\beta}{2}\right]=\frac{1}{65}\)
(b) \(\cos ^{2}\left[\frac{\alpha+\beta}{2}\right]=\frac{16}{65}\)
Solution:

(ii) If A is not an integral multiple of π, prove that cos A . cos 2A . cos 4A . cos 8A = \(\frac{\sin 16 A}{16 \sin A}\) and hence deduce that \(\cos \frac{2 \pi}{15} \cdot \cos \frac{4 \pi}{15} \cdot \cos \frac{8 \pi}{15} \cdot \cos \frac{16 \pi}{15}=\frac{1}{16}\)
Solution:
Take 16 sin A {cos A . cos 2A . cos 4A . cos 8A}
= 8(2 sin A . cos A) cos 2A . cos 4A . cos 8A
= 8 sin 2A . cos 2A . cos 4A . cos 8A
= 4(2 sin 2A . cos 2A) . cos 4A . cos 8A
= 4 sin 4A . cos 4A . cos 8A
= 2 (2 sin 4A . cos 4A) . cos 8A
= 2 sin 8A . cos 8A
= sin (16A)
∴ 16 sin A {cos A . cos 2A . cos 4A . cos 8A} = sin (16A)
∴ cos A . cos 2A . cos 4A . cos 8A = \(\frac{\sin (16 A)}{16 \sin A}\)

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Exercise 6(b) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Exercise 6(b)

I. Find the periods for the given 1 – 5 functions.

Question 1.
cos(3x + 5) + 7
Solution:
f(x) = cos(3x + 5) + 7
We know that the function g(x) = cos x for all x ∈ R has the period 2π.
Now f(x) = cos(3x + 5) + 7
We get that f(x) is periodic and the period of f is \(\frac{2 \pi}{|3|}=\frac{2 \pi}{3}\)

Question 2.
tan 5x
Solution:
The function g(x) = tan x periodic and π is the period.
∴ f(x) = tan 5x periodic and its period is \(\frac{\pi}{|5|}=\frac{\pi}{5}\)

Question 3.
\(\cos \left(\frac{4 x+9}{5}\right)\)
Solution:
The function h(x) = cos x for all x ∈ R has the period 2π.
Now f(x) = \(\cos \left(\frac{4 x}{5}+\frac{9}{5}\right)\) is periodic and period of f is \(\frac{2 \pi}{\left(\frac{4}{5}\right)}=\frac{5 \pi}{2}\)

Question 4.
|sin x|
Solution:
The function h(x) = sin x for all x ∈ R has the period 2π.
But f(x) = |sin x| is periodic and its period is π.
∵ f(x + π) = |sin(x + π)|
= |-sin x|
= sin x

Question 5.
tan(x + 4x + 9x + …… + n²x) (n any positive integer)
Solution:
tan(1² + 2² + 3² + …… + n²) x = \(\tan \left[\frac{n(n+1)(2 n+1)}{6}\right] x\)
period = \(\frac{6 \pi}{n(n+1)(2 n+1)}\)

Question 6.
Find a sine function whose period is \(\frac{2}{3}\)
Solution:
\(\frac{2 \pi}{|k|}=\frac{2}{3}\)
3π = |k|
∴ sin kx = sin 3πx

Question 7.
Find a cosine function whose period is 7.
Solution:
\(\frac{2 \pi}{|k|}\) = 7
\(\frac{2 \pi}{7}\) = |k|
∴ cos kx = cos \(\frac{2 \pi}{7}\) x

II. Sketch the graph of the following functions.

Question 1.
tan x between 0 and \(\frac{\pi}{4}\)
Solution:

Question 2.
cos 2x in [0, π]
Solution:

Question 3.
sin 2x in the interval (0, π)
Solution:

Question 4.
sin x in the interval [-π, +π]
Solution:

Question 5.
cos²x in [0, π]
Solution:

III.

Question 1.
Sketch the region enclosed by y = sin x, y = cos x and X-axis in the interval [0, π].
Solution:

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Exercise 6(a) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Exercise 6(a)

I. Convert the following into the simplest form.

Question 1.
(i) tan(θ – 14π)
Solution:
= tan(14π – θ)
= tan(2 . (7π) – θ)
= tan θ

(ii) \(\cot \left(\frac{21 \pi}{2}-\theta\right)\)
Solution:
\(\cot \left(\frac{21 \pi}{2}-\theta\right)\)
= \(\cot \left(10 \pi+\left(\frac{\pi}{2}-\theta\right)\right)\)
= \(\cot \left(\frac{\pi}{2}-\theta\right)\)
= tan θ

(iii) cosec(5π + θ)
Solution:
cosec(5π + θ) = cosec(2π + (3π + θ))
= cosec(3π + θ)
= cosec(2π + (π + θ))
= cosec(π + θ)
= -cosec θ

(iv) sec(4π – θ?)
Solution:
sec(4π – θ)
= sec(2π + (2π – θ))
= sec(2π – θ)
= sec θ

Question 2.
Find the value of each of the following.
(i) sin(-405°)
Solution:
sin(-405°) = -sin(360° + 45)
= -sin 45°
= \(-\frac{1}{\sqrt{2}}\)

(ii) \(\cos \left(-\frac{7 \pi}{2}\right)\)
Solution:
\(\cos \left(-\frac{7 \pi}{2}\right)\)
= -cos 630°
= -cos(360° + 270°)
= -cos 270°
= -cos(180° + 90°)
= -cos 90°
= 0

(iii) sec(2100°)
Solution:
sec(2100°) = sec (5 × 360° + 300°)
= sec 300°
= sec(360 – 60°)
= sec 60°
= 2

(iv) cot(-315°)
Solution:
cot(-315°) = -cot 315°
= -cot(360° – 45°)
= -cot 45°
= 1

Question 3.
Evaluate.
(i) cos² 45° + cos² 135° + cos² 225° + cos² 315°
Solution:

(ii) \(\sin ^{2} \frac{2 \pi}{3}+\cos ^{2} \frac{5 \pi}{6}-\tan ^{2} \frac{3 \pi}{4}\)
Solution:

(iii) cos 225° – sin 225° + tan 495° – cot 495°
Solution:
cos (180° + 45°) – sin(180° + 45°) + tan(360° + 135°) – cot(360° + 135°)
= -cos 45° + sin 45° – tan 135° + cot 135°
= \(-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}+1-1\)
= 0

(iv) (cos θ – sin θ) if (a) θ = \(\frac{7 \pi}{4}\) (b) θ = \(\frac{11 \pi}{3}\)
Solution:

Question 4.
(i) If sin θ = \(\frac{-1}{3}\) and θ does not lie in the third quadrant, find the values of (a) cos θ (b) cot θ
Solution:
∵ sin θ = \(\frac{-1}{3}\) and sin θ is negative and θ does not lie in the IIIrd quadrant.
⇒ θ lies in IV quadrant.
∴ In the IV quadrant, cos θ is positive and cot θ is negative.

(ii) If cos θ = t (0 < t < 1) and θ does not lie in the first quadrant, find the values of (a) sin θ (b) tan θ.
Solution:
cos θ = t, (0 < t < 1)
⇒ cos θ is positive and θ does not lie in the first quadrant.
⇒ θ lies in IV quadrant.
(a) sin θ = \(-\sqrt{1-\cos ^{2} \theta}=-\sqrt{1-t^{2}}\)
(b) tan θ = \(\frac{\sin \theta}{\cos \theta}=\frac{-\sqrt{1-t^{2}}}{t}\)

(iii) Find the value of sin 330°. cos 120° + cos 210°. sin 300°.
Solution:
sin 330°. cos 120° + cos 210°. sin 300°
= sin(360° – 30°) . cos(180° – 60°) + cos(180° + 30°) . sin(360° – 60°)
= (-sin 30°) (-cos 60°) + (-cos 30°) (-sin 60°)
= sin 30° cos 60° + cos 30° sin 60°
= sin(30° + 60°) [∵ sin A cos B + cos A sin B = sin(A + B)]
= sin (90°)
= 1

(iv) If cosec θ + cot θ = \(\frac{1}{3}\), find cos θ and determine the quadrant in which θ lies.
Solution:
cosec θ + cot θ = \(\frac{1}{3}\)
⇒ cosec θ – cot θ = 3 (∵ cosec²θ – cot²θ = 1)
∴ 2 cosec θ = \(\frac{1}{3}\) + 3 = \(\frac{10}{3}\)
∴ cosec θ = \(\frac{5}{3}\) and sin θ = \(\frac{3}{5}\)
2 cot θ = \(\frac{1}{3}\) – 3 = \(\frac{-8}{3}\)
∴ cot θ = \(\frac{-4}{3}\), tan θ = \(\frac{-3}{4}\)
cos θ = (cot θ) . sin θ = \(\left(\frac{-4}{3}\right)\left(\frac{3}{5}\right)=\frac{-4}{5}\)
∵ sin θ is +ve and cos θ is -ve
⇒ θ lies in II quadrant.

Question 5.
(i) If sin α + cosec α = 2. Find the value of sinⁿα + cosecⁿα, n ∈ z.
Solution:
Given sin α + cosec α = 2
Squaring on both sides
sin²α + cosec²α + 2 = 4
sin²α + cosec²α = 2
sin α + cosec α = 2
Cubing on both sides
sin³α + cosec³α + 3 sin α . cosec α (sin α + cosec α) = 8
sin³α + cosec³α + 3(2) = 8
sin³α + cosec³α = 8 – 6
sin³α + cosec³α = 2
similarly sinⁿα + cosecⁿα = -2

(ii) If sec θ + tan θ = 5. Find the quadrant in which θ lies and find the value of sin θ.
Solution:
sec θ + tan θ = 5
⇒ sec θ – tan θ = \(\frac{1}{5}\) (∵ sec²θ – tan²θ = 1)
2 sec θ = 5 + \(\frac{1}{5}\) = \(\frac{26}{5}\)
sec θ = \(\frac{26}{10}=\frac{13}{5}\)
Aquir 2 tan θ = 5 – \(\frac{1}{5}\) = \(\frac{24}{5}\)
tan θ = \(\frac{24}{10}=\frac{12}{5}\)
Now sin θ = \(\frac{\tan \theta}{\sec \theta}=\frac{\frac{12}{5}}{\frac{13}{5}}=\frac{12}{13}\)
tan θ is +ve, sec θ is +ve
⇒ θ lies in the first quadrant.

II.

Question 1.
Prove that
(i) \(\frac{\cos (\pi-A) \cdot \cot \left(\frac{\pi}{2}+A\right) \cos (-A)}{\tan (\pi+A) \tan \left[\frac{3 \pi}{2}+A\right] \sin (2 \pi-A)}\) = cos A
Solution:
\(\frac{\cos (\pi-A) \cdot \cot \left(\frac{\pi}{2}+A\right) \cos (-A)}{\tan (\pi+A) \tan \left[\frac{3 \pi}{2}+A\right] \sin (2 \pi-A)}\)
= \(\frac{-\cos A(-\tan A) \cos A}{\tan A(-\cot A)(-\sin A)}\)
= cos A

(ii) \(\frac{\sin (3 \pi-A) \cos \left(A-\frac{\pi}{2}\right) \tan \left(\frac{3 \pi}{2}-A\right)}{{cosec}\left(\frac{13 \pi}{2}+A\right) \sec (3 \pi+A) \cot \left(A-\frac{\pi}{2}\right)}\) = cos⁴A
Solution:

(iii) sin 780° sin 480° + cos 240°. cos 300° = \(\frac{1}{2}\)
Solution:
sin(2(360°) + 60°) . sin(360° + 120°) + cos(270° – 30°) . cos(360° – 60°)
= sin 60° . sin 120° – sin 30° . cos 60°
= \(\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2}-\frac{1}{2} \cdot \frac{1}{2}\)
= \(\frac{3}{4}-\frac{1}{4}\)
= \(\frac{1}{2}\)

(iv) \(\frac{\sin 150^{\circ}-5 \cos 300^{\circ}+7 \tan 225^{\circ}}{\tan 135^{\circ}+3 \sin 210^{\circ}}\) = -2
Solution:

(v) cot(\(\frac{\pi}{20}\)) . cot(\(\frac{3\pi}{20}\)) . cot(\(\frac{5\pi}{20}\)) . cot(\(\frac{7\pi}{20}\)) . cot(\(\frac{9\pi}{20}\)) = 1
Solution:
L.H.S. = cot(\(\frac{\pi}{20}\)) . cot(\(\frac{3\pi}{20}\)) . cot(\(\frac{5\pi}{20}\)) . cot(\(\frac{7\pi}{20}\)) . cot(\(\frac{9\pi}{20}\))
= cot (9°) cot (27°) cot (45°) cot (63°) cot (81°)
= cot (9°) cot (27°) (1) cot (90° – 27°) cot (90° – 9°)
= cot (9°) cot (27°) tan 27° tan 9°
= (tan 9° cot 9°) (tan 27° cot 27°)
= (1) (1)
= 1

Question 2.
Simplify.
(i) \(\frac{\sin \left(-\frac{11 \pi}{3}\right) \tan \left(\frac{35 \pi}{6}\right) \sec \left(-\frac{7 \pi}{3}\right)}{\cot \left(\frac{5 \pi}{4}\right) {cosec}\left(\frac{7 \pi}{4}\right) \cos \left(\frac{17 \pi}{6}\right)}\)
Solution:

(ii) If tan 20° = p, prove that \(\frac{\tan 610^{\circ}+\tan 700^{\circ}}{\tan 560^{\circ}-\tan 470^{\circ}}=\frac{1-p^{2}}{1+p^{2}}\)
Solution:
Given tan 20° = p
L.H.S. = \(\frac{\tan 610^{\circ}+\tan 700^{\circ}}{\tan 560^{\circ}-\tan 470^{\circ}}\)

(iii) If α, β are complementary angles such that b sin α = a, then find the value of (sin α cos β – cos α sin β).
Solution:
∵ α, β are complementary angles
⇒ α + β = 90°
⇒ β = 90° – α
Now sin α cos β – cos α sin β
= sin (α – β)
= sin [(α – (90° – α)]
= sin [2α – 90°]
= -sin (90° – 2α)
= -cos 2α
= -(1 – 2 sin²α) (∵ cos 2α = 1 – 2 sin²α)
= \(-1+2\left(\frac{a}{b}\right)^{2}\) (∵ sin α = \(\frac{a}{b}\))
= \(\frac{-b^{2}+2 a^{2}}{b^{2}}\)
= \(\frac{2 a^{2}-b^{2}}{b^{2}}\)

Question 3.
(i) If cos A = cos B = \(-\frac{1}{2}\), A does not lie in the second quadrant and B does not lie in the third quadrant, then find the value of \(\frac{4 \sin B-3 \tan A}{\tan B+\sin A}\)
Solution:
∵ cos A = \(-\frac{1}{2}\) and A does not lie in second quadrant.
⇒ A lies in the third quadrant, (∵ cos A is -ve)
and cos B = \(-\frac{1}{2}\) and B does not lie in third quadrant.
⇒ B lies in second quadrant.
∵ cos A = \(-\frac{1}{2}\) and A lies in third quadrant.
⇒ A = 240°
∵ cos B = \(-\frac{1}{2}\) and B lies in second quadrant.
⇒ B = 120°

(ii) If 8 tan A = -15 and 25 sin B = -7 and neither A nor B is in the fourth quadrant, then show that sin A cos B + cos A sin B = \(\frac{-304}{425}\)
Solution:
8 tan A = -15 ⇒ tan A = \(\frac{-15}{8}\)
25 sin B = -7 ⇒ sin B = \(-\frac{7}{25}\)
Given neither A nor B is in the fourth quadrant.
Clearly, A is the second quadrant B is the third quadrant
sin A cos B + cos A sin B = \(\left(\frac{15}{17}\right)\left(\frac{-24}{25}\right)+\left(\frac{-8}{17}\right)\left(\frac{-7}{25}\right)\)
= \(\frac{-360}{425}+\frac{56}{425}\)
= \(\frac{-304}{425}\)

(iii) If A, B, C, D are angles of a cyclic quadrilateral, then prove that
(a) sin A – sin C = sin D – sin B
(b) cos A + cos B + cos C + cos D = 0
Solution:
∵ A, B, C, D are angles of a cyclic quadrilateral.
⇒ A + C = 180° and B + D = 180° ……..(1)
C = 180° – A and D = 180° – B
(a) L.H.S. = sin A – sin C
= sin A – sin(180° – A)
= sin A – sin A
= 0
R.H.S. = sin D – sin B
= sin(180°- B) – sin B
= sin B – sin B
= 0
∴ L.H.S. = R.H.S.
i.e., sin A – sin C = sin D – sin B
(b) L.H.S. = cos A + cos B + cos C + cos D
= cos A + cos B + cos(180° – A) + cos(180° – B)
= cos A + cos B – cos A – cos B
= 0
∴ cos A + cos B + cos C + cos D = 0

Question 4.
(i) If a cos θ – b sin θ = c, then show that a sin θ + b sin θ = \(\pm \sqrt{a^{2}+b^{2}-c^{2}}\).
Solution:
a cos θ – b sin θ = c
let a sin θ + b cos θ = x
by squaring and adding
(a cos θ – b sin θ)² + (a sin θ + b cos θ)² = c² + x²
a² cos²θ + b² sin²θ – 2ab sin θ cos θ + a² sin²θ + b² cos²θ + 2ab sin θ cos θ = c² + x²
a² + b² = c² + x²
a² + b² – c² = x²
x = \(\pm \sqrt{a^{2}+b^{2}-c^{2}}\)
∴ a sin θ + b cos θ = \(\pm \sqrt{a^{2}+b^{2}-c^{2}}\)

(ii) If 3 sin A + 5 cos A = 5, then show that 5 sin A – 3 cos A = ±3.
Solution:
3 sin A + 5 cos A = 5
Let 5 sec A – 3 cos A = x
by squaring and adding
(3 sin A + 5 cos A)² + (5 sin A – 3 cos A)² = 5² + x²
9 sin² A + 25 cos² A + 30 sin A cos A + 25 sin² A + 9 cos² A – 30 sin A cos A = 25
9 + 25 = 25 + x²
x² = 9
x = ±3
∴ 5 sin A – 3 cos A = ±3

(iii) If tan²θ = (1 – e²), show that sec θ + tan³θ . cosec θ = (2 – e²)^3/2.
Solution:
tan²θ = 1 – e²
sec²θ = 1 + tan²θ = 2 – e²

III. Prove the following.

Question 1.
(i) \(\frac{(\tan \theta+\sec \theta-1)}{(\tan \theta-\sec \theta+1)}=\frac{1+\sin \theta}{\cos \theta}\)
Solution:

(ii) (1 + cot θ – cosec θ) (1 + tan θ + sec θ) = 2.
Solution:

(iii) 3(sin θ – cos θ)⁴ + 6(sin θ + cos θ)² + 4(sin⁶θ + cos⁶θ) = 13.
Solution:
(sin θ – cos θ)² = sin²θ + cos²θ – 2 sin θ . cos θ
= 1 – 2 sin θ cos θ
(sin θ – cos θ)⁴ = (1 – 2 sin θ cos θ)²
= 1 + 4 sin²θ cos²θ – 4 sin θ cos θ ……(1)
(sin θ + cos θ)² = sin²θ + cos²θ + 2 sin θ cos θ
= 1 + 2 sin θ cos θ …….(2)
sin⁶θ + cos⁶θ = (sin²θ +cos²θ)³ – 3 sin²θ cos²θ (sin²θ + cos²θ)
= 1 – 3 sin²θ cos²θ ……..(3)
L.H.S. = 3(1 + 4 sin²θ cos²θ – 4 sin θ cos θ) + 6(1 + 2 sin θ cos θ) + 4(1 – 3 sin²θ cos²θ)
= 3 + 12 sin²θ cos²θ – 12 sin θ cos θ + 6 + 12 sin θ cos θ + 4 – 12 sin²θ cos²θ
= 3 + 6 + 4
= 13
= R.H.S.

Question 2.
(i) Prove that (sin θ + cosec θ)² + (cos θ + sec θ)² – (tan²θ + cot²θ) = 7.
Solution:
L.H.S. = (sin θ + cosec θ)² + (cos θ + sec θ)² – (tan²θ + cot²θ)
= (sin²θ + cosec²θ + 2 sin θ cosec θ) + (cos²θ + sec²θ + 2 cos θ sec θ) – (tan²θ + cot²θ)
= (sin²θ + cos²θ) + (1 + cot²θ) + (1 + tan²θ) + 4 – tan²θ – cot²θ
= 1 + 1 + 1 + 4
= 7

(ii) cos⁴α + 2 cos²α \(\left(1-\frac{1}{\sec ^{2} \alpha}\right)\) = (1 – sin⁴α)
Solution:
L.H.S. = cos⁴α + 2 cos²α \(\left(1-\frac{1}{\sec ^{2} \alpha}\right)\)
= cos⁴α + 2 cos²α (1 – cos²α)
= cos²α [cos²α + 2 sin²α]
= (1 – sin²α) [cos²α + sin²α + sin²α]
= (1 – sin²α) (1 + sin²α)
= 1 – sin⁴α

(iii) \(\frac{(1+\sin \theta-\cos \theta)^{2}}{(1+\sin \theta+\cos \theta)^{2}}=\frac{1-\cos \theta}{1+\cos \theta}\)
Solution:

(iv) If \(\frac{2 \sin \theta}{(1+\cos \theta+\sin \theta)}\) = x, then find the value of \(\frac{(1-\cos \theta+\sin \theta)}{(1+\sin \theta)}\)
Solution:

Question 3.
Eliminate θ from the following.
(i) x = a cos³θ; y = b sin³θ
Solution:

(ii) x = a cos⁴θ; y = b sin⁴θ
Solution:

(iii) x = a(sec θ + tan θ); y = b(sec θ – tan θ)
Solution:
\(\frac{x}{a}\) = sec θ + tan θ; \(\frac{y}{b}\) = (sec θ – tan θ)
\(\frac{x}{a} \times \frac{y}{b}\) = (sec θ + tan θ) (sec θ – tan θ)
\(\frac{x y}{a b}\) = sec²θ – tan²θ
\(\frac{x y}{a b}\) = 1
xy = ab

(iv) x = cot θ + tan θ; y = sec θ – cos θ
Solution:

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Products of Vectors Solutions Exercise 5(c) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Products of Vectors Solutions Exercise 5(c)

I.

Question 1.
Compute \([\overline{\mathbf{i}}-\overline{\mathbf{j}} \overline{\mathbf{j}}-\overline{\mathbf{k}} \overline{\mathbf{k}}-\overline{\mathbf{i}}]\)
Solution:

Question 2.
If \(\overline{\mathbf{a}}=\overline{\mathbf{i}}-2 \overline{\mathbf{j}}-3 \overline{\mathbf{k}}, \overline{\mathbf{b}}=2 \overline{\mathbf{i}}+\overline{\mathbf{j}}-\overline{\mathbf{k}}\), \(\bar{c}=\bar{i}+3 \bar{j}-2 \bar{k}\), then compute \(\overline{\mathbf{a}} \cdot(\overline{\mathbf{b}} \times \overline{\mathbf{c}})\).
Solution:

Question 3.
If \(\bar{a}\) = (1, -1, -6), \(\bar{b}\) = (1, -3, 4) and \(\bar{c}\) = (2, -5, 3), then compute the following
(i) \(\overline{\mathbf{a}} \cdot(\bar{b} \times \bar{c})\)
(ii) \(\overline{\mathbf{a}} \times(\overline{\mathbf{b}} \times \overline{\mathbf{c}})\)
(iii) \((\overline{\mathbf{a}} \times \overline{\mathbf{b}}) \times \overline{\mathbf{c}}\)
Solution:

Question 4.
Simplify the following.
(i) \((\bar{i}-2 \bar{j}+3 \bar{k}) \times(2 i+j-\bar{k}) \cdot(\bar{j}+\bar{k})\)
(ii) \((2 \bar{i}-3 \bar{j}+\bar{k}) \cdot(\bar{i}-\bar{j}+2 \bar{k}) \cdot(2 \bar{i}+\bar{j}+\bar{k})\)
Solution:

Question 5.
Find the volume of the parallelopiped having coterminous edges.
\(\overline{\mathbf{i}}+\overline{\mathbf{j}}+\overline{\mathbf{k}}, \overline{\mathbf{i}}-\overline{\mathbf{j}}\) and \(\overline{\mathbf{i}}+\mathbf{2} \overline{\mathbf{j}}-\overline{\mathbf{k}}\)
Solution:
Let \(\overline{\mathrm{a}}=\overline{\mathrm{i}}+\overline{\mathrm{j}}+\overline{\mathrm{k}}, \overline{\mathrm{b}}=\overline{\mathrm{i}}-\overline{\mathrm{j}}\) and \(\overline{\mathrm{c}}=\overline{\mathrm{i}}+2 \overline{\mathrm{j}}-\overline{\mathrm{k}}\)
Volume of the parallelopiped = \([(\bar{a} \bar{b} \bar{c})]\)
= \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & -1 & 0 \\
1 & 2 & -1
\end{array}\right|\)
= 1(1 – 0) – 1(-1 – 0) + 1(2 + 1)
= 1 + 1 + 3
= 5 cubic units.

Question 6.
Find t for which the vectors \(\mathbf{2} \overline{\mathbf{i}}-\mathbf{3} \overline{\mathbf{j}}+\overline{\mathbf{k}}\), \(\bar{i}+2 \mathbf{j}-3 \bar{k}\) and \(\bar{j}-t \bar{k}\) are coplanar.
Solution:

Question 7.
For non-coplanar vectors, \(\bar{a}\), \(\bar{b}\) and \(\bar{c}\) determine p for which the vector \(\overline{\mathbf{a}}+\overline{\mathbf{b}}+\overline{\mathbf{c}}, \overline{\mathbf{a}}+\mathbf{p} \overline{\mathbf{b}}+\mathbf{2} \overline{\mathbf{c}}\) and \(-\overline{\mathbf{a}}+\overline{\mathbf{b}}+\overline{\mathbf{c}}\) are coplanar.
Solution:

Question 8.
Determine λ, for which the volume of the parallelopiped having coterminous edges \(\bar{i}+\bar{j}\), \(3 \overline{\mathbf{i}}-\overline{\mathbf{j}}\) and \(3 \bar{j}+\lambda \bar{k}\) is 16 cubic units.
Solution:

Question 9.
Find the volume of the tetrahedron having the edges \(\overline{\mathbf{i}}+\overline{\mathbf{j}}+\overline{\mathbf{k}}, \quad \mathbf{i}-\overline{\mathbf{j}}\) and \(\bar{i}+2 \bar{j}+\bar{k}\).
Solution:

Question 10.
Let \(\bar{a}\), \(\bar{b}\) and \(\bar{c}\) be non-coplanar vectors and \(\bar{\alpha}=\bar{a}+2 \bar{b}+3 c, \quad \bar{\beta}=2 \bar{a}+\bar{b}-2 c\) and \(\bar{\gamma}=3 \bar{a}-7 \bar{c}\), then find \(\left[\begin{array}{lll}
\bar{\alpha} & \bar{\beta} & \bar{\gamma}
\end{array}\right]\).
Solution:

Question 11.
Let \(\bar{a}\), \(\bar{b}\) and \(\bar{c}\) be non-coplanar vectors. If \(|2 \bar{a}-\bar{b}+3 \bar{c}|, \bar{a}+\bar{b}-2 \bar{c},|\bar{a}+\bar{b}-3 \bar{c}|\) = \(\lambda[\overline{\mathbf{a}} \overline{\mathbf{b}} \overline{\mathbf{c}}]\), then find the value of λ.
Solution:

Question 12.
Let \(\bar{a}\), \(\bar{b}\) and \(\bar{c}\) be non-coplanar vectors, if \(\left[\begin{array}{lll}
\bar{a}+2 \bar{b} & 2 \bar{b}+\bar{c} & 5 \bar{c}+\bar{a}
\end{array}\right]\) = \(\lambda\left[\begin{array}{lll}
\bar{a} & \bar{b} & \bar{c}
\end{array}\right]\), then find λ.
Solution:

Question 13.
If \(\bar{a}\), \(\bar{b}\) and \(\bar{c}\) are non-coplanar vectors, then find the value of \(\frac{(\bar{a}+2 \bar{b}-\bar{c}) \cdot[(\bar{a}-\bar{b}) \times(\bar{a}-\bar{b}-\bar{c})]}{[\bar{a} \bar{b} \bar{c}]}\)
Solution:

Question 14.
If \(\bar{a}\), \(\bar{b}\) and \(\bar{c}\) are mutually perpendicular unit vectors, then find the value of \(\left[\begin{array}{lll}
\bar{a} & \bar{b} & \bar{c}
\end{array}\right]^{2}\).
Solution:

Question 15.
\(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) are non-zero vectors and \(\overline{\mathbf{a}}\) is perpendicular to both \(\overline{\mathbf{b}}\) and \(\overline{\mathbf{c}}\). If \(|\overline{\mathbf{a}}|\) = 2, \(|\overline{\mathbf{b}}|\) = 3, \(|\overline{\mathbf{c}}|\) = 4 and \((\bar{b}, \bar{c})=\frac{2 \pi}{3}\), then find \(|[\bar{a} \bar{b} \bar{c}]|\).
Solution:

Question 16.
If \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) are unit coplanar vectors, then find \(\left[\begin{array}{lll}
2 \bar{a}-\bar{b} & 2 \bar{b}-\bar{c} & 2 \bar{c}-\bar{a}
\end{array}\right]\)
Solution:

II.

Question 1.
If \(\left[\begin{array}{lll}
\bar{b} & \bar{c} & \bar{d}
\end{array}\right]+\left[\begin{array}{lll}
\bar{c} & \bar{a} & \bar{d}
\end{array}\right]+\left[\begin{array}{lll}
\bar{a} & \bar{b} & \bar{d}
\end{array}\right]\) = \(\left[\begin{array}{lll}
\overline{\mathbf{a}} & \overline{\mathbf{b}} & \overline{\mathbf{c}}
\end{array}\right]\) then show that the points with position vectors \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) and \(\bar{d}\) are coplanar.
Solution:

Question 2.
If \(\overline{\mathbf{a}}, \overline{\mathbf{b}}\) and \(\overline{\mathbf{c}}\) non-coplanar vectors, then prove that the four points with position vectors \(2 \bar{a}+3 \bar{b}-\bar{c}\), \(\overline{\mathrm{a}}-2 \overline{\mathrm{b}}+3 \overline{\mathrm{c}}, 3 \overline{\mathrm{a}}+4 \overline{\mathrm{b}}-2 \overline{\mathrm{c}}\) and \(\bar{a}-6 \bar{b}+6 \bar{c}\) are coplanar.
Solution:
Suppose A, B, C, D are the given points.

The vectors \(\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{AC}}, \overrightarrow{\mathrm{AD}}\) are coplanar.
The given points A, B, C, D are coplanar.

Question 3.
\(\overline{\mathbf{a}}, \overline{\mathbf{b}}\) and \(\overline{\mathbf{c}}\) are non-zero and non- collinear vectors and θ ≠ 0, is the angle between \(\overline{\mathbf{b}}\) and \(\overline{\mathbf{c}}\). If \((\overline{\mathbf{a}} \times \overline{\mathbf{b}}) \times \overline{\mathbf{c}}\) = \(\frac{1}{3}|\bar{b}||\bar{c}|\bar{a}|\), then find sin θ.
Solution:

Question 4.
Find the volume of the tetrahedron whose vertices are (1, 2, 1), (3, 2, 5), (2, -1, 0) and (-1, 0, 1).
Solution:
Let ‘O’ be the given A, B, C, D be the vertices of the ten tetrahedrons. Then

Question 5.
Show that \((\bar{a}+\bar{b}) \cdot(\bar{b}+\bar{c}) \times(\bar{c}+\bar{a})\) = \(2\left[\begin{array}{lll}
\bar{a} & \bar{b} & \bar{c}
\end{array}\right]\)
Solution:

Question 6.
Show that equation of the plane passing through the points with position vectors. \(3 \bar{i}-5 \bar{j}-\overline{\mathbf{k}},-\overline{\mathbf{i}}+5 \bar{j}+7 \overline{\mathbf{k}}\) and parallel to the vector \(3 \overline{\mathbf{i}}-\overline{\mathbf{j}}+7 \overline{\mathbf{k}}\) is 3x + 2y – z = 0.
Solution:
The given plane passes through the points A, B (i.e.,) \(3 \bar{i}-5 \bar{j}-\overline{\mathbf{k}},-\overline{\mathbf{i}}+5 \bar{j}+7 \overline{\mathbf{k}}\) and parallel to the vector \(3 \overline{\mathbf{i}}-\overline{\mathbf{j}}+7 \overline{\mathbf{k}}\)

= x(70 + 8) – y(-28 – 24) + z(4 – 30)
= 78x + 52y – 26z
= 26(3x + 2y – z)
\(\left[\begin{array}{lll}
\bar{a} & \bar{b} & \bar{c}
\end{array}\right]\) = \(\left|\begin{array}{rrr}
3 & -5 & -1 \\
-4 & 10 & 8 \\
3 & -1 & 7
\end{array}\right|\)
= 3(70 + 8) + 5(-28 – 24) – 1(4 – 30)
= 234 – 260 + 26
= 0
Equation of the required plane is 26(3x + 2y – z) = 0
i.e., 3x + 2y – z = 0

Question 7.
Prove that \(\overline{\mathbf{a}} \times[\overline{\mathbf{a}} \times(\overline{\mathbf{a}} \times \overline{\mathbf{b}})]\) = \((\overline{\mathbf{a}} \cdot \overline{\mathbf{a}})(\overline{\mathbf{b}} \times \overline{\mathbf{a}})\)
Solution:

Question 8.
If \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) and \(\bar{d}\) are coplanar vectors, then show that \((\bar{a} \times \bar{b}) \times(\bar{c} \times \bar{d})=0\).
Solution:
\(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) and \(\bar{d}\) are coplanar
⇒ \(\overline{\mathbf{a}} \times \overline{\mathbf{b}}\) is perpendicular to the plane π.
similarly \(\bar{c} \times \bar{d}\) is perpendicular to the plane π.
\(\bar{a} \times \bar{b}\) and \(\bar{c} \times \bar{d}\) are parallel vectors.
⇒ \((\bar{a} \times \bar{b}) \times(\bar{c} \times \bar{d})\) = 0.

Question 9.
Show that \((\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times(\overline{\mathrm{a}} \times \overline{\mathrm{c}}) \cdot \overline{\mathrm{d}}=[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}](\overline{\mathrm{a}} \cdot \overline{\mathrm{d}})\).
Solution:

Question 10.
Show that \(\bar{a} \cdot[(\bar{b}+\bar{c}) \times(\bar{a}+\bar{b}+\bar{c})]=0\)
Solution:

Question 11.
Find λ in order that the four points A(3, 2, 1), B(4, λ, 5), C(4, 2, -2) and D(6, 5, -1) be coplanar.
Solution:

Question 12.
Find the vector equation of the plane passing through the intersection of planes \(\bar{r} \cdot(2 \bar{i}+2 \bar{j}-3 \bar{k})=7, \bar{r} \cdot(2 \bar{i}+5 \bar{j}+3 \bar{k})=9\) and through the point (2, 1, 3)
Solution:

Question 13.
Find the equation of the plane passing through (a, b, c) and parallel to the plane \(\bar{r} \cdot(\bar{i}+\bar{i}+\bar{k})=2\).
Solution:
Given equation plane is \(\bar{r} \cdot(\bar{i}+\bar{i}+\bar{k})=2\)
Let \(\bar{r}=x \bar{i}+y \bar{j}+z \bar{k}\)
∴ \(\bar{r} \cdot(\bar{i}+\bar{i}+\bar{k})=2\)
\((x \bar{i}+y \bar{j}+z \bar{k}) \cdot(i+j+k)=2\)
x + y + z = 2
Required plane equation is x + y + z = k …….(1)
Equation (1) passes through (a, b, c)
∴ a + b + c = k
Substitute ‘k’ in equation (1)
∴ x + y + z = a + b + c

Question 14.
Find the shortest distance between the lines \(\bar{r}=6 \bar{i}+2 \bar{j}+2 \bar{k}+\lambda, \bar{i}-2 \bar{j}+2 \bar{k}\) and \(\bar{r}=-4 \bar{j}-\bar{k}+\mu=3 \bar{j}-2 \bar{j}-2 \bar{k}\).
Solution:
The first line passes through 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 15.
Find the equation of the plane passing through the line of intersection of the planes \(\bar{r} \cdot(\bar{i}+\bar{j}+\bar{k})=1\) and \(\bar{r} \cdot(2 \bar{i}+3 \bar{i}-\bar{k})+4=0\) and parallel to X-axis.
Solution:
Given the equation of planes are

Since it is parallel to X-axis.

Question 16.
Prove that the four points \(4 \bar{i}+5 \bar{j}+\bar{k}\), \(-(\overline{\mathbf{j}}+\overline{\mathbf{k}}), 3 \overline{\mathbf{i}}+9 \overline{\mathbf{j}}+4 \overline{\mathbf{k}}\) and \(-4 \bar{i}+4 \bar{j}+4 \bar{k}\) are coplanar.
Solution:
Let ‘O’ be the origin. A, B, C, D be the given points. Then

Question 17.
If \(\bar{a}, \bar{b}, \bar{c}\) are non – copianar, then show that the vectors \(\overline{\mathbf{a}}-\overline{\mathbf{b}}, \overline{\mathbf{b}}+\overline{\mathbf{c}}\), \(\overline{\mathbf{c}}+\overline{\mathbf{a}}\) are coplanar.
Solution:

Question 18.
If \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) are the position vectors of the points A, B and C respectively, then prove that the vector \(\overline{\mathbf{a}} \times \overline{\mathbf{b}}+\overline{\mathbf{b}} \times \overline{\mathbf{c}}+\overline{\mathbf{c}} \times \overline{\mathbf{a}}\) is perpendicular to the plane of ∆ABC.
Solution:

III.

Question 1.
Show that \((\bar{a} \times(\bar{b} \times \bar{c}) \times \bar{c})=(\bar{a} \cdot \bar{c})(\bar{b} \times \bar{c})\) and \((\overline{\mathbf{a}} \times \overline{\mathbf{b}}) \cdot(\overline{\mathbf{a}} \times \overline{\mathbf{c}})+(\overline{\mathbf{a}} \cdot \overline{\mathbf{b}})(\overline{\mathbf{a}} \cdot \overline{\mathbf{c}})\) = \((\overline{\mathbf{a}} \cdot \overline{\mathbf{a}})(\overline{\mathbf{b}} \cdot \overline{\mathbf{c}})\)
Solution:

Question 2.
If A = (1, -2, -1), B = (4, 0, -3), C = (1, 2, -1) and D = (2, -4, -5), find the distance between AB and CD.
Solution:

Question 3.
If \(\overline{\mathbf{a}}=\overline{\mathbf{i}}-2 \overline{\mathbf{j}}+\overline{\mathbf{k}}, \overline{\mathbf{b}}=2 \overline{\mathbf{i}}+\overline{\mathbf{j}}+\overline{\mathbf{k}}\) and \(\overline{\mathbf{c}}=\overline{\mathbf{i}}+2 \overline{\mathbf{j}}-\overline{\mathbf{k}}\), find \(\bar{a} \times(\bar{b} \times \bar{c})\) and \(|(\overline{\mathbf{a}} \times \overline{\mathbf{b}}) \times \overline{\mathbf{c}}|\).
Solution:

Question 4.
If \(\overline{\mathbf{a}}=\overline{\mathbf{i}}-2 \overline{\mathbf{j}}-3 \overline{\mathbf{k}}, \overline{\mathbf{b}}=2 \overline{\mathbf{i}}+\overline{\mathbf{j}}-\overline{\mathbf{k}}\) and \(\bar{c}=\bar{i}+3 \bar{j}-2 \bar{k}\), verift that \(\overline{\mathbf{a}} \times(\overline{\mathbf{b}} \times \overline{\mathbf{c}}) \neq(\overline{\mathbf{a}} \times \overline{\mathbf{b}}) \times \overline{\mathbf{c}}\)
Solution:

From (1) and (2), we get
\(\overline{\mathbf{a}} \times(\overline{\mathbf{b}} \times \overline{\mathbf{c}}) \neq(\overline{\mathbf{a}} \times \overline{\mathbf{b}}) \times \overline{\mathbf{c}}\)
i.e., vector multiplication is not associative.

Question 5.
If \(\overline{\mathbf{a}}=2 \overline{\mathbf{i}}+\overline{\mathbf{j}}-3 \overline{\mathbf{k}}, \overline{\mathbf{b}}=\overline{\mathbf{i}}-\mathbf{2} \mathbf{j}+\overline{\mathbf{k}}\), \(\bar{c}=-\bar{i}+\bar{j}-4 \bar{k}\) and \(\overline{\mathbf{d}}=\overline{\mathbf{i}}+\overline{\mathbf{j}}+\overline{\mathbf{k}}\), then compute \(|(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times(\overline{\mathrm{c}} \times \overline{\mathrm{d}})|\)
Solution:

Question 6.
If A = (1, a, a²), B = (1, b, b²) and C = (1, c, c²) are non-coplanar vectors and \(\left|\begin{array}{lll}
a & a^{2} & 1+a^{3} \\
b & b^{2} & 1+b^{3} \\
c & c^{2} & 1+c^{3}
\end{array}\right|\) = 0, then show that abc + 1 = 0
Solution:
\(\bar{A}, \bar{B}, \bar{C}\) are non-coplanar vectors.
∆ = \(\left|\begin{array}{lll}
1 & a & a^{2} \\
1 & b & b^{2} \\
1 & c & c^{2}
\end{array}\right|\) ≠ 0 ………..(1)

∆ + (abc) ∆ = 0; ∆(1 + abc) = 0
∆ ≠ 0 ⇒ 1 + abc = 0

Question 7.
If \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) are non-zero vectors, then \(|(\overline{\mathbf{a}} \times \mathbf{b} \cdot \overline{\mathbf{c}})|=|\overline{\mathbf{a}}||\mathbf{b}||\overline{\mathbf{c}}|\) \(\Leftrightarrow \overline{\mathbf{a}} \cdot \overline{\mathbf{b}}=\overline{\mathbf{b}} \cdot \overline{\mathbf{c}}=\overline{\mathbf{c}} \cdot \overline{\mathbf{a}}=\mathbf{0}\)
Solution:

Question 8.
If \(\overline{\mathbf{a}}=\overline{\mathbf{i}}-\mathbf{2} \overline{\mathbf{j}}+3 \overline{\mathbf{k}}, \quad \mathbf{b}=2 \overline{\mathbf{i}}+\overline{\mathbf{j}}+\overline{\mathbf{k}}\), \(\bar{c}=\overline{\mathbf{i}}+\overline{\mathbf{j}}+2 \overline{\mathbf{k}}\) then find \(|(\bar{a} \times \bar{b}) \times \bar{c}|\) and \(|\overline{\mathbf{a}} \times(\mathbf{b} \times \mathbf{c})|\).
Solution:

Question 9.
If \(|\bar{a}|=1,|\bar{b}|=1,|\bar{c}|=2\) and \(\bar{a} \times(a \times \bar{c})+\bar{b}=0\) then find the angle between \(\bar{a}\) and \(\bar{c}\).
Solution:

Question 10.
Let \(\overline{\mathbf{a}}=\overline{\mathbf{i}}-\overline{\mathbf{k}}, \quad \overline{\mathbf{b}}=\mathbf{x} \overline{\mathbf{i}}+\overline{\mathbf{j}}+(\mathbf{1}-\mathbf{x}) \overline{\mathbf{k}}\) and \(\bar{c}=y \bar{i}+x \bar{j}+(1+x-y) \bar{k}\), prove that the scalar triple product \(\left[\begin{array}{lll}
\bar{a} & \bar{b} & \bar{c}
\end{array}\right]\) is independent of both x and y.
Solution:

Question 11.
Let \(\overline{\mathbf{b}}=\mathbf{2} \overline{\mathbf{i}}+\overline{\mathbf{j}}-\overline{\mathbf{k}}, \overline{\mathbf{c}}=\overline{\mathbf{i}}+\mathbf{3} \overline{\mathbf{k}}\). If \(\overline{\mathrm{a}}\) is a unit vector then find the maximum value of \(\left[\begin{array}{lll}
\overline{\mathbf{a}} & \overline{\mathbf{b}} & \bar{c}
\end{array}\right]\).
Solution:
Let \(\bar{a}=x \bar{i}+y \bar{j}+z \bar{k}\) and x² + y² + z² = 1
∵ \(\overline{\mathrm{a}}\) unit vector

Question 12.
Let \(\overline{\mathrm{a}}=\overline{\mathrm{i}}-\overline{\mathrm{i}}, \overline{\mathrm{b}}=\overline{\mathrm{i}}-\overline{\mathrm{k}}, \overline{\mathrm{c}}=\overline{\mathrm{k}}-\overline{\mathrm{i}}\) Find unit vector \(\bar{d}\) such that \(\overline{\mathrm{a}} \cdot \overline{\mathrm{d}}=0=[\bar{b} \overline{\mathrm{c}} \overline{\mathrm{d}}]\)
Solution:

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Products of Vectors Solutions Exercise 5(b) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Products of Vectors Solutions Exercise 5(b)

I.

Question 1.
If \(|\overline{\mathbf{p}}|=2,|\overline{\mathbf{q}}|=3\) and \((\bar{p}, \bar{q})=\frac{\pi}{6}\), then find \(|\bar{p} \times \bar{q}|^{2}\).
Solution:

Question 2.
If \(\overline{\mathbf{a}}=2 \overline{\mathbf{i}}-\overline{\mathbf{j}}+\overline{\mathbf{k}}\) and \(\overline{\mathbf{b}}=\overline{\mathbf{i}}-3 \overline{\mathbf{j}}-5 \overline{\mathbf{k}}\), then find \(|\overline{\mathbf{a}} \times \overline{\mathbf{b}}|\).
Solution:

Question 3.
If \(\bar{a}=2 \bar{i}-3 \bar{j}+\overline{\mathbf{k}}\) and \(\bar{b}=\bar{i}+4 \bar{j}-2 \bar{k}\), then find \((\overline{\mathbf{a}}+\overline{\mathbf{b}}) \times(\overline{\mathbf{a}}-\overline{\mathbf{b}})\).
Solution:

Question 4.
If \(4 \bar{i}+\frac{2 p}{3} \bar{j}+p \bar{k}\) is parallel to the vector \(\overline{\mathbf{i}}+2 \overline{\mathbf{j}}+3 \overline{\mathbf{k}}\), find p.
Solution:

Question 5.
Compute \(\overline{\mathbf{a}} \times(\overline{\mathbf{b}}+\overline{\mathbf{c}})+\overline{\mathbf{b}} \times(\overline{\mathbf{c}}+\overline{\mathbf{a}})+\overline{\mathbf{c}} \times(\overline{\mathbf{a}}+\overline{\mathbf{b}})\)
Solution:

Question 6.
If \(\overline{\mathbf{p}}=\mathbf{x} \overline{\mathbf{i}}+\mathbf{y} \overline{\mathbf{j}}+\mathbf{z} \overline{\mathbf{k}}\), find the value of \(|\overline{\boldsymbol{p}} \times \overline{\mathbf{k}}|^{2}\)
Solution:

Question 7.
Compute \(2 \bar{j} \times(3 \bar{i}-4 \bar{k})+(\bar{i}+2 \hat{j}) \times \bar{k}\).
Solution:

Question 8.
Find a unit vector perpendicular to both \(\overline{\mathbf{i}}+\overline{\mathbf{j}}+\overline{\mathbf{k}}\) and \(2 \bar{i}+\bar{j}+3 \bar{k}\).
Solution:

Question 9.
If θ is the angle between the vectors \(\overline{\mathbf{i}}+\overline{\mathbf{j}}\) and \(\overline{\mathbf{j}}+\overline{\mathbf{k}}\), then find sin ?.
Solution:

Question 10.
Find the area of the parallelogram having \(\bar{a}=2 \bar{j}-\bar{k}\) and \(\overline{\mathbf{b}}=-\overline{\mathbf{i}}+\overline{\mathbf{k}}\) as adjacent sides.
Solution:
Vector area of the parallelogram having \(\bar{a}=2 \bar{j}-\bar{k}\) and \(\overline{\mathbf{b}}=-\overline{\mathbf{i}}+\overline{\mathbf{k}}\) as adjacent sides.

Question 11.
Find the area of the parallelogram, whose diagonals are \(3 \overline{\mathbf{i}}+\overline{\mathbf{j}}-2 \overline{\mathbf{k}}\) and \(\bar{i}-3 \bar{j}+4 \bar{k}\).
Solution:

Question 12.
Find the area of the triangle having \(3 \bar{i}+4 \bar{j}\) and \(-5 \bar{i}+7 \bar{j}\) as two of its sides.
Solution:

Question 13.
Find unit vector perpendicular to the plane determined by the vectors \(\bar{a}=4 \bar{i}+3 \bar{j}-\bar{k}\) and \(\overline{\mathbf{b}}=2 \tilde{i}-6 \overline{\mathbf{j}}-3 \overline{\mathbf{k}}\).
Solution:

Question 14.
Find the area of the triangle whose vertices are A(1, 2, 3), B(2, 3, 1) and C(3, 1, 2).
Solution:
Suppose \(\bar{i}, \bar{j}, \bar{k}\) are unit vectors along the co-ordinate axes.
Position vectors of A, B, C are

II.

Question 1.
If \(\overline{\mathbf{a}}+\overline{\mathbf{b}}+\overline{\mathbf{c}}=\overline{\mathbf{0}}\), then prove that \(\overline{\mathbf{a}} \times \overline{\mathbf{b}}=\overline{\mathbf{b}} \times \overline{\mathbf{c}}=\overline{\mathbf{c}} \times \overline{\mathbf{a}}\).
Solution:

Question 2.
If \(\overline{\mathbf{a}}=2 \bar{i}+\bar{j}-\bar{k}, \quad \bar{b}=-\bar{i}+2 \bar{j}-4 \bar{k}\) and \(\overline{\mathbf{c}}=\overline{\mathbf{i}}+\mathbf{j}+\overline{\mathbf{k}}\), then find \((\bar{a} \times \bar{b}) \cdot(\bar{b} \times \bar{c})\).
Solution:

Question 3.
Find the vector area and the area of the parallelogram having \(\overline{\mathbf{a}}=\overline{\mathbf{i}}+2 \overline{\mathbf{j}}-\overline{\mathbf{k}}\) and \(\overline{\mathbf{b}}=2 \overline{\mathbf{i}}-\overline{\mathbf{j}}+\mathbf{2} \overline{\mathbf{k}}\) as adjacent sides.
Solution:

Question 4.
If \(\overline{\mathbf{a}} \times \overline{\mathbf{b}}=\overline{\mathbf{b}} \times \overline{\mathbf{c}} \neq \overline{\mathbf{0}}\), show that, \(\overline{\mathbf{a}}+\overline{\mathbf{c}}=\mathbf{p} \overline{\mathbf{b}}\), where p is some scalar.
Solution:

Question 5.
Let \(\overline{\mathbf{a}}\) and \(\overline{\mathbf{b}}\) be vectors, satisfying \(|\overline{\mathbf{a}}|=|\overline{\mathbf{b}}|=5\) and \((\bar{a}, \bar{b})=45^{\circ}\). Find the area of the triangle having \(\overline{\mathbf{a}}-\mathbf{2} \overline{\mathbf{b}}\) and \(3 \bar{a}+2 \bar{b}\) as two of its sides.
Solution:

Question 6.
Find the vector having magnitude ?6 units and perpendicular to both \(\mathbf{2} \overline{\mathbf{i}}-\overline{\mathbf{k}}\) and \(\mathbf{3} \overline{\mathbf{i}}-\overline{\mathbf{j}}-\overline{\mathbf{k}}\).
Solution:

Question 7.
Find a unit vector perpendicular to the plane determined by the points P(1, -1, 2), Q(2, 0, -1) and R(0, 2, 1).
Solution:

Question 8.
If \(\overline{\mathbf{a}} \cdot \overline{\mathbf{b}}=\overline{\mathbf{a}} \cdot \overline{\mathbf{c}}\) and \(\overline{\mathbf{a}} \times \overline{\mathrm{b}}=\overline{\mathrm{a}} \times \overline{\mathrm{c}}, \overline{\mathbf{a}} \neq 0\), then show that \(\overline{\mathbf{b}}=\overline{\mathbf{c}}\).
Solution:

Question 9.
Find a vector of magnitude 3 and perpendicular to both the vector \(\overline{\mathbf{b}}=2 \bar{i}-2 \bar{j}+\bar{k}\) and \(\bar{c}=2 \bar{i}+2 \bar{j}+3 \bar{k}\).
Solution:

Question 10.
If \(|\overline{\mathbf{a}}|\) = 13, \(|\overline{\mathbf{b}}|\) = 5 and \(\overline{\mathbf{a}} \cdot \overline{\mathbf{b}}=\mathbf{6 0}\), then find \(|\overline{\mathbf{a}} \times \overline{\mathbf{b}}|\).
Solution:

Question 11.
Find a unit vector perpendicular to the plane passing through the points (1, 2, 3), (2, -1, 1) and (1, 2, -4).
Solution:
Let ‘O’ be the origin and let A, B, C be the given points.

III.

Question 1.
If \(\overline{\mathbf{a}}\), \(\overline{\mathbf{b}}\) and \(\overline{\mathbf{c}}\) represent the vertices A, B and C respectively of ∆ABC, then prove that \(|(\overline{\mathbf{a}} \times \overline{\mathbf{b}})+(\overline{\mathbf{b}} \times \overline{\mathbf{c}})+(\overline{\mathbf{c}} \times \overline{\mathbf{a}})|\) is twice the area of ∆ABC.
Solution:

Question 2.
If \(\overline{\mathbf{a}}=2 \overline{\mathbf{i}}+3 \overline{\mathbf{j}}+4 \overline{\mathbf{k}}, \overline{\mathbf{b}}=\overline{\mathbf{i}}+\overline{\mathbf{j}}-\overline{\mathbf{k}}\) and \(\overline{\mathbf{c}}=\overline{\mathbf{i}}-\overline{\mathbf{j}}+\overline{\mathbf{k}}\), then compute \(\overline{\mathbf{a}} \times(\overline{\mathbf{b}} \times \overline{\mathbf{c}})\) and verify that it is perpendicular to \(\bar{a}\).
Solution:

Question 3.
If \(\overline{\mathbf{a}}=7 \overline{\mathbf{i}}-2 \overline{\mathbf{j}}+3 \overline{\mathbf{k}}, \overline{\mathbf{b}}=2 \overline{\mathbf{i}}+8 \overline{\mathbf{k}}\) and \(\overline{\mathbf{c}}=\overline{\mathbf{i}}+\overline{\mathbf{j}}+\overline{\mathbf{k}}\) then compute \(\overline{\mathbf{a}} \times \mathbf{b}, \overline{\mathbf{a}} \times \overline{\mathbf{c}}\) and \(\bar{a} \times(\bar{b}+\bar{c})\). Verify whether cross product is distributive over vector addition.
Solution:

Question 4.
If \(\overline{\mathbf{a}}=\overline{\mathbf{i}}+\overline{\mathbf{j}}+\overline{\mathbf{k}}, \overline{\mathbf{c}}=\overline{\mathbf{j}}-\overline{\mathbf{k}}\), then find vector \(\overline{\mathbf{b}}\) such that \(\overline{\mathbf{a}} \times \overline{\mathbf{b}}=\overline{\mathbf{c}}\) and \(\bar{a} \cdot \bar{b}=3\).
Solution:

Question 5.
\(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) are three vectors of equal magnitudes and each of them is inclined at an angle of 60° to the others. If \(|\bar{a}+\bar{b}+\bar{c}|=\sqrt{6}\), then find \(|\overline{\mathbf{a}}|\).
Solution:

Question 6.
For any two vectors \(\bar{a}\) and \(\bar{b}\), show that \(\left(1+|\bar{a}|^{2}\right)\left(1+|\bar{b}|^{2}\right)\) = \(|\mathbf{1}-\overline{\mathbf{a}} \cdot \overline{\mathbf{b}}|^{2}+|\overline{\mathbf{a}}+\overline{\mathbf{b}}+\overline{\mathbf{a}} \times \overline{\mathbf{b}}|^{2}\)
Solution:

Question 7.
If \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) are unit vectors such that \(\overline{\mathbf{a}}\) is perpendicular to the plane of \(\overline{\mathbf{b}}, \overline{\mathbf{c}}\) and the angle between \(\overline{\mathbf{b}}\) and \(\overline{\mathbf{c}}\) is \(\frac{\pi}{3}\), then find \(|\bar{a}+\bar{b}+\bar{c}|\).
Solution:
Given that \(|\bar{a}|=|\bar{b}|=|\bar{c}|=1\)

Question 8.
\(\overline{\mathbf{a}}=3 \overline{\mathbf{i}}-\overline{\mathbf{j}}+2 \overline{\mathbf{k}}, \overline{\mathbf{b}}=-\overline{\mathbf{i}}+3 \overline{\mathbf{j}}+2 \overline{\mathbf{k}}\), \(\bar{c}=4 \bar{i}+5 \bar{j}-2 \bar{k}\) and \(\bar{d}=\bar{i}+3 \bar{j}+5 \bar{k}\) then compute the following.
(i) \((\overline{\mathbf{a}} \times \overline{\mathbf{b}}) \times(\bar{c} \times \bar{d})\) and
(ii) \((\overline{\mathbf{a}} \times \overline{\mathbf{b}}) \cdot \overline{\mathbf{c}}-(\overline{\mathbf{a}} \times \overline{\mathbf{d}}) \cdot \overline{\mathbf{b}}\)
Solution: