AP Inter 2nd Year

Practice AP Inter 2nd Year Maths Study Material Chapter 5 Continuity and Differentiability MCQ to identify your strengths and weak areas.

AP Inter 2nd Year Maths Continuity and Differentiability MCQ

I. Select the correct option from the given choices.

Question 1.
Which of the following statement is true?
1) Every polynomial function is continuous
2) The function f(x) = 5x + 3 is continuous at x = 0
3) The function f(x) = |x| is continuous at x = 0
4) All of the options are correct
Solution:
4) All of the options are correct
By definition, all are correct

Question 2.
If f(x) = \(\begin{cases}3 a x-2 b, & x>1 \\ a x+b+1, & x<1\end{cases}\) and \(\underset{x \rightarrow 1}{\mathrm{Lt}}\) f(x) exists.
Then the relation between a and b is
1) 3a – 2b = 1
2) 2a – 3b = 1
3) 2a + 3b = 1
4) 2a + 3b = 1
Solution:
2) 2a – 3b = 1
\(\underset{x \rightarrow 1}{\mathrm{Lim}}\) f(x) exists ⇒ LHL = RHL ⇒ \(\underset{{x \rightarrow 1-\\(x<1)}}{{Lim}}\) f(x) = \(\underset{{x \rightarrow 1+\\(x>1)}}{{Lim}}\) f(x)
⇒ a + b + 1 = 3a – 2b ⇒ 2a – 3b = 1

Question 3.
The function f(x) = \(\begin{cases}\frac{2}{5-x}, & x<3 \\ 5-x, & x \geq 3\end{cases}\) is
1) Left discontinuous at x = 3
2) Left continuous at x = 3
3) Right discontinuous at x = 5
4) Discontinuous at x = 5
Solution:
1) Left discontinuous at x = 3
At x = 3, f(3) = 5 – 3 = 2

LHL ≠ f(3) ⇒ f(x) is Left discontinuous at x = 3

Question 4.
If the function f(x) = \(\frac{\sqrt{1+x}-1}{x}\) is continuous at x = 0. Then f(0) =
1) \(-\frac{1}{2}\)
2) \(\frac{1}{3}\)
3) \(\frac{1}{2}\)
4) \(-\frac{1}{3}\)
Solution:
3) \(\frac{1}{2}\)
f(x) is continuous at x = 0

Question 5.
If a function f(x) defined on [a, b] is discontinuous at x = α ∈ [a, b] . Then

Solution:
f(x) is discontinuous at x = α ⇒ \(\underset{x \rightarrow \alpha}{\mathrm{Lim}}\) f(x) ≠ f(x) [by definition]

Question 6.
If the function f defined by f(x) = \(\begin{cases}\cos x, & \text { if } x \leq 0 \\ 3 x+\alpha, & \text { if } 0<x<2 \\ \beta x+3, & \text { if } 2 \leq x \leq 4 \\ 11, & \text { if } x>4\end{cases}\)
where α,β are real constants is continuous on R. Then α² + β² =
1) 3
2) 9
3) 5
4) 4
Solution:
3) 5
Given f is continuouson R f is continuous at every real number.
Consider continuity of f(x) ar x = 0

⇒ cos 0° = 3(0) + α ⇒ 1 = 0 + α ⇒ α = 1
Now, consider continuity at x = 4

⇒ β(4) + 3 = 11 ⇒ 4β = 8 ⇒ β = 2 Now, α² + β² = 1² + 2² = 5

Question 7.
In the interval [0, 3]. The function f(x) = |x – 1| + |x – 2| is
1) discontinuous
2) differentiable
3) continuous but not differentiable at x = 2 only
4) continuous but not differentiable at x = 1 and x = 2.
Solution:
4) continuous but not differentiable at x = 1 and x = 2.

Graph f(x) = |x – 1| + |x – 2| is
f(x) is continuous on [0, 3]
bot not differentiable at x = 1 and x = 2 (turning points)

Question 8.
If y = \(\sqrt{\mathbf{x}+\sqrt{\mathbf{x}+\sqrt{\mathbf{x}+\ldots . . \infty}}}\). Then \(\frac{d y}{d x}\) is equal to
1) \(\frac{1}{y}\)
2) \(\frac{1}{x}\)
3) \(\frac{1}{2x-1}\)
4) \(\frac{1}{2y-1}\)
Solution:
4) \(\frac{1}{2y-1}\)
Formula: If y = \(\sqrt{f(x)+\sqrt{f(x)+\sqrt{f(x)+\ldots}}}\) ∞, then \(\frac{d y}{d x}=\frac{f^{\prime}(x)}{2 y-1}\)
Given f(x) = x ⇒ f'(x) = 1 ∴ \(\frac{d y}{d x}=\frac{1}{2 y-1}\)

Question 9.
The set of all points where the function f(x) = 2x|x| is differentiable is
1) (-∞, ∞)
2) (-∞, 0) ∪ (0, ∞)
3) (0, ∞)
4) (-∞, 0)
Solution:
1) (-∞, ∞)

f(x) = 2x |x| = \(\begin{cases}-2 x^2 & \forall x \leq 0 \\ 2 x^2 & \forall x>0\end{cases}\)
f'(x) = \(\left\{\begin{aligned}
-4 \mathrm{x} & \forall \mathrm{x} \leq 0 \\
4 \mathrm{x} & \forall \mathrm{x}>0
\end{aligned}\right.\) exists ∀x ∈ R ⇒ f(x) is differentiable ∀x ∈ R ⇒ x ∈ (-∞, ∞)

Question 10.
Differentiation of (x² – 5x + 8) (x³ + 7x + 9) can be done
1) only by using product rule
2) only by obtaining a single polynomial expanding it
3) only by using logarithmic differentiation
4) All of the options are correct
Solution:
4) All of the options are correct
All are correct.

Question 11.
If y = cos^-1(cos x) the find \(\frac{d y}{d x}\) at x = \(\frac{5 \pi}{4}\)
1) 1
2) -1
3) 0
4) \(-\frac{1}{\sqrt{2}}\)
Solution:
2) -1
\(\frac{d}{d x}\left(\cos ^{-1} x\right)=\frac{-1}{\sqrt{1-x^2}}\)
y = \(\cos ^{-1}(\cos x) \Rightarrow \frac{d y}{d x}=\frac{-1}{\sqrt{1-(\cos x)^2}} \cdot \frac{d}{d x}(\cos x)=\frac{(-1)(-\sin x)}{\sqrt{-1(\cos x)^2}}=\frac{\sin x}{\sqrt{1-(\cos x)^2}}\)

Question 12.
If f(x) = x⁴ – x³ + 7x² + 14, then what is the value of f¹(5)?
1) 594
2) 549
3) 954
4) 495
Solution:
4) 495
f(x) = x⁴ – x³ + 7x² + 14 ⇒ f'(x) = 4x³ – 3x² + 14x
at x = 5; f'(5) = 4(5)³ – 3(5)² + 14(5) = 4(125) – 3 × 25 + 70 = 500 – 75 + 70 = 495

Question 13.
If y = x + \(\frac{1}{\mathbf{x}}\) then which among the following holds?
1) x²y¹ + xy = 0
2) x²y¹ + xy + 2 = 0
3) x²y¹ – xy + 2 = 0
4) x²y¹ + xy – 2 = 0
Solution:
3) x²y¹ – xy + 2 = 0
Given y = x + \(\frac{1}{x}\) ..(1); y = 1 – \(\frac{1}{x^2}\)
⇒ x²y¹ = x² – 1 …….(2) ⇒ x²y¹ – ⇒ x² + 1 = 0 x²y¹ – [xy – 1] + 1 = 0
⇒ x²y¹ – xy + 1 + 1 = 0 ⇒ x²y¹ – xy + 2 = 0

Question 14.
\(\frac{d}{d x}\left(e^{\log _e \sqrt{1+\tan ^2 x}}\right)\) when x ∈ Q₁
1) sec²(x) tan x
2) sec x tan²(x)
3) sec x tan x
4) tan² (x)
Solution:
3) sec x tan x
Given y = \(e^{\log _e \sqrt{1+\tan ^2 x}}\) [∵ e^logN_e = N]
y = \(\sqrt{1+\tan ^2 x}\) = sec x ∴ \(\frac{d y}{d x}=\frac{d}{d x}(\sec x)\)= sec x tan x

Question 15.
If y = log(cosh x) then \(\frac{d^2 y}{d x^2}\) =
1) sech² x
2) -sech² x
3) sinh x
4) -sinh x
Solution:
1) sech² x
y = log(cosh x)
⇒ \(\frac{d y}{d x}=\frac{1}{\cosh x}(\sinh x)=\tanh x \Rightarrow \frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}(\tanh x) \Rightarrow \frac{d^2 y}{d x^2}=\operatorname{sech}^2 x\)

Question 16.
If f(x) = \(\begin{cases}\frac{\sin ^2(a x)}{x^2} ; & x \neq 0 \\ 1 ; & x=0\end{cases}\) is continuous at x = 0, then the value of ‘a’ is
1) -1
2) 1
3) 0
4) ±1
Solution:
4) ±1

Question 17.
If y = sinh^-1\(\left[\frac{1-\mathbf{x}}{1+\mathbf{x}}\right]\). Then \(\frac{d y}{d x}\) is equal to
1) \(\frac{-\sqrt{2}}{(1+x) \sqrt{1+x^2}}\)
2) \(\frac{-1}{(1+x) \sqrt{x}}\)
3) \(\frac{1}{\left(1+x^2\right) \sqrt{1+x}}\)
4) \(\frac{\sqrt{2}}{1+x \sqrt{1-x^2}}\)
Solution:
1) \(\frac{-\sqrt{2}}{(1+x) \sqrt{1+x^2}}\)
Formula: \(\frac{d}{d x}\left(\sinh ^{-1} x\right)=\frac{1}{\sqrt{x^2+1}}\)
Given y = \(\sinh ^{-1}\left[\frac{1-x}{1+x}\right] \Rightarrow \frac{d y}{d x}=\frac{1}{\sqrt{\left(\frac{1-x}{1+x}\right)^2+\frac{1}{1}}} \cdot \frac{d}{d x}\left(\frac{1-x}{1+x}\right)\)
= \(\frac{1+x}{\sqrt{(1-x)^2+(1+x)^2}}\left[\frac{(1+x)[-1]-[(1-x)(1)]}{(1+x)^2}\right]\)
= \(\frac{-1-x-1+x}{\sqrt{2\left(1^2+x^2\right)}(1+x)}=\frac{-2}{\sqrt{2} \sqrt{1+x^2}(1+x)}=\frac{-2}{\sqrt{1+x^2}(1+x)}\)

Question 18.
[x] represents the greatest integer function of x. At x = \(-1 \frac{\mathrm{~d}}{\mathrm{dx}}(\sin \pi|\mathrm{x}|)\) =
1) 0
2) 2
3) -2
4) 1/2
Solution:
1) 0
Let y = sin π[x] \(\frac{\mathrm{d}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}[\sin \pi[\mathrm{x}]]=\frac{\mathrm{d}}{\mathrm{dx}}[\sin (\mathrm{n} \pi)]=\frac{\mathrm{d}}{\mathrm{dx}}(0)=0\)
where n = [x] = An integer ∈ Z ∀x ∈ R
G.S of θ = nπ ∀n ∈ Z ⇒ sin (nπ) = 0 ∀n ∈ Z

Question 19.
If 3.sin(xy) + 4.cos(xy) = 5, then \(\frac{d y}{d x}\) is equal to
1) \(\frac{3 \sin x y+4 \cos x y}{3 \cos x y-4 \sin x y}\)
2) \(\frac{3 \cos x y+4 \sin x y}{4 \cos x y-3 \sin x y}\)
3) \(\frac{-y}{x}\)
4) \(\frac{x}{y}\)
Solution:
3) \(\frac{-y}{x}\)

Given 3 sin(xy) + 4 cos(xy) = 5
⇒ \(\frac{3}{5}\)sin (xy) + \(\frac{4}{5}\)cos(xy) = \(\frac{5}{5}\) ⇒ sin (xy)cos α + cos (xy)sin α = 1
⇒ sin(xy + α) = sin 90° ⇒ xy = \(\frac{\pi}{2}\) – α = A constant
Diff. w.r.t x
⇒ \(x \frac{d y}{d x}+y(1)=0 \Rightarrow x \frac{d y}{d x}=-y\)
\(\frac{d y}{d x}=-\frac{y}{x}\)

Question 20.
If y = log^x_y then \(\frac{d y}{d x}\) is equal to
1) \(\frac{1}{x \log y}\)
2) \(\frac{\log y}{x(1+\log y)}\)
3) \(\frac{1}{x(1+\log y)}\)
4) \(\frac{1}{1+\log y}\)
Solution:
3) \(\frac{1}{x(1+\log y)}\)
Formula: \(\log _{\mathrm{b}}^{\mathrm{a}}=\frac{\log \mathrm{a}}{\log \mathrm{~b}}, \frac{\mathrm{~d}}{\mathrm{dx}}(\mathrm{U} \cdot \mathrm{~V})=\mathrm{U} \cdot \frac{\mathrm{dU}}{\mathrm{dx}}+\mathrm{V} \frac{\mathrm{dV}}{\mathrm{dx}}\)
Given y = \(\log _y^x \Rightarrow y=\frac{\log x}{\log y} \Rightarrow y \cdot(\log y)=\log x\)
⇒ \(y\left(\frac{1}{y} ; \frac{d y}{d x}\right)+(\log y) \frac{d y}{d x}=\frac{1}{x} \Rightarrow(1+\log y) \frac{d y}{d x}=\frac{1}{x} \Rightarrow \frac{d y}{d x}=\frac{1}{x(1+\log y)}\)

Practice AP Inter 2nd Year Maths Study Material Chapter 4 Determinants MCQ to identify your strengths and weak areas.

AP Inter 2nd Year Maths Determinants MCQ

Question 1.
If \(\left|\begin{array}{cc}
x & 2 \\
18 & x
\end{array}\right|=\left|\begin{array}{cc}
6 & 2 \\
18 & 6
\end{array}\right|\), then x is equal to
1) 6
2) ±6
3) -6
4) 0
Solution:
2) ±6
G.E = x² – 36 = 36 – 36 ⇒ x² – 36 = 0 ⇒ x² = 36 ⇒ x = ±6

Question 2.
If A is 3 × 3 matrix and det (3A) = k (deta A), then k =
1) 9
2) 6
3) 1
4) 27
Solution:
4) 27
Given A_{3 × 3} ∴ |3A| = 3³|A| = 27 (det A) = k (det A) ⇒ k = 27

Question 3.
Value of k, which \(\) is singular
1) 4
2) -4
3) ±4
4) 0
Solution:
3) ±4
If A is singular |A| = 0 ⇒ \(\left|\begin{array}{ll}
\mathrm{k} & 2 \\
8 & \mathrm{k}
\end{array}\right|\) = 0 ⇒ k² – 16 = 0 ⇒ k = ±4

Question 4.
The area of a triangle with vertices (-3, 0), (0, 3) and (0, k) is 9 sq. units the value of k will be
1) 9
2) 3
3) -9
4) 6
Solution:
1) 9
Area of the triangle formed by (x₁, y₁) (x₂, y₂) (x₃, y₃) is
= \(\frac{1}{2}\left|\begin{array}{lll}
1 & x_1 & y_1 \\
1 & x_2 & y_2 \\
1 & x_3 & y_3
\end{array}\right|=9 \Rightarrow \frac{1}{2}\left|\begin{array}{ccc}
1 & -3 & 0 \\
1 & 0 & 3 \\
1 & 0 & k
\end{array}\right|=9 \Rightarrow\left|\begin{array}{ccc}
1 & -3 & 0 \\
1 & 0 & 3 \\
1 & 0 & k
\end{array}\right|\) = 18 ⇒ 3|k – 3| = 18 ⇒ |k – 3| = 6
k – 3 = 6 ⇒ k = 9; k – 3 = -6 ⇒ k = -3

Question 5.
If A is square matrix of order 3 and |A| = -4, then |adj A| is equal to
1) -4
2) 4
3) -16
4) 16
Solution:
4) 16
If A_n×n, then |AdjA| = |A|^n-1
We have A_3×3 ∴ |AdjA| = (|A|)² = (-4)² = 16

Question 6.
If area of triangle is 35 sq units with vertices (2, -6), (5, 4) and (k, 4). Then k is
1) 12
2) -2
3) -12, -2
4) 12, -2
Solution:
4) 12, -2
Area of triangle = 35
⇒ \(\frac{1}{2}\left|\begin{array}{ccc}
1 & 2 & -6 \\
1 & 5 & 4 \\
1 & \mathrm{k} & 4
\end{array}\right|\) = 35 ⇒ |1(20 – 4k) – 2(4 – 4) – 6(k – 5)| = 70
⇒ 20 – 4k + 0 – 6k + 30 = ±70 ⇒ 50 – 10k = ±70 ⇒ 5 – k = ±7
5 – k = 7 ⇒ k = -2; 5 – k = -7 ⇒ k = 12

Question 7.
If A = \(\left|\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right|\) and A_ij is Cofactors of a_ij then value of ∆ is given by
1) a₁₁A₃₁ + a₁₁A₃₂ + a₁₃A₃₃
2) a₁₁A₁₁ + a₁₂A₂₁ + a₁₃A₃₁
3) a₂₁A₁₁ + a₂₂A₁₂ + a₂₃A₁₃
4) a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁
Solution:
4) a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁
∆ = Determinant of a mathix
= sum of the products of the elements of a row (or) column with the corresponding co-factors = (a₁₁)A₁₁ + (a₁₂)A₁₂ + (a₃₁)A₃₁ [using 1^st row]

Question 8.
Let A be a nonsingular square matrix of order 3 × 3. Then |adj A| is equal to
1) |A|
2) |A|²
3) |A|³
4) 3|A|
Solution:
2) |A|²
If A_n×n, then |AdjA| = |A|^n-1
We have A_3×3 ∴ |AdjA| = |A|²

Question 9.
If A is an invertible matrix of order 2, then det (A^-1) is equal to
1) det(A)
2) \(\frac{1}{{det}(\mathrm{~A})}\)
3) 1
4) 0
Solution:
2) \(\frac{1}{{det}(\mathrm{~A})}\)
A_2×2, and A, exists ∵ AA^-1 = A^-1A = I
Consider AA^-1 = I
|AA^-1| = |I| ⇒ |A||A^-1| = I ⇒ |A^-1| = \(\frac{1}{|\mathrm{~A}|}\) ⇒ det(A^-1| = \(\frac{1}{{det} A}\)

Question 10.
If x, y, z are nonzero real numbers, then the inverse of matrix A = \(\left[\begin{array}{lll}
x & 0 & 0 \\
0 & y & 0 \\
0 & 0 & z
\end{array}\right]\) is
1) \(\left[\begin{array}{ccc}
x^{-1} & 0 & 0 \\
0 & y^{-1} & 0 \\
0 & 0 & z^{-1}
\end{array}\right]\)
2) \(x y z\left[\begin{array}{ccc}
x^{-1} & 0 & 0 \\
0 & y^{-1} & 0 \\
0 & 0 & z^{-1}
\end{array}\right]\)
3) \(\frac{1}{\mathrm{xyz}}\left[\begin{array}{ccc}
\mathrm{x} & 0 & 0 \\
0 & \mathrm{y} & 0 \\
0 & 0 & \mathrm{z}
\end{array}\right]\)
4) \(\frac{1}{\mathrm{xyz}}\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
Solution:
1) \(\left[\begin{array}{ccc}
x^{-1} & 0 & 0 \\
0 & y^{-1} & 0 \\
0 & 0 & z^{-1}
\end{array}\right]\)
If A = diag[a b c] then A^-1 = \(\frac{1}{|\mathrm{~A}|}\) AdjA = \(\left[\begin{array}{lll}
a^{-1} & b^{-1} & c^{-1}
\end{array}\right]=\left[\begin{array}{ccc}
1 / a & 0 & 0 \\
0 & 1 / b & 0 \\
0 & 0 & 1 / c
\end{array}\right]\)
∴ A = \(\left[\begin{array}{lll}
x & 0 & 0 \\
0 & y & 0 \\
0 & 0 & z
\end{array}\right]\) = diag [x y z]
⇒ A^-1 = diag\(\left[\begin{array}{lll}
x^{-1} & y^{-1} & z^{-1}
\end{array}\right]=\left[\begin{array}{ccc}
x^{-1} & 0 & 0 \\
0 & y^{-1} & 0 \\
0 & 0 & z^{-1}
\end{array}\right]\)

Question 11.
Let A = \(\left[\begin{array}{ccc}
1 & \sin \theta & 1 \\
-\sin \theta & 1 & \sin \theta \\
-1 & -\sin \theta & 1
\end{array}\right]\), where 0 ≤ θ ≤ 2π. Then
1) Det(A) = 0
2) Det(A) ∈ (2, ∞)
3) Det(A) ∈ (2, 4)
4) Det(A) ∈ [2, 4]
Solution:
4) Det(A) ∈ [2, 4]
|A| = 1(1 + sin²θ) – sinθ[-sin θ + sin θ] + 1[sin² θ + 1]
= 1 + sin² θ – 0 + sin² θ + 1 = 2 + 2sin² θ
⇒ det A = 2 + 2sin² θ
∵ 0 ≤ sin² θ ≤ 1 ⇒ 0 ≤ 2sin² θ ≤ 2 ⇒ 0 + 2 ≤ (2 + 2sin² θ) ≤ 2 + 2 ⇒ 2 ≤ |A| ≤ 4
⇒ |A| ∈ [2, 4]

Practice AP Inter 2nd Year Maths Study Material Chapter 3 Matrices MCQ to identify your strengths and weak areas.

AP Inter 2nd Year Maths Matrices MCQ

I. Select the correct option from the given choices.

Question 1.
A = [a_ij]_m×n is a square matrix, if
1) m < n
2) m > n
3) m = n
4) m = n²
Solution:
3) m = n
In a square matrix, the number of rows is equal to the number of columns.
∴ A = [a_ij]_m×n is a square matrix, if m = n.

Question 2.
Which of the given values of x and y make the following pair of matrices equal \(\left[\begin{array}{cc}
3 x+2 & 5 \\
y+1 & 2-3 x
\end{array}\right]=\left[\begin{array}{cc}
0 & y-2 \\
8 & 4
\end{array}\right]\)
1) x = \(\frac{-1}{3}\), y = 7
2) y = 7, x = \(\frac{2}{3}\)
3) y = 7, x = \(\frac{-2}{3}\)
4) x = \(\frac{-1}{3}\), y = \(\frac{-2}{3}\)
Solution:
3) y = 7, x = \(\frac{-2}{3}\)
Equating the corresponding elements, in the given matrices, we get
3x + 2 = 0 ⇒ 3x = -2 ⇒ x = \(\frac{-2}{3}\); 2 – 3x = 4 ⇒ x = \(\frac{-2}{3}\)
y – 2 = 5 ⇒ y = 7; y + 1 = 8 ⇒ y = 7

Question 3.
The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is:
1) 27
2) 18
3) 81
4) 512
Solution:
4) 512
Matrix of the order 3 × 3 has 9 elements and each of these elements can be either 0 or 1.
Now, each of the 9 elements can be filled in two possible ways.
Hence, by the multiplication principle, the required number of possible matrices is 2⁹ = 512.

Question 4.
Assume Y, W, P are matrices of order 3 × k, n ×3, p × k respectively.
The restriction on n, k and p so that PY + WY will be defined are:
1) k = 3, p = n
2) k is arbitrary, p = 2
3) p is arbitrary, k = 3
4) k = 2, p = 3
Solution:
1) k = 3, p = n
\(\underset{p \times k}{P} \underset{3 \times k}{Y}+\underset{n \times 3}{W} \underset{3 \times k}{Y}\) has to be deflned
i) If k = 3 then PY is defined (PY_p×k)
ii) If n = p then WY is defined (WY_p×k )
⇒ PY + WY (addition of matrices of same order) will be defined when k = 3, p = n.

Question 5.
Assume X, Z are matrices of order 2 × n, 2 × p respectively.
If n = p, then the order of the matrix 7X – 5Z is
1) p ×2
2) 2 × n
3) n × 3
4) p × n
Solution:
2) 2 × n
Given X_2×n ⇒ order of 7X is 2 × n
Given Z_2×p ⇒ order of 5Z is 2 × p.
Also p = n ⇒ order of 7X – 5Z is 2 × p (or) 2 × n

Question 6.
If A, B are symmetric matrices of same order, then AB – BA is a
1) Skew symmetric matrix
2) Symmetric matrix
3) Zero matrix
4) Identity matrix
Solution:
1) Skew symmetric matrix
If A, B are symmetric matrices of same order, then A’ =A and B’ = B ………(1)
Now consider(AB – BA)’ = (AB)’ – (BA)’ [∵ (A – B)’ = A’ – B’ ]
=B’A’ – A’B’ [∵ (AB)’ = B’A’]
= BA – AB [from(1)] = -(AB – BA) ∴ (AB – BA)’ = -(AB – BA)
Thus, AB – BA is a skew symmetric matrix.

Question 7.
If A = \(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\) and A + A’ = I, then the value of α is
1) \(\frac{\pi}{6}\)
2) \(\frac{\pi}{3}\)
3) π
4) \(\frac{3\pi}{2}\)
Solution:
2) \(\frac{\pi}{3}\)
Given that A = \(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right] \Rightarrow A^{\prime}=\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\)
Now, A + A’ = I
∴ \(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]+\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \Rightarrow\left[\begin{array}{cc}
2 \cos \alpha & 0 \\
0 & 2 \cos \alpha
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Equating the corresponding elements of the two matrices, we get
2cosα = 1 ⇒ cos α = \(\frac{1}{2}\) ⇒ α = cos^-1\(\left(\frac{1}{2}\right)\) ⇒ α = \(\frac{\pi}{3}\)

Question 8.
Matrices A and B will be inverse of each other only if
1) AB = BA
2) AB = BA = O
3) AB = O, BA = I
4) AB = BA = I
Solution:
4) AB = BA = I
From the definition of Inverse of a matrix, two matrices A and B are inverses of each other only when AB = BA = I.

Question 9.
If A = \(\left[\begin{array}{cc}
\alpha & \beta \\
-\gamma & \alpha
\end{array}\right]\) is such that A² = I, then
1) 1 + α² + βγ = 0
2) 1 – α² + βγ = 0
3) 1 – α² – βγ = 0
4) 1 + α² – βγ = 0
Solution:
3) 1 – α² – βγ = 0
Given that A = \(\left[\begin{array}{cc}
\alpha & \beta \\
\gamma & -\alpha
\end{array}\right]\)
A² = A.A = \(\left[\begin{array}{cc}
\alpha & \beta \\
\gamma & -\alpha
\end{array}\right]\left[\begin{array}{cc}
\alpha & \beta \\
\gamma & -\alpha
\end{array}\right]=\left[\begin{array}{cc}
\alpha^2+\beta \gamma & \alpha \beta-\alpha \beta \\
\alpha \gamma-\alpha \gamma & \beta \gamma+\alpha^2
\end{array}\right]=\left[\begin{array}{cc}
\alpha^2+\beta \gamma & 0 \\
0 & \beta \gamma+\alpha^2
\end{array}\right]\)
Now A² = I. Hence \(\left[\begin{array}{cc}
\alpha^2+\beta \gamma & 0 \\
0 & \beta \gamma+\alpha^2
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Equating the corresponding elements, we get
α² + βγ = 1 ⇒ α² + βγ – 1 = 0 ⇒ 1 – α² – βγ = 0

Question 10.
If the matrix A is both symmetric and skew symmetric, then
1) A is a diagonal matrix
2) A is a zero matrix
3) A is a square matrix
4) None of these
Solution:
2) A is a zero matrix
Given that the matrix A is both symmetric and skew symmetric.
So A’ = A and A’ = -A
∴ A + A = O ⇒ 2A = O ⇒ A = O ∴ A is a zero matrix

Question 11.
If A is square matrix such that A² = A, then (I + A)³ – 7 A is equal to
1) A
2) I – A
3) I
4) 3A
Solution:
3) 1
Given that the matrix A is a square matrix such that A² = A
Now, (I + A)³ – 7A = I³ + A³ + 3I²A + 3A²I – 7A
= I + A².A + 3A + 3A² – 7A
= I + A.A + 3A + 3A – 7A [A² = A]
= I + A² – A = I + A – A = I [∵ A² = A]
Hence, (I + A)³ – 7A = I

Practice AP Inter 2nd Year Maths Study Material Chapter 2 Inverse Trigonometric Functions MCQ to identify your strengths and weak areas.

AP Inter 2nd Year Maths Inverse Trigonometric Functions MCQ

I. Select the correct option from the given choices.

Question 1.
If sin^-1x = y , then
1) 0 ≤ y ≤ π
2) \(-\frac{\pi}{2}\) ≤ y ≤ \(\frac{\pi}{2}\)
3) 0 < y < π
4) \(-\frac{\pi}{2}\) < y < \(\frac{\pi}{2}\)
Solution:
2) \(-\frac{\pi}{2}\) ≤ y ≤ \(\frac{\pi}{2}\)
We know that range of the principle value of \(\sin ^{-1} x=\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)
Given that sin^-1 x = y, ∴ \(-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}\)

Question 2.
tan^-1\(\sqrt{3}\) – sec^-1(-2) is equal to
1) π
2) \(-\frac{\pi}{3}\)
3) \(\frac{\pi}{3}\)
4) \(\frac{2 \pi}{3}\)
Solution:
2) \(-\frac{\pi}{3}\)
Formula: sec^-1(-x) = -sec^-1x;
tan^-1\(\sqrt{3}\) – sec^-1(-2) = tan^-1\(\sqrt{3}\) – (π – sec^-12) = 60° – 180° = -60° = \(\frac{-\pi}{3}\)

Question 3.
cos^-1(cos\(\frac{7 \pi}{6}\)) is equal to
1) \(\frac{7 \pi}{6}\)
2) \(\frac{5 \pi}{6}\)
3) \(\frac{\pi}{6}\)
4) \(\frac{\pi}{6}\)
Solution:
2) \(\frac{5 \pi}{6}\)
\(\frac{7 \pi}{6}=\pi+\frac{\pi}{6}\) and cos^-1(-x) = π – cos^-1x
∴ cos^-1 \(\cos \frac{7 \pi}{6}=\cos ^{-1}\left[\cos \left(\pi+\frac{\pi}{6}\right)\right]=\cos ^{-1}\left[-\cos \frac{\pi}{6}\right]=\pi-\cos ^{-1}\left(\cos \frac{\pi}{6}\right)=\pi-\frac{\pi}{6}=\frac{5 \pi}{6}\)

Question 4.
sin\(\left(\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right)\) is equal to
1) \(\frac{1}{2}\)
2) \(\frac{1}{3}\)
3) \(\frac{1}{4}\)
4) 1
Solution:
4) 1
Formula: sin^-1(-x) = -sin^-1(x)
\(\sin \left(\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right)=\sin \left[60^{\circ}+\sin \left(\frac{1}{2}\right)\right]\) = sin[60° + 30°] = sin 90° = 1

Question 5.
tan^-1\(\sqrt{3}\) – cot^-1(\(-\sqrt{3}\)) is equal to
1) π
2) \(\frac{-\pi}{2}\)
3) 0
4) 2\(\sqrt{3}\)
Solution:
2) \(\frac{-\pi}{2}\)
Formula: cot^-1(-x) = π – cot^-1(x)
tan^-1\(\sqrt{3}\) – cot^-1(\(-\sqrt{3}\)) = tan^-1(\(\sqrt{3}\)) – (π – cot^-1(\(\sqrt{3}\)\frac{-\pi}{2})) = 60° – 180° + 30° = -90° = \(\frac{-\pi}{2}\)

Question 6.
sin(tan^-1 x), |x| < 1 is equal to
1) \(\frac{x}{\sqrt{1-x^2}}\)
2) \(\frac{1}{\sqrt{1-x^2}}\)
3) \(\frac{1}{\sqrt{1+x^2}}\)
4) \(\frac{x}{\sqrt{1+x^2}}\)
Solution:
4) \(\frac{x}{\sqrt{1+x^2}}\)
tan^-1 x = θ ⇒ \(\frac{x}{1}=\tan \theta \Rightarrow \sin \theta=\frac{A B}{B C}=\frac{x}{\sqrt{1+x^2}}\)

Question 7.
If sin^-1(1 – x) – 2sin^-1 x = \(\frac{\pi}{2}\), then x is equal to
1) 0, \(\frac{1}{2}\)
2) 1, \(\frac{1}{2}\)
3) 0
4) \(\frac{1}{2}\)
Solution:
3) 0
Formula: sin(90° + θ) = cosθ; cos2θ = 1 – 2sin²θ
G.E = \(\sin ^{-1}(1-x)=\frac{\pi}{2}+2 \sin ^{-1} x \Rightarrow(1-x)=\sin \left[\frac{\pi}{2}+2 \sin ^{-1} x\right]\)
⇒ 1 – x = cos(2sin^-1 x) = 1 – 2(sin(sin^-1 x))² = 1 – 2x²
∴ 1 – x = 1 – 2x² ⇒ 2x² – x = 0 ⇒ x(2x – 1) = 0 x = 1,\(\frac{1}{2}\)

Question 8.
sin\(\left[\frac{\pi}{3}+\sin ^{-1}\left(\frac{-1}{2}\right)\right]\) is equal to:
1) 1
2) \(\frac{1}{2}\)
3) \(\frac{1}{3}\)
4) \(\frac{1}{4}\)
Solution:
2) \(\frac{1}{2}\)
sin^-1\(\left(-\frac{1}{2}\right)\) = -30 and cos^-1(-x) = π – cos^-1x
sin(60° – 30°) = sin30° = \(\frac{1}{2}\)

Question 9.
The principle value of \(\cos ^{-1}\left(\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{\sqrt{2}}\right)\) is
1) \(\frac{\pi}{12}\)
2) π
3) \(\frac{\pi}{3}\)
4) \(\frac{\pi}{6}\)
Solution:
1) \(\frac{\pi}{12}\)
cos^-1\(\left(\frac{1}{2}\right)\) – sin^-1\(\left(\frac{1}{\sqrt{2}}\right)\) = 60° – 45° = 15° = \(\frac{\pi}{12}\)

Question 10.
The principle value of \(\tan ^{-1}\left(\tan \frac{9 \pi}{8}\right)\)
1) \(\frac{\pi}{8}\)
2) \(\frac{3\pi}{8}\)
3) \(\frac{-\pi}{8}\)
4) \(\frac{-3\pi}{8}\)
Solution:
1) \(\frac{\pi}{8}\)
\(\tan ^{-1}\left(\tan \left(\pi+\frac{\pi}{8}\right)\right)=\tan ^{-1}\left(\tan \left(\frac{\pi}{8}\right)\right)=\frac{\pi}{8}\)

Question 11.
The domain of the function cos^-1(2x – 3) is
1) [-1, 1]
2) (1, 2)
3) (-1, 1)
4) [1, 2]
Solution:
4) [1, 2]
Formula: cos^-1x is defined for x ∈ (-1, 1)
-1 ≤ (2x – 3) ≤ 1 ⇒ (3 – 1) ≤ 2x ≤ (3 + 1) ⇒ 2 ≤ 2x ≤ 4 ⇒ 1 ≤ x ≤ 2 x ∈ [1, 2]

Question 12.
The value of \(\sin ^{-1}\left(\cos \frac{3 \pi}{5}\right)\)
1) \(\frac{\pi}{10}\)
2) \(\frac{3 \pi}{5}\)
3) \(-\frac{\pi}{10}\)
4) \(-\frac{3 \pi}{5}\)
Solution:
3) \(-\frac{\pi}{10}\)
Formula: cosθ = sin(90° – θ)
\(\sin ^{-1}\left(\cos \frac{3 \pi}{5}\right)=\sin ^{-1}\left[\sin \left(\frac{\pi}{2}-\frac{3 \pi}{5}\right)\right]=\sin ^{-1}\left[\sin \left(\frac{-\pi}{10}\right)\right]=-\frac{\pi}{10}\)

Question 13.
The principle value of tan^-1(-1) is
1) \(\frac{\pi}{4}\)
2) \(-\frac{\pi}{4}\)
3) \(\frac{\pi}{2}\)
4) \(\frac{\pi}{3}\)
Solution:
2) \(-\frac{\pi}{4}\)
Formula: tan^-1(-x) = -tan^-1x
tan^-1(-1) = -tan^-1(1) = \(-\frac{\pi}{4}\)

Question 14.
The principle value of \(\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)\) is
1) \(\frac{13 \pi}{6}\)
2) \(\frac{\pi}{2}\)
3) \(\frac{\pi}{3}\)
4) \(\frac{\pi}{6}\)
Solution:
4) \(\frac{\pi}{6}\)
\(\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)=\cos ^{-1}\left[\cos \left(2 \pi+\frac{\pi}{6}\right)\right]=\cos ^{-1}\left[\cos \left(\frac{\pi}{6}\right)\right]=\frac{\pi}{6}\)

Question 15.
The simplest form of \(\tan ^{-1}\left[\frac{\sqrt{1+\mathrm{x}}-\sqrt{1-\mathrm{x}}}{\sqrt{1+\mathrm{x}}+\sqrt{1-\mathrm{x}}}\right.\) is
1) \(\frac{\pi}{4}-\frac{\pi}{2}\)
2) \(\frac{\pi}{4}+\frac{\pi}{2}\)
3) \(\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x\)
4) \(\frac{\pi}{4}+\frac{\pi}{2} \cos ^{-1} x\)
Solution:
3) \(\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x\)
Put x = cos2θ ⇒ 2θ = cos^-1x ⇒ θ = \(\frac{1}{2}\)cos^-1x
\(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}=\frac{\sqrt{1+\cos 2 \theta}-\sqrt{1-\cos 2 \theta}}{\sqrt{1+\cos 2 \theta}+\sqrt{1-\cos 2 \theta}}=\frac{\sqrt{2 \cos ^2 \theta}-\sqrt{2 \sin ^2 \theta}}{\sqrt{2 \cos ^2 \theta}+\sqrt{2 \sin ^2 \theta}}\)
= \(\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}=\frac{1-\tan \theta}{1+\tan \theta}=\tan \left(\frac{\pi}{4}-\theta\right)\)
∴ \(\tan ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)=\frac{\pi}{4}-\theta=\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x\)

Practice AP Inter 2nd Year Maths Study Material Chapter 1 Relations and Functions MCQ to identify your strengths and weak areas.

AP Inter 2nd Year Maths Relations and Functions MCQ

I. Select the correct option from the given choices.

Question 1.
Let R he the relation in the set {1, 2, 3, 4} given by I
R = {(1, 2), (2, 2), (1, 1), (4,4), (1, 3), (3, 3), (3, 2)). Choose the correct answer.
1) R is reflexive and symmetric but not transitive.
2) R Is reflexive and transitive but not symmetric.
3) R is symmetric and transitive but not reflexive.
4)R is an equivalence relation.
Solution:
2) R Is reflexive and transitive but not symmetric.
Let A= {1, 2, 3, 4} and R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}
(i) (a, a) ∈ R for every a ∈A. So, R is reflexive.
(ii) (1, 2) ∈ R but (2, 1)∉ R. So, R is not symmetric.
(iii) (a, b), (b, c) ∈ R⇒ (a, c) ∈ R ∀ a, b, c ∈ A. So, R is transitive.
R is reflexive and transitive but not symmetric.

Question 2.
Let R be the relation in the set N given by R = {(a, b): a = b – 2. b > 6}.
Choose the correct answer.
1) (2, 4) ∈ R
2) (3, 8) ∈ R
3) (6, 8) ∈ R
4) (8, 7) ∈ R
Solution:
3) (6, 8) ∈ R
Given set is the set of Natural numbers N
Given that b > 6 ⇒ b = 7, 8, 9, … and a = b – 2,
Now, b = 7 ⇒ a = 7 – 2 = 5; (5, 7) ∈ R
b = 8 ⇒ a = 8 – 2 = 6; (6, 8) ∈ R

Question 3.
Let f : R → R be defined as f(x) = x⁴. Choose the correct answer.
1) f is one-one onto
2) f is many-one onto
3) f is one-one but not onto
4) f is neither one-one nor onto.
Solution:
4) f is neither one-one nor onto.
f: R → R and f(x) = x⁴
⇒ f(-1) = (-1)⁴ =1; f(1) = 1⁴ = 1
∴ f is not – one – one
Also x⁴ ≥ 0 (Domain) ⇒ Range = [0, ∞) co-domain ∀ x ∈ R
⇒ f is not onto ∴ f is neither one-one nor onto

Question 4.
Let f : R → R be defined as f (x) = 3x. Choose the correct answer.
1) f is one-one onto
2) f is many-one onto
3) f is one-one but not onto
4) f is neither one-one nor onto.
Solution:
1) f is one-one onto
f: R → R and f(x) = 3x . This is a linear function of the form f(x) = ax + b
⇒ f is both one-one and onto

Question 5.
Let A = {1, 2, 3}. Then number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is
1) 1
2) 2
3) 3
4) 4
Solution:
1) 1
The given set is A= {1, 2, 3}
The smallest relation containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is given by R={(1, 1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 1), (3, 1)}
(i) {(1, 1), (2, 2), (3, 3)} ∈ R ⇒R is reflexive.
(ii) {(1, 2), (2, l)} ∈ R and {(1, 3)(3, 1)} ∈ R ⇒ R is symmetric.
(iii) {(3, 1), (1, 2)} ∈ R but (3, 2) ∉ R ⇒ R is not transitive.
Now, if we add only any of the two pairs (3, 2) and (2, 3) (or both) to relation R,
then relation R will become transitive.
Hence, the total number of desired relations is one.

Question 6.
Let A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is
1) 1
2) 2
3) 3
4) 4
Solution:
2) 2
The given set is A-{1, 2, 3}
The smallest equivalence relation containing (1, 2) is given by:
R₁ = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}
Now, we are left with only four other pairs (2, 3), (3, 2), (1, 3) and (3, 1).
If we add any one pair say(2, 3) to R₁, then for symmetry we must add (3, 2), and for transitivity we are required to add (1, 3) and (3, 1).
Hence, the only equivalence relation (bigger than R₁) is the universal relation. This shows that the total number of equivalence relations containing (1.2) is two.

Question 7.
Set A has 3 elements and the set B has 4 elements. Then the number of injective mappings that can be defined from A to B :
1) 144
2) 12
3) 24
4) 64
Solution:
3) 24
n(A) = 3, n(B) = 4
no.of injective mappings from A to B = ^n(B)P_n(A) = ⁴P₃ = 4 × 3 × 2 = 24

Question 8.
Let A = {3, 5}. Then the number of reflexive relations on set A is
1) 2
2) 4
3) 0
4) 8
Solution:
2) 4
A = {3, 5}
A × A ={3, 5} × {3, 5} == {(3, 3), (5, 5), (3, 5), (5, 3)}
Here, reflexive relations are
R₁ = {(3, 3) (5, 5)}; R₂ = {(3, 3) (5, 5) (3, 5)};
R₃ = {(3, 3) (5, 5) (5, 3)}; R₄ = {(3, 3) (5, 5) (3, 5) (5, 3)};
∴ Number of reflexive relations on set A is 4.

Question 9.
Let f : R → R be the function defined by f(x) = \(\frac{2 x-1}{2}\) and g: R → R be another function defined by g(x) = x + 2, then (gof) \(\left(\frac{3}{2}\right)\) is
1) 2
2) 3
3) \(\frac{7}{2}\)
4) \(\frac{5}{2}\)
Solution:
2) 3
g of \(\left(\frac{3}{2}\right)=g\left[f\left(\frac{3}{2}\right)\right]=g\left[\frac{2 \times \frac{3}{2}-1}{2}\right]=g\left[\frac{2}{2}\right]\)
= g(1) = 1 + 2 = 3

Question 10.
If f(x) = \(\) then f^-1(x) is .
1) \(\frac{4}{2 x+3}\)
2) \(\frac{2 x-3}{4}\)
3) 2x + 3
4) \(\frac{4 x-3}{2}\)
Solution:
4) \(\frac{4 x-3}{2}\)
Put f^-1(x) = y …………. (1) ⇒ x = f(y)
x = \(\frac{2 y+3}{4}\) ⇒ 4x = 2y + 3 ⇒ 2y = 4x – 3 ⇒ y = \(\frac{4 x-3}{2}\)
∴ f^-1(x) = \(\frac{4 x-3}{2}\) (∵ from(1))

Referring to the AP Inter 2nd Year Maths Study Material Chapter 10 Vector Algebra Exercise 10e Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Vector Algebra Solutions Exercise 10e

I.

Question 1.
Write down a unit vector in XY-plane, making an angle of 30° with the positive direction of x-axis.
Solution:
Let the unit vector be taken as \(\overrightarrow{\mathrm{r}}=\cos \theta \hat{\mathrm{i}}+\sin \theta \hat{\mathrm{j}}\), where θ is angle with positive x-axis.
[∵ cos²θ + sin²θ = 1]
∴ \(\overrightarrow{\mathrm{r}}=\cos 30^{\circ} \hat{\mathrm{i}}+\sin 30^{\circ} \hat{\mathrm{j}}=\frac{\sqrt{3}}{2} \hat{\mathrm{i}}+\frac{1}{2} \hat{\mathrm{j}}\)

Question 2.
Find the scalar components and magnitude of the vector joining the points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂).
Solution:
Given that P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) ⇒ \(\overrightarrow{O P}=x_1 \hat{i}+y_1 \hat{j}+z_1 \hat{k} \text { and } \overrightarrow{O Q}=x_2 \hat{i}+y_2 \hat{j}+z_2 \hat{k}\)
∴ \(\overrightarrow{\mathrm{PQ}}=\overrightarrow{\mathrm{OQ}}-\overrightarrow{\mathrm{OP}}=\left(\mathrm{x}_2-\mathrm{x}_1\right) \hat{\mathrm{i}}+\left(\mathrm{y}_2-\mathrm{y}_1\right) \hat{\mathrm{j}}+\left(\mathrm{z}_2-\mathrm{z}_1\right) \hat{\mathrm{k}}\)
⇒ \(|\overrightarrow{\mathrm{PQ}}|=\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2+\left(\mathrm{z}_2-\mathrm{z}_1\right)^2}\)
Hence, the scalar components of the vectors are (x₂ – x₁), (y₂ – y₁), (z₂ – z₁)
and magnitude of the vector is \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}\)

Question 3.
A girl walks 4 km towards west, then she walks 3 km in a direction 30° east of north and stops. Determine the girl’s displacement from her initial point of departure.
Solution:
Let O and B be the initial and final positions of the girl, respectively. Then, the girl’s position can be shown by the adjacent diagram. \(\overrightarrow{\mathrm{OA}}=-4 \hat{\mathrm{i}}\)

Hence, the girl’s displacement from her intial point of departure is \(\overrightarrow{\mathrm{d}}=\frac{-5}{2} \hat{\mathrm{i}}+\frac{3 \sqrt{3}}{2} \hat{\mathrm{j}}\)

Question 4.
If \(\vec{a}=\vec{b}+\vec{c}\), then is it true that \(|\vec{a}|=|\vec{b}|+|\vec{c}|\)? Justify your answer.
Solution:
No. If \(\vec{a}=\vec{b}+\vec{c}\), then they form a triangle.
In ABC, \(\overrightarrow{\mathrm{CB}}=\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{CA}}=\overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{c}}\)
From triangle law of addition of vectors we have \(\vec{a}=\vec{b}+\vec{c}\)
From triangle inequality we have \(|\vec{a}|<|\vec{b}|+|\vec{c}|\)
Hence, it is not true that \(|\vec{a}|=|\vec{b}|+|\vec{c}|\)

Question 5.
Find the value of x for which \(x(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})\) is a unit vector.
Solution:
If \(x(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})\) is a unit vector then
\(|x(\hat{i}+\hat{j}+\hat{k})|=1 \Rightarrow \sqrt{x^2+x^2+x^2}=1 \Rightarrow \sqrt{3 x^2}=1 \Rightarrow \sqrt{3} x=1 \Rightarrow x= \pm \frac{1}{\sqrt{3}}\)

Question 6.
Find a vector of magnitude 5 units, and parallel to the resultant of the vectors \(\overrightarrow{\mathbf{a}}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}} \text { and } \overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}\)
Solution:
Given vectors \(\overrightarrow{\mathbf{a}}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}} \text { and } \overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}\)
Let vectors \(\overrightarrow{\mathbf{c}}\) be the resultant of vectors \(\vec{a} \text { and } \vec{b}\)
∴ \(\vec{c}=\vec{a}+\vec{b}=2 \hat{i}+3 \hat{j}-\hat{k}+\hat{i}-2 \hat{j}+\hat{k}=3 \hat{i}+\hat{j}+0 \hat{k}\)
∴ Required vector of magnitude 5 units and parallel to the resultant of the vectors \(\overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}\)
\(5 \hat{\mathrm{c}}=5 \frac{\overrightarrow{\mathrm{c}}}{|\overrightarrow{\mathrm{c}}|}=5\left(\frac{3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+0 \hat{\mathrm{k}}}{\sqrt{9+1+0}}\right)=\frac{5}{\sqrt{10}}(3 \hat{\mathrm{i}}+\hat{\mathrm{j}})=\frac{5}{\sqrt{10}} \frac{\sqrt{10}}{\sqrt{10}}(3 \hat{\mathrm{i}}+\hat{\mathrm{j}})=\frac{5}{10} \sqrt{10}(3 \hat{\mathrm{i}}+\hat{\mathrm{j}})\)
= \(\frac{\sqrt{10}}{2}(3 \hat{\mathrm{i}}+\hat{\mathrm{j}})=\frac{3}{2} \sqrt{10 \mathrm{i}}+\frac{\sqrt{10}}{2} \hat{\mathrm{j}} .\)

Question 7.
If \(\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \overrightarrow{\mathbf{b}}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}} \text { and } \overrightarrow{\mathbf{c}}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}\) find a unit vector parallel to the vector \(2 \vec{a}-\vec{b}+3 \vec{c}\).
Solution:
Given vectors are \(\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}-\hat{j}+3 \hat{k}, \vec{c}=\hat{i}-2 \hat{j}+\hat{k}\)
Let \(\vec{d}=2 \vec{a}-\vec{b}+3 \vec{c}=2(\hat{i}+\hat{j}+\hat{k})-(2 \hat{i}-\hat{j}+3 \hat{k})+3(\hat{i}-2 \hat{j}+\hat{k})=2 \hat{i}+2 \hat{j}+2 \hat{k}-2 \hat{i}+\hat{j}-3 \hat{k}+3 \hat{i}-6 \hat{j}+3 \hat{k}\)
∴ \(\hat{d}=3 \hat{i}-3 \hat{j}+2 \hat{k}\)
A unit vector parallel to the vector \(\overrightarrow{\mathrm{d}}=3 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}} \text { is } \hat{\mathrm{d}}=\frac{\overrightarrow{\mathrm{d}}}{|\overrightarrow{\mathrm{~d}}|}\)
= \(\frac{3 \hat{i}-3 \hat{j}+2 \hat{k}}{\sqrt{9+9+4}}=\frac{3 \hat{i}-3 \hat{j}+2 \hat{k}}{\sqrt{22}}=\frac{3}{\sqrt{22}} \hat{i}-\frac{3}{\sqrt{22}} \hat{j}+\frac{2}{\sqrt{22}} \hat{k}\)

Question 8.
Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are \(\pm\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)\)
Solution:
Let a vector be equally inclined to OX, OY and OZ at an angle α
So, the DCs of the vectors are cosα, cosα, and cosα
∴ cos²α + cos²α + cos²α = 1 ⇒ 3 cos²α = 1 ⇒ cos²α = \(\frac{1}{3}\) ⇒ cosα = \(\pm \frac{1}{\sqrt{3}}\)

II.

Question 1.
Show that the points A(1, -2, -8), B(5, 0, -2) and C(11, 3, 7) are collinear, and find the ratio in which B divides AC.
Solution:
Given points are A(1, -2, -8), B(5, 0, -2) C(11, 3, 7)

On equating the corresponding components, we get
⇒ 5(λ + 1) = (11λ + 1) ⇒ 5λ + 5 = 11λ + 1 ⇒ 6λ = 4 ⇒ λ = \(\frac{2}{3}\). Thus, the ratio is 2 : 3

Question 2.
Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are \((2 \vec{a}+\vec{b}) \text { and }(\vec{a}-3 \vec{b})\) externally in the ratio 1 : 2. Also, show that P is the mid point of the line segment RQ.
Solution:
We have \(\overrightarrow{\mathrm{OP}}=2 \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}} \text { and } \overrightarrow{\mathrm{OQ}}=\overrightarrow{\mathrm{a}}-3 \overrightarrow{\mathrm{~b}}\)
It is given that point R divides a line segment joining two points P and Q externally in the ratio 1 : 2.
Then, by using the section formula, we get
\(\overrightarrow{\mathrm{OR}}=\frac{2(\overrightarrow{\mathrm{OP}})-1 \overrightarrow{\mathrm{OQ}}}{2-1}=\frac{2(2 \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})-(\overrightarrow{\mathrm{a}}-3 \overrightarrow{\mathrm{~b}})}{2-1}=\frac{4 \overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{~b}}-\overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{~b}}}{1}=3 \overrightarrow{\mathrm{a}}+5 \overrightarrow{\mathrm{~b}}\)
Hence, the position vector of R is \(3 \overrightarrow{\mathrm{a}}+5 \overrightarrow{\mathrm{~b}}\)
Thus, the position vector of midpoint of RQ = \(\frac{\overrightarrow{\mathrm{OQ}}+\overrightarrow{\mathrm{OR}}}{2}\)
\(R Q=\frac{\overrightarrow{\mathrm{OQ}}+\overrightarrow{\mathrm{OR}}}{2}=\frac{(\overrightarrow{\mathrm{a}}-3 \overrightarrow{\mathrm{~b}})+(3 \overrightarrow{\mathrm{a}}+5 \overrightarrow{\mathrm{~b}})}{2}=2 \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}\)
Thus, P is the midpoint of line segment RQ.

Question 3.
The two adjacent sidt s of a parallelogram are \(2 \hat{i}-4 \hat{j}+5 \hat{k} \text { and } \hat{i}-2 \hat{j}-3 \hat{k}\). Find the unit vector parallel to its diagonal. Also, find its area.
Solution:
Let \(\vec{a}=2 \hat{i}-4 \hat{j}+5 \hat{k}, \quad \vec{b}=\hat{i}-2 \hat{j}-3 \hat{k}\)
Diagonal of the parallelogram is \(\vec{a}+\vec{b}\)
⇒ \(\vec{a}+\vec{b}=2 \hat{i}-4 \hat{j}+5 \hat{k}+\hat{i}-2 \hat{j}-3 \hat{k}=(2+1) \hat{i}+(-4-2) \hat{j}+(5-3) \hat{k}=3 \hat{i}-6 \hat{j}+2 \hat{k}\)
So the unit vector parallel to the diagonal is \(\frac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|}\)

Question 4.
Let \(\vec{i}=\hat{i}+4 \hat{j}+2 \hat{k}, \vec{b}=3 \hat{i}-2 \hat{j}+7 \hat{k} \text { and } \vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}\). Find a vector d which is perpendicular to both \(\text { \vec{a} and } \vec{b}, \text { and } \vec{c} . \vec{d} =15\)
Solution:
Let \(\vec{d}=d_1 \hat{i}+d_2 \hat{j}+d_3 \hat{k}\) Since \(\overrightarrow{\mathrm{d}}\) is perpendicular to both \(\vec{a} \text { and } \vec{b}\) we have
\(\overrightarrow{\mathrm{d}} \cdot \overrightarrow{\mathrm{a}}=0 \Rightarrow \mathrm{~d}_1+4 \mathrm{~d}_2+2 \mathrm{~d}_3=0\) ……(1)
\(\overrightarrow{\mathrm{d}} \cdot \overrightarrow{\mathrm{~b}}=0 \Rightarrow 3 \mathrm{~d}_1-2 \mathrm{~d}_2+7 \mathrm{~d}_3=0\) …………(2)
Also it is given that \(\vec{c} \cdot \vec{d}=15 \Rightarrow 2 d_1-d_2+4 d_3=15\) ……..(3)
On solving (1),(2) and (3) we get \(d_1=\frac{160}{3}, d_2=-\frac{5}{3}, d_3=-\frac{70}{3}\)
∴ \(\overrightarrow{\mathrm{d}}=\frac{160}{3} \hat{\mathrm{i}}-\frac{5}{3} \hat{\mathrm{j}}-\frac{70}{3} \hat{\mathrm{k}}=\frac{1}{3}(160 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}-70 \hat{\mathrm{k}})\)

Question 5.
The scalar product of the vector \(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}\) with a unit vector along the sum of vectors \(2 \hat{i}+4 \hat{j}-5 \hat{k} \text { and } \lambda \hat{i}+2 \hat{j}+3 \hat{k}\) is equsil to one. Find the value of λ.
Solution:
We have \((2 \hat{i}+4 \hat{j}-5 \hat{k})+(\lambda \hat{i}+2 \hat{j}+3 \hat{k})=(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}\)
Unit vector along \((2 \hat{i}+4 \hat{j}-5 \hat{k})+(\lambda \hat{i}+2 \hat{j}+3 \hat{k})\) is

⇒ λ² + 4λ + 44 = (λ + 6)² ⇒ λ² + 4x + 44 = λ² + 12λ + 36 ⇒ 8λ = 8 ⇒ λ = 1

Question 6.
If \(\vec{a}, \vec{b}, \vec{c}\) are mutually perpendicular vectors of equal magnitudes, show that the vector \(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}\) is equally inclined to \(\vec{a}, \vec{b}, \vec{c}\)
Solution:
Given that \(\vec{a}, \vec{b}, \vec{c}\) are mutually perpendicular vectors of equal magnitudes
⇒ \(\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=\vec{c} \cdot \vec{a}=0\)
Let \(\vec{a}+\vec{b}+\vec{c}\) be inclined to \(\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{c}}\) at angles θ₁, θ₂, θ₃ respectively.

Given that \(\vec{a}, \vec{b}, \vec{c}\) are of equal magnitude ⇒ \(|\overrightarrow{\mathrm{a}}|=|\overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{c}}|\)
Hence from (1), (2), (3) we get cosθ₁= cos θ₂ = cos θ₃ Thus, θ₁ = θ₂ = θ₃

Question 7.
Prove that \((\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{a}|^2+|\vec{b}|^2\), if and only if \(\vec{a} , \vec{b}\) are perpendicular, given \(a \neq 0, b \neq 0\)
Solution:

Referring to the AP Inter 2nd Year Maths Study Material Chapter 10 Vector Algebra Exercise 10d Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Vector Algebra Solutions Exercise 10d

I.

Question 1.
Find \(|\mathbf{a} \times \mathbf{b}| \text {, if } \overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}-7 \hat{\mathbf{j}}+7 \hat{\mathbf{k}} \text { and } \mathbf{b}=3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}\)
Solution:
Given that \(\vec{a}=\hat{i}-7 \hat{j}+7 \hat{k} \text { and } \vec{b}=3 \hat{i}-2 \hat{j}+2 \hat{k}\)
\(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -7 & 7 \\
3 & -2 & 2
\end{array}\right|=\hat{i}(-14+14)-\hat{j}(2-21)+\hat{k}(-2+21)=19 \hat{j}+19 \hat{k}\)
∴ \(|\vec{a} \times \vec{b}|=\sqrt{19^2+19^2}=\sqrt{2 \times(19)^2}=19 \sqrt{2}\)

Question 2.
If a unit vector \(\vec{a}\) makes angles \(\frac{\pi}{3}\) with \(\hat{\mathbf{i}}, \frac{\pi}{4} \text { with } \hat{\mathbf{j}}\) and an acute angle θ with \(\hat{\mathbf{k}}\), then find 6 and hence, the components of \(\vec{a}\).
Solution:
Let us take the unit vector as \(\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k} \text {. Then, }|\bar{a}|=1\)
Now, cos\(\frac{\pi}{3}=\frac{a_1}{|\vec{a}|} \Rightarrow a_1=\frac{1}{2} ; \cos \frac{\pi}{4}=\frac{a_2}{|\vec{a}|} \Rightarrow a_2=\frac{1}{\sqrt{2}}\)
\(\cos \theta=\frac{a_3}{|\vec{a}|} \Rightarrow a_3=\cos \theta\)
Since \(\vec{a}\) is a unit vector, we have \(\sqrt{a_1^2+a_2^2+a_3^2}=1 \Rightarrow\left(\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{2}}\right)^2+\cos ^2 \theta=1\)
⇒ \(\frac{1}{4}+\frac{1}{2}+\cos ^2 \theta=1 \Rightarrow \frac{3}{4}+\cos ^2 \theta=1 \Rightarrow \cos ^2 \theta=1-\frac{3}{4}=\frac{1}{4} \Rightarrow \cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3}\)
Hence, \(a_3=\cos \frac{\pi}{3}=\frac{1}{2}\) (∵ θ = \(\frac{\pi}{3}\)) Components of \(\hat{\mathbf{a}}\) are \(\left(\frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2}\right)\)

Question 3.
Show that \((\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})=2(\vec{a} \times \vec{b})\)
Solution:
L.H.S = \((\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})=(\vec{a}-\vec{b}) \times \vec{a}+(\vec{a}-\vec{b}) \times \vec{b}\)
= \(\vec{a} \times \vec{a}-\vec{b} \times \vec{a}+\vec{a} \times \vec{b}-\vec{b} \times \vec{b}=0+\vec{a} \times \vec{b}+\vec{a} \times \vec{b}-0=2(\vec{a} \times \vec{b})\) = R.H.S

Question 4.
Find λ and µ if \((2 \hat{i}+6 \hat{j}+27 \hat{k}) \times(\hat{i}+\lambda \hat{j}+\mu \hat{k})=0\) = 0
Solution:
Given that \((2 \hat{i}+6 \hat{j}+27 \hat{k}) \times(\hat{i}+\lambda \hat{j}+\mu \hat{k})=\overrightarrow{0} \Rightarrow(2 \hat{i}+6 \hat{j}+27 \hat{k}) \times(\hat{i}+\lambda \hat{j}+\mu \hat{k})=\overrightarrow{0}\)
⇒ \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 6 & 27 \\
1 & \lambda & \mu
\end{array}\right|=0 \hat{i}+0 \hat{j}+0 \hat{k} \Rightarrow \hat{i}(6 \mu-27 \lambda)-\hat{j}(2 \mu-27)+\hat{k}(2 \lambda-6)=0 \hat{i}+0 \hat{j}+0 \hat{k}\)
On Comparing the corresponding components, we have
6µ – 27λ = 0, 2λ – 6 = 0
2λ – 6 = 0 ⇒ λ = 3 2µ – 27 = 0 ⇒ 2µ – 27 = 0 ⇒ µ = \(\frac{27}{2}\)

Question 5.
Given that \(\vec{a} . \vec{b}=0 \text { and } \vec{a} \times \vec{b}=0\). What can you conclude about the vectors \(\vec{a} \text { and } \vec{b}\)?
Solution:
When \(\vec{a} \cdot \vec{b}=0\) then either \(|\overrightarrow{\mathrm{a}}|=0 \text { or }|\overrightarrow{\mathrm{b}}|=0\) or a JL b (if \(\overrightarrow{\mathrm{a}} \mid \neq 0 \text { and }|\overrightarrow{\mathrm{b}}| \neq 0\))
When \(\vec{a} \times \vec{b}=\overrightarrow{0}\) then either \(|\overrightarrow{\mathrm{a}}|=0 \text { or }|\overrightarrow{\mathrm{b}}|=0\) or \(\overrightarrow{\mathrm{a}} \| \overrightarrow{\mathrm{b}}\) (if \(\vec{a} \mid \neq 0 \text { and }|\vec{b}| \neq 0\))
But \(\vec{a} \text { and } \vec{b}\) cannot be perpendicular and parallel simultaneously.
We conclude that \(\overrightarrow{\mathrm{a}}=0 \text { or } \overrightarrow{\mathrm{b}}=0\)

Question 6.
If either \(\vec{a}=0 \text { or } \vec{b}=0 \text {, then } \vec{a} \times \vec{b}=\overrightarrow{0}\). Is the converse true? Justify your answer with an example.
Solution:
Let \(\vec{a}=2 \vec{i}+3 \vec{j}+4 \vec{k} \text { and } \vec{b}=4 \vec{i}+6 \vec{j}+8 \vec{k}\)
∴ \(\vec{a} \times \vec{b}=\left|\begin{array}{lll}
\hat{i} & \hat{j} & \hat{k} \\
2 & 3 & 4 \\
4 & 6 & 8
\end{array}\right|=\hat{i}(24-24)-\hat{j}(16-16)+\hat{k}(12-12)=\overrightarrow{0}\)
Here \(\vec{a}, \vec{b}\) are two non-zero collinear vectors
So, the converse of the statement is not true.

Question 7.
Find the area of the parallelogram whose adjacent sides are determined by the vectors \(\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}} \text { and } \overrightarrow{\mathbf{b}}=2 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}+\hat{\mathbf{k}}\)
Solution:
Given that \(\vec{a}=\hat{i}-\hat{j}+3 \hat{k} \text { and } \vec{b}=2 \hat{i}-7 \hat{j}+\hat{k}\)
∴ \(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -1 & 3 \\
2 & -7 & 1
\end{array}\right|=\hat{i}(-1+21)-\hat{j}(1-6)+\hat{k}(-7+2)=20 \hat{i}+5 \hat{j}-5 \hat{k}\)
We know that area of parallelogram = \(|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} \mid\)
= \(\sqrt{400+25+25}=\sqrt{450}=\sqrt{25 \times 9 \times 2}=5(3) \sqrt{2}=15 \sqrt{2} \text { Sq.units. }\)

II.

Question 1.
Find a unit vector perpendicular to each of the vector \(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}} \text { and } \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}\), where \(\vec{a}=3 \hat{i}+2 \hat{j}+2 \hat{k} \text { and } \vec{b}=\hat{i}+2 \hat{j}-2 \hat{k}\)
Solution:
Given \(\vec{a}=3 \hat{i}+2 \hat{j}+2 \hat{k} \text { and } \vec{b}=\hat{i}+2 \hat{j}-2 \hat{k}\)
Adding we have \(\vec{a}+\vec{b}=4 \hat{i}+4 \hat{j}+0 \hat{k}\)
Substracting \(\vec{a}-\vec{b}=2 \hat{i}+0 \hat{j}+4 \hat{k}\)
∴ \((\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
4 & 4 & 0 \\
2 & 0 & 4
\end{array}\right|=\hat{i}(16-0)-\hat{j}(16-0)+\hat{k}(0-8)=16 \hat{i}-16 \hat{j}-8 \hat{k}=\vec{c}(\text { say })\)
∴ \(|\vec{c} \mid=\sqrt{16^2+(-16)^2+(-8)^2}=\sqrt{256+256+64}=\sqrt{576}=24 .\)
∴ a unit vector perpendicular to both \(\vec{a} \text { and } \vec{b} \text { is } \hat{c}= \pm \frac{\vec{c}}{|\vec{c}|}\)
= \(\pm \frac{(16 \hat{\mathrm{i}}-16 \hat{\mathrm{j}}-8 \hat{\mathrm{k}})}{24}= \pm\left(\frac{16}{24} \hat{\mathrm{i}}-\frac{16}{24} \hat{\mathrm{j}}-\frac{8}{24} \hat{\mathrm{k}}\right)= \pm\left(\frac{2}{3} \hat{\mathrm{i}}-\frac{2}{3} \hat{\mathrm{j}}-\frac{1}{3} \hat{\mathrm{k}}\right) .\)

Question 2.
Let the vectors \(\vec{a}, \vec{b}, \vec{c} \text { be given as } a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}, b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}, c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}\) given as \(a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}, b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}, c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}\). Then show that \(\overrightarrow{\mathbf{a}} \times(\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}})=\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{c}}\).
Solution:
Given vectors \(\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}, \vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}, \vec{c}=c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}\)
∴ \(\vec{b}+\vec{c}=\left(b_1+c_1\right) \hat{i}+\left(b_2+c_2\right) \hat{j}+\left(b_3+c_3\right) \hat{k}\)

Question 3.
Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5)
Solution:
Vertices of ∆ABC are A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5)
Position Vector (P.V) of point A(1, 1, 2) is \(\overrightarrow{O A}=\hat{i}+\hat{j}+2 \hat{k}\)
Position Vector (P.V) of point B(2, 3, 5) is \(\overrightarrow{\mathrm{OB}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\)
Position Vector (P.V) of point C(1, 5, 5) is \(\overrightarrow{\mathrm{OC}}=\hat{\mathrm{i}}+5 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\)
\(\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=(2 \hat{i}+3 \hat{j}+5 \hat{k})-(\hat{i}+\hat{j}+2 \hat{k})=2 \hat{i}+3 \hat{j}+5 \hat{k}-\hat{i}-\hat{j}-2 \hat{k}=\hat{i}+2 \hat{j}+3 \hat{k}\)
\(\overrightarrow{A C}=\overrightarrow{O C}-\overrightarrow{O A}=(\hat{i}+5 \hat{j}+5 \hat{k})-(\hat{i}+\hat{j}+2 \hat{k})=\hat{i}+5 \hat{j}+5 \hat{k}-\hat{i}-\hat{j}-2 \hat{k}=0 \hat{i}+4 \hat{j}+3 \hat{k}\)
∴ \(\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=\left|\begin{array}{lll}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & 3 \\
0 & 4 & 3
\end{array}\right|=\hat{i}(6-12)-\hat{j}(3-0)+\hat{k}(4-0)=-6 \hat{i}-3 \hat{j}+4 \hat{k}\)
Area of triangle ABC = \(\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=\frac{1}{2} \sqrt{36+9+16}=\frac{\sqrt{61}}{2} \text { sq. units. }\)

III.

Question 1.
If \(\vec{a}=2 \vec{i}+\vec{j}-3 \vec{k}, \vec{b}=\vec{i}-2 \vec{j}+\vec{k}, \vec{c}=-\vec{i}+\vec{j}-4 \vec{k} \text { and } \vec{d}=\vec{i}+\vec{j}+\vec{k}\) then compute \(|(\overline{\mathbf{a}} \times \overline{\mathbf{b}}) \times(\overline{\mathbf{c}}+\overline{\mathbf{d}})|\)
Solution:
Given vectors \(\vec{a}=2 \vec{i}+\vec{j}-3 \vec{k}, \vec{b}=\vec{i}-2 \vec{j}+\vec{k}, \vec{c}=-\vec{i}+\vec{j}-4 \vec{k} \text { and } \vec{d}=\vec{i}+\vec{j}+\vec{k}\)

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

Question 3.
If \(\overline{\mathrm{a}}=7 \overline{\mathrm{i}}-\overline{2 \mathrm{j}}+\overline{\mathrm{k}}, \overline{\mathrm{~b}}=2 \overline{\mathrm{i}}+8 \overline{\mathrm{k}} \text { and } \overline{\mathrm{c}}=\overline{\mathrm{i}}+\overline{\mathrm{j}}+\overline{\mathrm{k}}\), then compute \(\mathbf{a} \times \mathbf{b}, \overline{\mathbf{a}} \times \overline{\mathbf{c}} \text { and } \overline{\mathbf{a}} \times(\overline{\mathbf{b}}+\mathbf{c})\). Verify whether cross product is distributive over vector addiion.
Solution:
Given that \(\overline{\mathrm{a}}=7 \overline{\mathrm{i}}-\overline{2 \mathrm{j}}+\overline{\mathrm{k}}, \overline{\mathrm{~b}}=2 \overline{\mathrm{i}}+8 \overline{\mathrm{k}} \text { and } \overline{\mathrm{c}}=\overline{\mathrm{i}}+\overline{\mathrm{j}}+\overline{\mathrm{k}}\)
Now \(\bar{a} \times \bar{b}=\left|\begin{array}{ccc}
\bar{i} & \bar{j} & \bar{k} \\
7 & -2 & 3 \\
2 & 0 & 8
\end{array}\right|=\bar{i}(-16-0)-\bar{j}(56-6)+\bar{k}(0+4)=-16 \bar{i}-50 \bar{j}+4 \bar{k}\)
Also, \(\bar{a} \times \bar{c}=\left|\begin{array}{ccc}
\bar{i} & \bar{j} & \bar{k} \\
7 & -2 & 3 \\
1 & 1 & 1
\end{array}\right|=\bar{i}(-2-3)-\bar{j}(7-3)+\bar{k}(7+2)=-5 \bar{i}-4 \bar{j}+9 \bar{k}\)
Now \(\bar{b}+\bar{c}=(2 \bar{i}+8 \bar{k})+(\bar{i}+\bar{j}+\bar{k})=3 \bar{i}+\bar{j}+9 \bar{k}\)
\(\bar{a} \times(\bar{b}+\bar{c})=\left|\begin{array}{ccc}
\bar{i} & \bar{j} & \bar{k} \\
7 & -2 & 3 \\
3 & 1 & 9
\end{array}\right|==\bar{i}(-18-3)-\bar{j}(63-9)+\bar{k}(7+6)=-21 \bar{i}-54 \bar{j}+13 \bar{k}\) …….(1)
Now \((\overline{\mathrm{a}} \times \overline{\mathrm{b}})+(\overline{\mathrm{a}} \times \overline{\mathrm{c}})=(-16 \overline{\mathrm{i}}-50 \overline{\mathrm{j}}+4 \overline{\mathrm{k}})+(-5 \overline{\mathrm{i}}-4 \overline{\mathrm{j}}+9 \overline{\mathrm{k}})=-21 \overline{\mathrm{i}}-54 \overline{\mathrm{j}}+13 \overline{\mathrm{k}}\) ……(2)
From (1) and (2) we have \(\bar{a} \times(\bar{b}+\bar{c})=(\bar{a} \times \bar{b})+(\bar{a} \times \bar{c})\)
∴ Vector product is distributive over vector addition.

Question 4.
If \(\bar{a}=2 \bar{i}+3 \bar{j}+4 \bar{k}, \bar{b}=\bar{i}+\bar{j}-\bar{k}, \bar{c}=\bar{i}-\bar{j}+\bar{k}\), compute \(\overline{\mathbf{a}} \times(\overline{\mathbf{b}} \times \overline{\mathbf{c}})\) and verify that it is perpendicular to \(\overline{\mathbf{a}}\)
Solution:
Given \(\bar{a}=2 \bar{i}+3 \bar{j}+4 \bar{k}, \bar{b}=\bar{i}+\bar{j}-\bar{k}, \bar{c}=\bar{i}-\bar{j}+\bar{k}\)
\(\bar{b} \times \bar{c}=\left|\begin{array}{ccc}
\bar{i} & \bar{j} & \bar{k} \\
1 & 1 & -1 \\
1 & -1 & 1
\end{array}\right|=\bar{i}(1-1)-\bar{j}(1+1)+\bar{k}(-1-1)=-2 \bar{j}-2 \bar{k}\)
∴ \(\overline{\mathrm{a}} \times(\overline{\mathrm{b}} \times \overline{\mathrm{c}})=\left|\begin{array}{ccc}
\overline{\mathrm{i}} & \overline{\mathrm{j}} & \overline{\mathrm{k}} \\
2 & 3 & 4 \\
0 & -2 & -2
\end{array}\right|=\overline{\mathrm{i}}(\cdot 6+8)-\overline{\mathrm{j}}(-4-0)+\overline{\mathrm{k}}(-4-0)=2 \overline{\mathrm{i}}+4 \overline{\mathrm{j}}-4 \overline{\mathrm{k}}\)
Now \([(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}})] \cdot \overrightarrow{\mathrm{a}}=(2 \overline{\mathrm{i}}+4 \cdot \overline{\mathrm{j}}-4 \overline{\mathrm{k}}) \cdot(2 \overline{\mathrm{i}}+3 \overline{\mathrm{j}}+4 \overline{\mathrm{k}})\) = 4 + 12 – 16 = 0
∴ \((\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}})\) is perpendicular to \(\overline{\mathbf{a}}\)

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

Referring to the AP Inter 2nd Year Maths Study Material Chapter 11 Three Dimensional Geometry Exercise 11b Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Three Dimensional Geometry Solutions Exercise 11b

I.

Question 1.
Show that the three lines with direction are mutually perpendicular.
Solution:
Lines with dc’s l₁, m₁, n₁ and l₂, m₂, n₂ are perpendicular if l₁l₂ + m₁m₂ + n₁n₂ = 0
(i) l₁, m₁, n₁ = \(\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}\) and l₂, m₂, n₂ = \(\frac{4}{13}, \frac{12}{13}, \frac{3}{13}\)
∴ l₁l₂ + m₁m₂ + n₁n₂ = \(\frac{12}{13} \times \frac{4}{13}+\left(\frac{-3}{13}\right) \times \frac{12}{13}+\left(\frac{-4}{13}\right) \times \frac{3}{13}\)
= \(\frac{48}{169}-\frac{36}{169}-\frac{12}{169}\) = 0
Hence, the lines are perpendicular.

(ii) l₁, m₁, n₁ = \(\frac{4}{13}, \frac{12}{13}, \frac{3}{13}\) and l₂, m₂, n₂ = \(\frac{3}{13}, \frac{-4}{13}, \frac{12}{13}\)
∴ l₁l₂ + m₁m₂ + n₁n₂ = \(\frac{4}{13} \times \frac{3}{13}+\frac{12}{13} \times\left(\frac{-4}{13}\right)+\frac{3}{13} \times \frac{12}{13}\)
= \(\frac{12}{169}-\frac{48}{169}+\frac{36}{169}\) = 0
Hence, the lines are perpendicular.

(iii) l₁, m₁, n₁ = \(\frac{3}{13}, \frac{-4}{13}, \frac{12}{13}\) and l₂, m₂, n₂ = \(\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}\)
∴ l₁l₂ + m₁m₂ + n₁n₂ = \(\left(\frac{3}{13}\right) \times\left(\frac{12}{13}\right)+\left(\frac{-4}{13}\right) \times\left(\frac{-3}{13}\right)+\left(\frac{12}{13}\right) \times\left(\frac{-4}{13}\right)\)
= \(\frac{36}{169}+\frac{12}{169}-\frac{48}{169}\) = 0
Hence, the lines are perpendicular.
So, the all three lines are mutually perpendicular.

Question 2.
Show that the line through the points (1, -1, 2), (3, 4, -2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).
Solution:
Let AB be the line joining the points (1, -1, 2) and (3, 4, -2)and;
CD be the line through the points (0, 3, 2) and (3, 5, 6).
Hence, a₁ = 3 – 1 = 2, b₁ = 4 – (-1) = 5, c₁ = -2 – 2 = -4
a₂ = 3 – 0 = 3, b₂ = 5 – 3 = 2, c₂ = 6 – 2 = 4
If AB ⊥ CD then a₁a₂ + b₁b₂ + c₁c₂ = 0
⇒ (2)(3) + 5(2) + (-4)(4) = 6 + 10 – 16 = 16 – 16 = 0
Hence, AB and CD are perpendicular to each other.

Question 3.
If A(5, 6, 4), B(3, 5, 2) C(4, 3, x) are vertices of a triangle such that ∠ABC = \(\frac{\pi}{2}\). then
Solution:
Given A = (5, 6, 4), B = (3, 5, 2) C = (4, 3, x)
D.r’s of AB = (a₁, b₁, c₁) = (5-3, 6-5, 4-2) = (2, 1, 2)
D.r’s of BC = (a₂, b₂, c₂) = (4-3, 3-5, x-2) = (1, -2, x-2)
If AB ⊥ BC then a₁a₂ + b₁b₂ + c₁c₂ = 0
⇒ 2(1) + 1(-2) + 2(x-2) = 0
⇒ 2(1) – 2 + 2x – 4 = 0
⇒ 2x = 4
⇒ x = 2

Question 4.
Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (- 1, – 2, 1), (1, 2, 5).
Solution:
Let AB be the line joining the points (4, 7, 8) and (2,3,4) ;
and CD be the line through the points (- 1, -2, 1) and (1, 2, 5).
Hence, a₁, = 2 – 4 = -2, b₁ = 3 – 7 = -4, c₁ = 4 – 8 = -4
a₂ = 1 -(-1) = 2, b₂ = 2-(-2) = 4, c₂ = 5 – 1 = 4
If AB || CD then \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
Here, \(\frac{a_1}{a_2}=\frac{-2}{2}\) = -1; \(\frac{b_1}{b_2}=\frac{-4}{4}\) = -1; \(\frac{c_1}{c_2}=\frac{-4}{4}\) = -1
Hence, AB is parallel to CD.

Question 5.
If the line through the poInts (1, 3, 4), (3, 1, 6) is perpendicular to the line through the points (0, -1, 3), (2, λ, -1), then find A.
Solution:
Let As (1, 3, 4), B = (3, 1, 6), C(0, -1, 3), and D (2, λ, -1)
D.r’s of AB = (a₁, b₁, c₁) = (3-1, 1-3, 6-4) = (2, -2, 2)
D.r’s of CD = (a₂, b₂, c₂) = (2-0, λ+1, -1-3) = (2, λ+1, -4)
If AB ⊥ CD then a₁a₂ + b₁b₂ + c₁c₂ = 0
⇒ 2(2) + (-2)(λ+1) + 2(-4) = 4 – 2λ – 2 – 8 = 0
⇒ -2λ = 6
⇒ λ = -3

Question 6.
Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector \(3 \hat{i}+2 \hat{j}-2 \hat{k}\).
Solution:
Given that the line passes through the point A(1, 2, 3).
∴ the position vector through A( 1, 2, 3) is \(\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}\), \(\vec{b}=3 \hat{i}+2 \hat{j}-2 \hat{k} .\)
So, line passing through point A(1, 2, 3) and parallel to b is given by \(\vec{r}=\vec{a}+\lambda \vec{b}\), λ is a real
Hence required equation of the line is \(\vec{r}=\hat{i}+2 \hat{j}+3 \hat{k}+\lambda(3 \hat{i}+2 \hat{j}-2 \hat{k})\)
The Cartesian form is with (xi,yj,zj) = (1,2,3) and the direction ratios (a, b, c) = (3, 2, -2) is
\(\frac{\mathrm{x}-\mathrm{x}_1}{\mathrm{a}}=\frac{\mathrm{y}-\mathrm{y}_1}{\mathrm{~b}}=\frac{\mathrm{z}-\mathrm{z}_1}{\mathrm{c}}\)
⇒ \(\frac{x-1}{3}=\frac{y-2}{2}=\frac{z-3}{-2}\)

Question 7.
Find the equation of a line parallel to x-axis and passing through the origin.
Solution:
The line parallel to x-axis and passing through the origin O(0,0,0)is x-axis itself.
Let A be a point on x-axis.
∴ the coordinates of A are given by (a, 0, 0), where a∈R
Hence, the direction ratios of OA are (a, 0, 0)
The equation of OA is given by ⇒ \(\frac{x-0}{a}=\frac{y-0}{0}=\frac{z-0}{0}\) ⇒ \(\frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{0}=\frac{\mathrm{z}}{0}\) = a
Hence, the equation of line parallel to x-axis and passing origin is y = 0, z = 0

Question 8.
Find the equation of the line in vector and in cartesian form that passes through the point with position vector \(2 \hat{i}-\hat{j}+4 \hat{k}\) and is in the direction \(\hat{i}+2 \hat{j}-\hat{k}\)
Solution:
Given that \(\vec{a}=2 \hat{i}-\hat{j}+4 \hat{k}\), \(\vec{b}=\hat{i}+2 \hat{j}-\hat{k}\)
The vector equation of the line is given by \(\vec{r}=\vec{a}+\lambda \vec{b}\), where λ is some real number
Hence, \(\vec{r}=2 \hat{i}-\hat{j}+4 \hat{k}+\lambda(\hat{i}+2 \hat{j}-\hat{k})\)
The cartesian form is with (x₁, y₁, z₁) = (2, -1, 4) and the direction ratios (a, b, c) = (1, 2, -1) is
\(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\)
⇒ \(\frac{x-2}{1}=\frac{y+1}{2}=\frac{z-4}{-1}\)

Question 9.
Find the cartesian equation of the line which passes through the point (- 2, 4, – 5) and parallel to the line given by \(\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}\)
Solution:
It is given that the required line passes through the point (- 2,4, – 5) and is parallel to \(\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}\)
∴ its direction ratios are 3k, 5k and 6k, where k ≠ 0
It is known that the equation of the line through the point (x₁, y₁, z₁)and with direction ratios (a, b, c) is given by
\(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\)
Hence, the equation of the required line is \(\frac{x+2}{3 k}=\frac{y-4}{5 k}=\frac{z+5}{6 k}\) ⇒ \(\frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}\) = k
Thus, the cartesian equation of the line is \(\frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}\)

Question 10.
The cartesian equation of a line is \(\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}\). Write its vector form.
Solution:
It is given that the Cartesian equation of the line is \(\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}\)
Hence, the given line passes through the point (5, -4, 6)
∴ the position vector of the point is \(\vec{a}=5 \hat{i}-4 \hat{j}+6 \hat{k}\)
Also, the direction ratios of the given line are (3,7,2).
This means that the line is in the direction of the vector, \(\vec{b}=3 \hat{i}+7 \hat{j}+2 \hat{k}\)
As we known that the line through positive vector and in the direction of the vector \(\vec{b}\) is
given by the equation, \(\vec{r}=a+\lambda \vec{b}\), λ ∈ R
∴ \(\vec{r}=(5 \hat{i}-4 \hat{j}+6 \hat{k})+\lambda(3 \hat{i}+7 \hat{j}+2 \hat{k})\) is the required equation of the given line in vector form.

Question 11.
Find the angle between the lines whose direction ratios are a, b, c and b – c, c – a„ a – b.
Solution:
The angle θ between the lines with direction ratios a,b,c and (b – c), (c – a), (a – b) is given by,
cos θ = \(\left|\frac{a(b-c)+b(c-a)+c(a-b)}{\sqrt{a^2+b^2+c^2} \sqrt{(b-c)^2+(c-a)^2+(a-b)^2}}\right|\)
θ = cos^-1\(\frac{ab-ac+bc-ab+ac-bc}{\sqrt{a^2+b^2+c^2} \sqrt{(b-c)^2+(c-a)^2+(a-b)^2}}\) = cos^-1 0 = 90°
∴ the required angle is 90°

Question 12.
Find the angle between the following pairs of lines:
\(\vec{r}=2 \hat{i}-5 \hat{j}+\hat{k}+\lambda(3 \hat{i}+2 \hat{j}+6 \hat{k})\) and \(\vec{r}=7 \hat{i}-6 \hat{k}+\mu(\hat{i}+2 \hat{j}+2 \hat{k})\)
Solution:
Angle between the given pairs of lines is given by cos θ = \(\left|\frac{\overline{b_1} \cdot \overline{b_2}}{\left|\overline{b_1}\right|\left|\overline{b_2}\right|}\right|\)
The given lines are parallel to the vectors, \(\overrightarrow{b_1}=3 \hat{i}+2 \hat{j}+6 \hat{k}\) and \(\overrightarrow{b_2}=\hat{i}+2 \hat{j}+2 \hat{k}\)
\(\left|\vec{b}_1\right|=\sqrt{3^2+2^2+6^2}=\sqrt{49}\) = 7; \(\left|\overrightarrow{b_2}\right|=\sqrt{1^2+2^2+2^2}=\sqrt{9}\) = 3
\(\vec{b}_1 \cdot \vec{b}_2\) = (3i + 2j + 6k) . (i + 2j + 2k) = 3(1) + 2(2) + 6(2) = 3 + 4 + 12 = 19
∴ cos θ = \(\left|\frac{\overrightarrow{b_1} \cdot \overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right|\left|\overrightarrow{b_2}\right|}\right|=\left|\frac{19}{7 \times 3}\right|=\frac{19}{21}\)
⇒ θ = cos^-1\(\left(\frac{19}{21}\right)\)

Question 13.
Find the angle between the following pairs of lines:
\(\vec{\mathrm{r}}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}+\lambda(\hat{\mathrm{i}}-\hat{\mathrm{j}}-2 \hat{\mathrm{k}})\) and \(\vec{r}=2 \hat{i}-\hat{j}-56 \hat{k}+\mu(3 \hat{i}-5 \hat{j}-4 \hat{k})\)
Solution:

Question 14.
Find the angle between the following pair of lines:
\(\frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3}\) and \(\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}\)
Solution:

Question 15.
Find the angle between* the following pair of lines:
\(\frac{x}{2}=\frac{y}{2}=\frac{z}{1}\) and \(\frac{x-5}{4}=\frac{y-2}{1}=\frac{z-3}{8}\)
Solution:
Let \(\vec{\mathrm{b}_1}\) and \(\vec{\mathrm{b}_2}\) be the vectors parallel to the pair of lines
\(\frac{x}{2}=\frac{y}{2}=\frac{z}{1}\) and \(\frac{x-5}{4}=\frac{y-2}{1}=\frac{z-3}{8}\) respectively

Question 16.
Show that the lines \(\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}\) and \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) are perpendicular to each other.
Solution:
The equations of the given lines are \(\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}\) and \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\)
Here, a₁ = 7, b₁ = -5, c₁ = 1; a₂ = 1, b₂ = 2, c₂ = 3
Two lines with direction ratios, a₁, b₁, c₁ and a₂, b₂, c₂ are perpendicular to each other,
if a₁a₂ + b₁b₂+ c₁c₂ = 0.
Here, 7(1) + (-5)2 + 1(3) = 7 – 10 + 3 = 0
∴ the given lines are perpendicular to each other.

Question 17.
Show that the pair of lines \(\frac{4-x}{3}=\frac{2 y+1}{2}=\frac{z-1}{5}\) and \(\frac{2 x+1}{2}=\frac{3-y}{2}=\frac{3 z-1}{3}\) are perpendicular to each other.
Solution:
The equations of the given lines are \(\frac{4-x}{3}=\frac{2 y+1}{2}=\frac{z-1}{5}\) ⇒ \(\frac{x-4}{-3}=\frac{y+1 / 2}{1}=\frac{z-1}{5}\)
\(\frac{2 x+1}{2}=\frac{3-y}{2}=\frac{3 z-1}{3}\) ⇒ \(\frac{x+1 / 2}{1}=\frac{y-3}{-2}=\frac{z-1 / 3}{1}\)
Here, a₁ = -3, b₁ = 1, c₁ = 5
a₂ = 1, b₂ = -2, c₂ = 1
Two lines with direction ratios, a₁, b₁, c₁ and a₂, b₂, c₂ are perpendicular to each other,
if a₁a₂ + b₁b₂ + c₁c₂ = 0.
Here, (-3)1 + 1(-2) + 5(1) = -3 – 2 + 5 = 0
∴ the given lines are perpendicular to each other.

Question 18.
If the lines \(\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}\) and \(\frac{x-1}{3 k}=\frac{y-1}{1}=\frac{z-6}{-5}\) are perpendicular, find the value of k.
Solution:
From the given equations, we have a₁ = -3, b₁ = 2k, c₁ = 2; a₂ = 3k, b₂ = 1, c₂ = -5
Two lines with d.r’s a₁, b₁, c₁ and a₂, b₂, c₂ are perpendicular, if a₁a₂ + b₁b₂ + c₁c₂ = 0
⇒ -3(3k) + 2k(1) + 2(-5) = 0
⇒ -9k + 2k – 10 = 0
⇒ 7k = -10
⇒ k = \(\frac{-10}{7}\)

Question 19.
Find the values of p so that the lines
\(\frac{1-x}{3}=\frac{7 y-14}{2 p}=\frac{z-3}{2}\) and \(\frac{7-7 x}{3 p}=\frac{y-5}{1}=\frac{6-z}{5}\) are at right angles.
Solution:
Equation of the lines in the standard form are
\(\frac{1-x}{3}=\frac{7 y-14}{2 p}=\frac{z-3}{2}\)
⇒ \(\frac{x-1}{-3}=\frac{y-2}{\frac{2 p}{7}}=\frac{z-3}{2}\) and
\(\frac{7-7 x}{3 p}=\frac{y-5}{1}=\frac{6-z}{5}\)
⇒ \(\frac{x-1}{\frac{-3 p}{7}}=\frac{y-5}{1}=\frac{z-6}{-5}\)
The direction ratios of the lines are given by a₁ = -3, b₁ = \(\frac{2 p}{7}\), c₁ = 2; a₂ = \(\frac{-3 p}{7}\), b₂ = 1, c₂ = -5
Since, both the lines are perpendicular to each other, we have
a₁a₂ + b₁b₂ + c₁c₂ = 0
⇒ (-3) \(\left(\frac{-3 p}{7}\right)\) + \(\left(\frac{2 p}{7}\right)\)1 + 2(-5) = 0
⇒ \(\frac{9 p}{7}\) + \(\frac{2 p}{7}\) – 10 = 0
⇒ \(\frac{11}{7}\)p = 10
⇒ 11p = 10 × 7
⇒ p = \(\frac{70}{11}\)

Question 20.
Find the values of p so that the lines \(\frac{x-5}{5 p+2}=\frac{2-y}{5}=\frac{z-1}{1}\) and \(\frac{x}{1}=\frac{2 y+1}{4 p}=\frac{z-1}{3}\) arc at right angles.
Solution:
The given lines are \(\frac{x-5}{5 p+2}=\frac{2-y}{5}=\frac{z-1}{1}\) and \(\frac{x}{1}=\frac{2 y+1}{4 p}=\frac{z-1}{3}\)
The direction ratios of the lines are given by
a₁ = 5P + 2, b₁ = -5, c₁ = 1; a₂ = 1, b₂ = 2p, c₂ = 3
Since, both the lines are perpendicular to each other, we have a₁a₂ + b₁b₂ + c₁c₂ = 0
⇒ (5p + 2)(1)+(-5)2p + 1(3) = 0
⇒ -5p + 5 = 0
⇒ 5p = 5
⇒ p = 1

II.

Question 1.
Find the distance between the pair of parallel lines
\(\vec{r}=(\hat{i}+2 \hat{j}-4 \hat{k})+\lambda(4 \hat{i}+3 \hat{j}-12 \hat{k})\) and \(\vec{r}=(3 \hat{i}+3 \hat{j}-5 \hat{k})+\mu(4 \hat{i}+3 \hat{j}-12 \hat{k})\)
Solution:

Question 2.
Find the distance between the pair of parallel lines
\(\vec{r}=(\hat{i}+7 \hat{j}+4 \hat{k})+\lambda(3 \hat{i}+2 \hat{j}+5 \hat{k})\) and \(\vec{r}=(3 \hat{i}+6 \hat{j}+4 \hat{k})+\mu(3 \hat{i}+2 \hat{j}+5 \hat{k})\)
Solution:

Question 3.
Find the distance between the pair of parallel lines.
\(\frac{x+3}{-2}=\frac{y-5}{3}=\frac{z+2}{-1}\) and \(\frac{x+2}{2}=\frac{y+2}{-3}=\frac{z-3}{1}\)
Solution:

Question 4.
Find the distance between the pair of parallel lines.
\(\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{-3}\) and \(\frac{x+2}{-1}=\frac{y+1}{-2}=\frac{z+2}{3}\)
Solution:

Question 5.
If the lines whose direction ratios are (1, 2, z), (1, y, -1) and (x, -4, 1) are mutually perpendicular,then find the value of x+y+z .
Solution:
Let Dr’s of given lines are (a₁, b₁, c₁) = (1, 2, z) …………. (1)
(a₂, b₂, c₂) = (1, y, -1) ……………. (2)
(a₃, b₃, c₃) = (x, -4, 1) …………… (3)
Lines with Dys (a₁, b₁, c₁), (a₂, b₂, c₂) are mutually perpendicular if a₁a₂ + b₁b₂ + c₁c₂ = 0
Here, (1)1 + 2(y) + z(-1) = 0 ⇒ 1+ 2y – z = 0 ……………… (4)
a₂a₃ + b₂b₃ + c₂c₃ = 0 ⇒ x – 4y – 1 = 0 ……………… (5)
a₃a₁ + b₃b₁ + c₃c₁ = 0 ⇒ x – 8+ z = 0 …………… (6)
By solving (4), (5) & (6) we get the values of x, y, z
(5) – (6)
-4y – 1 + 8 – z = 0 ⇒ -4y – z+ 7 = 0 ………. (7)
(4) × 2 ⇒ 4y – 2z + 2 = 0
Adding we get -3z + 9 = 0
⇒ -3z = -9
⇒ z = 3
From (6), we have x – 8 + z = 0
⇒ x – 8 + 3 = 0 ⇒ x = 5
From (4), we have 2y = z – 1 = 0
⇒ 2y = 3 – 1 ⇒ 2y = 2
⇒ y = 1
∴ x + y + z = 5 + 1 + 3 = 9

III.

Question 1.
Find the shortest distance between the lines
\(\vec{r}=(\hat{i}+2 \hat{j}+\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})\) and \(\vec{r}=2 \hat{i}-\hat{j}-\hat{k}+\mu(2 \hat{i}+\hat{j}+2 \hat{k})\)
Solution:

Question 2.
Find the shortest distance between the lines whose vector equations are
\(\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-3 \hat{j}+2 \hat{k})\) and \(\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-3 \hat{j}+2 \hat{k})\)
Solution:

Question 3.
Find the shortest distance between the lines whose equations are
\(\vec{r}=(1-t) \hat{i}+(t-2) \hat{j}+(3-2 t) \hat{k}\) and \(\vec{r}=(s+1) \hat{i}+(2 s-1) \hat{j}-(2 s+1) \hat{k}\)
Solution:

Question 4.
Find the shortest distance between lines \(\vec{r}=6 \hat{i}+2 \hat{j}+2 \hat{k}+\lambda(\hat{i}-2 \hat{j}+2 \hat{k})\) and \(\vec{r}=-4 \hat{i}-\hat{k}+\mu(3 \hat{i}-2 \hat{j}-2 \hat{k})\)
Solution:

Question 5.
Find the shortest distance between the lines
\(\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}\) and \(\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}\)
Solution:
The shortest distance between the two lines,

Question 6.
Find the vector equation of the line passing through the point (1, 2, – 4) and perpendicular to the two lines: \(\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}\) and \(\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}\)
Solution:
Let \(\vec{a}=\hat{i}+2 \hat{j}-4 \hat{k}\), \(\vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\)
The equation of the line passing through (1, 2, -4) and parallel to vector \(\vec{b}\) is given by
⇒ \(\vec{\mathrm{r}}=\vec{\mathrm{a}}+\lambda \vec{\mathrm{b}}\) ⇒ \(\vec{r}=(\hat{i}+2 \hat{j}-4 \hat{k})+\lambda\left(b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\right)\) ………….. (1)
The equations of the given lines are
\(\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}\) …………… (2) and
\(\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}\) ………….. (3)
Since, lines of the equations (1) and (2) are perpendicular to each other we have
3b₁ – 16b₂ + 7b₃ = 0 ………….(4)
Also, Lines (1) and (3) are perpendicular to each other ⇒ 3b₁ + 8b₂ – 5b₃ = 0 …………… (5)
Solving equations (4) and (5), we have
\(\frac{b_1}{(-16) \times(-5)-8 \times 7}=\frac{b_2}{7 \times 3-3 \times(-5)}=\frac{b_3}{3 \times 8-3 \times(-16)}\)
⇒\(\frac{b_1}{24}=\frac{b_2}{36}=\frac{b_3}{72}\)
⇒ \(\frac{b_1}{2}=\frac{b_2}{3}=\frac{b_3}{6}\)
Hence, the direction ratios of \(\) are and 2, 3, 6
∴ \(\vec{b}=2 \hat{i}+3 \hat{j}+6 \hat{k}\)
Putting \(\vec{b}=2 \hat{i}+3 \hat{j}+6 \hat{k}\) in equation (1), we get \(\vec{r}=(\hat{i}+2 \hat{j}-4 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})\)
∴ the required vector equation of the line is \(\vec{r}=(\hat{i}+2 \hat{j}-4 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})\)

Referring to the AP Inter 2nd Year Maths Study Material Chapter 10 Vector Algebra Exercise 10c Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Vector Algebra Solutions Exercise 10c

I.

Question 1.
Find the angle between two vectors a and b with magnitude \(\sqrt{3}\) and 2 respectively having \(\vec{a} \cdot \vec{b}=\sqrt{6}\)
Solution:
Given that \(|\vec{a}|=\sqrt{3} ;|\vec{b}|=2\) and \(\vec{a} \cdot \vec{b}=\sqrt{6}\) Let θ be the ange between the vectors \(\overrightarrow{\mathrm{a}} \text { and } \overrightarrow{\mathrm{b}}\).
We have cos θ = \(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \Rightarrow \cos \theta=\frac{\sqrt{6}}{\sqrt{3}(2)}=\frac{\sqrt{6}}{\sqrt{3} \sqrt{4}}=\frac{\sqrt{6}}{\sqrt{12}}=\sqrt{\frac{6}{12}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}=\cos \frac{\pi}{4}\) ∴ θ = \(\frac{\pi}{4}\)

Question 2.
Find the angle between the vectors \(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \text { and } 3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}\).
Solution:
Given that \(\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k} \text { and } \vec{b}=3 \hat{i}-2 \hat{j}+\hat{k}\).
⇒ \(|\overrightarrow{\mathrm{a}}|=\sqrt{1+4+9}=\sqrt{14} \text { and }|\overrightarrow{\mathrm{b}}|=\sqrt{9+4+1}=\sqrt{14}\)
Also \(\vec{a} \cdot \vec{b}=(\hat{i}-2 \hat{j}+3 \hat{k}) \cdot(3 \hat{i}-2 \hat{j}+\hat{k})\) = 1(3) + (-2)(-2) + 3(1) = 3 + 4 + 3 = 10
Let θ be the angle between the vectors \(\overrightarrow{\mathrm{a}} \text { and } \overrightarrow{\mathrm{b}}\).
∴ cos θ = \(\frac{\vec{a} \cdot \vec{b}}{|\vec{a} \| \vec{b}|}=\frac{10}{\sqrt{14} \sqrt{14}}=\frac{10}{14}=\frac{5}{7} \Rightarrow \theta=\cos ^{-1}\left(\frac{5}{7}\right)\)

Question 3.
Find the projection of the vector \(\hat{\mathbf{i}}-\hat{\mathbf{j}}\) on the vector \(\hat{\mathbf{i}}+\hat{\mathbf{j}}\)
Solution:
Let \(\vec{a}=\hat{i}-\hat{j}=\hat{i}-\hat{j}+0 \hat{k} \text { and } \vec{b}=\hat{i}+\hat{j}=\hat{i}+\hat{j}+0 \hat{k}\)
Projection of vector \(\vec{a} \text { on } \vec{b}=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}=\frac{(1)(1)+(-1)(1)+0(0)}{\sqrt{(1)^2+(1)^2+0^2}}=\frac{1-1+0}{\sqrt{2}}=\frac{0}{\sqrt{2}}=0\)

Question 4.
Find the projection of the vector \(\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}\) on the vector \(7 \hat{\mathbf{i}}-\hat{\mathbf{j}}+8 \hat{\mathbf{k}}\)
Solution:
Let \(\vec{a}=i+3 \hat{j}+7 \hat{k} \text { and } \vec{b}=7 \hat{i}-\hat{j}+8 \hat{k}\)
Projection of vector \(\vec{a} \text { on } \vec{b}=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}=\frac{1(7)+3(-1)+7(8)}{\sqrt{7^2+(-1)^2+8^2}}=\frac{7-3+56}{\sqrt{49+1+64}}=\frac{60}{\sqrt{114}}\)

Question 5.
Evaluate the product \((3 \vec{a}-5 \vec{b}) \cdot(2 \vec{a}+7 \vec{b})\)
Solution:
The given dot product is \((3 \vec{a}-5 \vec{b}) \cdot(2 \vec{a}+7 \vec{b})\)
= \((3 \vec{a}) \cdot(2 \vec{a})+(3 \vec{a}) \cdot(7 \vec{b})-(5 \vec{b})(2 \vec{a})-(5 \vec{b})(7 \vec{b})=6 \vec{a} \cdot \vec{a}+21 \vec{a} \cdot \vec{b}-10 \vec{b} \cdot \vec{a}-35 \vec{b} \cdot \vec{b}\)
= \(6|\vec{a}|^2+21 \vec{a} \cdot \vec{b}-10 \vec{a} \cdot \vec{b}-35|\vec{b}|^2\) [∵ \(\vec{a} \cdot \vec{a}=|\vec{a}|^2 \text { and } \vec{b} \cdot \vec{b}=|\vec{b}|^2 \text { and } \vec{b} \cdot \vec{a}=\vec{a} \cdot \vec{b}\)]
= \(6|\vec{a}|^2+11 \vec{a} \cdot \vec{b}-35|\vec{b}|^2\)

Question 6.
Find \(|\overrightarrow{\mathbf{x}}|\), if for a unit vector a,\((\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})=12\)
Solution:
Given that \(\vec{a}\) is a unit vector ⇒ \(|\vec{a}|=1\) …..(i)
Also given that \((\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})=12 \Rightarrow \vec{x} \cdot \vec{x}+\vec{x} \cdot \vec{a}-\vec{a} \cdot \vec{x}-\vec{a} \cdot \vec{a}=12 \Rightarrow|\vec{x}|^2-|\vec{a}|^2=12\)
Putting \(|\vec{a}|=1 \text { from (i), }|\vec{x}|^2-1=12 \Rightarrow|\vec{x}|^2=13 \Rightarrow|\vec{x}|=\sqrt{13} \text {. }\)

Question 7.
If \(\vec{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}=-\hat{i}+2 \hat{j}+\hat{k} \text { and } \vec{c}=3 \hat{i}+\hat{j}\) are such that \(\vec{a}+\lambda \vec{b}\) is perpendicular to \(\overrightarrow{\mathrm{c}}\), then find the value of λ.
Solution:
Given that \(\vec{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}=-\hat{i}+2 \hat{j}+\hat{k} \text { and } \vec{c}=3 \hat{i}+\hat{j}\)
Now, \(\vec{a}+\lambda \vec{b}=2 \hat{i}+2 \hat{j}+3 \hat{k}+\lambda(-\hat{i}+2 \hat{j}+\hat{k})=2 \hat{i}+2 \hat{j}+3 \hat{k}-\lambda \hat{i}+2 \lambda \hat{j}+\lambda \hat{k}\)
⇒ \(\vec{a}+\lambda \vec{b}=(2-\lambda) \hat{i}+(2+2 \lambda) \hat{j}+(3+\lambda) \hat{k}\)
Also we have \(\vec{c}=3 \hat{i}+\hat{j}=3 \hat{i}+\hat{j}+0 \hat{k}\)
Given that \(\vec{a}+\lambda \vec{b}\) is perpendicular to \(\vec{c} \Rightarrow(\vec{a}+\lambda \vec{b}) \cdot \vec{c}=0\)
⇒ (2 – λ)3 + (2 + 2λ)1 + (3 + λ)0 = 0
⇒ 6 – 3λ + 2 + 2λ = 0 ⇒ -λ + 8 = 0 ⇒ -λ = -8 ⇒ λ = 8

Question 8.
Show that \(|\vec{a}| \vec{b}+|\vec{b}| \vec{a}\) is perpendicular to \(|\vec{a}|\vec{b}-|\vec{b}| \vec{a}/latex], for any two nonzero vectors [latex]\vec{a} \text { and } \vec{b}\)
Solution:
The dot product of the given vectors is \((|\vec{a}|\vec{b}+|\vec{b}| \vec{a}) \cdot(|\vec{a}| \vec{b}-|\vec{b}| \vec{a})\)
= \(|\vec{a}|^2 \overrightarrow{b . \mathrm{b}}-|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{a}}+|\overrightarrow{\mathrm{b}}||\overrightarrow{\mathrm{a}}| \overrightarrow{\mathrm{a} .} \overrightarrow{\mathrm{b}}-|\overrightarrow{\mathrm{b}}|^2 \overrightarrow{\mathrm{a} . \mathrm{a}}=|\overrightarrow{\mathrm{a}}|^2|\overrightarrow{\mathrm{~b}}|^2-|\overrightarrow{\mathrm{b}}|^2|\overrightarrow{\mathrm{a}}|^2=0\)
∴ The given two vectors are perpendicular.

Question 9.
If \(\overrightarrow{\mathbf{a}} . \overrightarrow{\mathbf{a}}=0\) and \(\overrightarrow{\mathbf{a}} . \overrightarrow{\mathbf{b}}=0\), then what can be concluded about the vector \(\overrightarrow{\mathbf{b}}\)?
Solution:
We have \(\vec{a} \cdot \vec{a}=0 \text { and } \vec{a} \cdot \vec{b}=0\)
Hence \(|\overrightarrow{\mathbf{a}}|^2=0 \Rightarrow|\overrightarrow{\mathbf{a}}|=0\)
∴ \(\overrightarrow{\mathbf{a}}\) is the zero vector
Thus, any vector \(\overrightarrow{\mathbf{b}}\) can satisfy \(\vec{a} . \vec{b}=0\)

Question 10.
If \(\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{c}}\) are unit vectors such that \(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}=\overrightarrow{\mathbf{0}}\), find the value of \(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}\)
Solution:
Consider \(|\vec{a}+\vec{b}+\vec{c}|^2=(\vec{a}+\vec{b}+\vec{c}) \cdot(\vec{a}+\vec{b}+\vec{c})\)
⇒ \(0=|\overrightarrow{\mathrm{a}}|^2+|\overrightarrow{\mathrm{b}}|^2+|\overrightarrow{\mathrm{c}}|^2+2(\overrightarrow{\mathrm{a}} \cdot \dot{\overrightarrow{\mathrm{~b}}}+\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}})\)
⇒ 0 = 1 + 1 + 1 + \(2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=-3\)
⇒ \((\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=\frac{-3}{2}\)

Question 11.
If either vector \(\vec{a}=\overrightarrow{0} \text { or } \vec{b}=\overrightarrow{0}\) then \(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}=0\). But the converse need not be true. Justify your answer with an example.
Solution:
Let \(\vec{a}=2 \hat{i}+4 \hat{j}+3 \hat{k} \text { and } \vec{b}=3 \hat{i}+3 \hat{j}-6 \hat{k}\)
∴ \(\vec{a} . \vec{b}\) = 2(3) + 4(3) + 3(-6) = 6 + 12 – 18 = 0
Here \(\vec{a} , \vec{b}\) are two non-zero perpendicular vectors
So, the converse of the statement need not to be true.

II.

Question 1.
Show that each of the given three vectors \(\frac{1}{7}(2 \hat{i}+3 \hat{j}+6 \hat{k}), \frac{1}{7}(3 \hat{i}-6 \hat{j}+2 \hat{k}), \frac{1}{7}(6 \hat{i}+2 \hat{j}-3 \hat{k})\) is a unit vector: Also, show that they are mutually perpendicular to each other.
Solution:
Let \(\frac{1}{7}(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})=\frac{2}{7} \hat{\mathrm{i}}+\frac{3}{7} \hat{\mathrm{j}}+\frac{6}{7} \hat{\mathrm{k}}\) ……(i)

So, the 3 vectors are mutually perpendicular to each other.

Question 2.
Find \(|\vec{a}| \text { and }|\vec{b}| \text { if }(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=8 \text { and }|\vec{a}|=8|\vec{b}|\)
Solution:

Question 3.
Find the magnitude of two vectors \(\vec{a} \text { and } \vec{b}\), having the same magnitude and such that the angle between them is 60° and their scalar product is 1/2.
Solution:
Given that \(\vec{a} \text { = } \vec{b}\); angle θ between \(\vec{a} \text { and } \vec{b}\) is 60°, their scalar product =1/2
Thus \(\vec{a} \cdot \vec{b}=\frac{1}{2} \Rightarrow|\vec{a} \| \vec{b}| \cos \theta=\frac{1}{2}\) [∵ \(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{~b}}=|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \cos \theta\)]
Putting \(|\vec{b}|=|\vec{a}|\)and θ = 60° (in the above), we have \(|\vec{a}| .|\vec{a}| \cos 60^{\circ}=\frac{1}{2} \Rightarrow|\vec{a}|^2\left(\frac{1}{2}\right)=\frac{1}{2}\)
⇒ \(|\left.\vec{a}\right|^2=1 \Rightarrow|\vec{a}|=1\) ……..(i) (∵ Length of a vector is never negative)
∴ \(|\vec{b}|=|\vec{a}|=1\) [By (i)]
∴ \(|\vec{a}|=1 \text { and }|\vec{b}|=1\)

Question 4.
If the vertices A, B, C of a triangle ABC are (1, 2, 3), (-1, 0, 0), (0, 1, 2), respectively, then find ∠ABC. [∠ABC is the angle between the vectors \(\overrightarrow{B A} \text { and } \overrightarrow{B C}\)].
Solution:
Given vertices A, B, C of a triangle ABC are (1, 2, 3), (-1, 0, 0), (0, 1, 2) respectively.
Position vector (P. V) of point A (1, 2, 3) is \(\overrightarrow{O A}=\hat{i}+2 \hat{j}+3 \hat{k}\)
Position vector (P. V) of point B(-1, 0, 0) is \(\overrightarrow{O B}=-\hat{i}+0 \hat{j}+0 \hat{k}\)
Position vector (P.V) of point C(0, 1, 2) is \(\overrightarrow{\mathrm{OC}}=0 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}\)
Now \(\overrightarrow{B A}=\overrightarrow{O A}-\overrightarrow{O B}=(\hat{i}+2 \hat{j}+3 \hat{k})-(-\hat{i}+0 \hat{j}+0 \hat{k})=\hat{i}+2 \hat{j}+3 \hat{k}+\hat{i}-0 \hat{j}-0 \hat{k}=2 \hat{i}+2 \hat{j}+3 \hat{k}\) ….(i)
and \(\overrightarrow{B C}=\overrightarrow{O C}-\overrightarrow{O B}=0 \hat{i}+\hat{j}+2 \hat{k}-(-\hat{i}+0 \hat{j}+0 \hat{k})=0 \hat{i}+\hat{j}+2 \hat{k}+\hat{i}-0 \hat{j}-0 \hat{k}=\hat{i}+\hat{j}+2 \hat{k}\) …………….(ii)
Using (i) and (ii) cos ∠ABC = \(\frac{\overrightarrow{\mathrm{BA}} \cdot \overrightarrow{\mathrm{BC}}}{|\overrightarrow{\mathrm{BA}}||\overrightarrow{\mathrm{BC}}|}=\frac{2(1)+2(1)+3(2)}{\sqrt{4+4+9} \sqrt{1+1+4}}=\frac{10}{\sqrt{17} \sqrt{6}}=\frac{10}{\sqrt{102}}\) [∵ \(\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\)]
∴ ∠ABC = \(\cos ^{-1}\left(\frac{10}{\sqrt{102}}\right)\)

Question 5.
Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, -1) are coliinear.
Solution:
Given points are A(1, 2, 7), B(2, 6, 3) and C(3, 10, -1)

Question 6.
Show that the vectors \(2 \bar{i}-\bar{j}+\bar{k}, \bar{i}-3 \bar{j}-5 \bar{k}, \text { and } 3 \bar{i}-4 \bar{j}-4 \bar{k}\) form the vertices of a right angled triangle.
Solution:

∴ Given vectors form a right angled triangle.

Referring to the AP Inter 2nd Year Maths Study Material Chapter 11 Three Dimensional Geometry Exercise 11a Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Three Dimensional Geometry Solutions Exercise 11a

I.

Question 1.
If a line makes angles 90°, 135°, 45° with the .v, y and z-axes respectively, find its direction cosines.
Solution:
Let l, m, n be the direction cosines of the given line.
Then l = cos90° =0
m = cos 135° = cos(90° +45°) = -sin 45° = –\(\frac{1}{\sqrt{2}}\)
n = cos 45° = \(\frac{1}{\sqrt{2}}\)
∴ the direction cosines of the line are 0, –\(\frac{1}{\sqrt{2}}\) and \(\frac{1}{\sqrt{2}}\)

Question 2.
If a line makes an angles of 60″ and 45″ with the direction of positive X and Z aixs respectively. Find the angle made by the line with the Y-aixs.
Solution:
Let l, m, n be the direction cosines of the given line.
Let β be the angle made by line with y-axis.
Then l = cos 60° = \(\frac{1}{2}\), m = cosβ, n = cos45°= \(\frac{1}{\sqrt{2}}\)
We know that l² + m² + n² = 1
⇒ \(\frac{1}{4}\) + m² + \(\frac{1}{2}\) = 1
⇒ m² + \(\frac{1+2}{4}\) = 1
⇒ m² = 1 – \(\frac{3}{4}\) = \(\frac{1}{4}\)
⇒ cos² β = \(\frac{1}{4}\)
⇒ cos β = ±\(\frac{1}{2}\)
∴ β = 60° (or) 120°

Question 3.
Find the direction cosines of a line which makes equal angles with the coordinate axes.
Solution:
Let the line makes an angle a with each of the coordinates axes.
Then l = cosα, m = cosα, n = cosα
We know that l² + m² + n² = 1
⇒ cos²α + cos²α + cos²α = 1
⇒ cos²α = \(\frac{1}{3}\)
⇒ cos α = ±\(\frac{1}{\sqrt{3}}\)
Thus, the direction cosines of the line are ±\(\frac{1}{\sqrt{3}}\), ±\(\frac{1}{\sqrt{3}}\) and ±\(\frac{1}{\sqrt{3}}\).

Question 4.
If a line makes angle-75 and 60° with the positive directions of Y and Z axis respectively and the angle made by the line with thc positive direction of X-axis is a.. then find the value cos²α ?
Solution:
Let l, m, n be the direction cosines of the given line.
Then l = cosα, m = cosβ = cos 75° = \(\frac{\sqrt{3}-1}{2 \sqrt{2}}\), n =cosy = cos60°= \(\frac{1}{2}\)
We know that l² + m² + n² = 1
⇒ cos²α + \(\frac{(\sqrt{3}-1)^2}{8}\) + \(\frac{1}{4}\) = 1
⇒ cos²α = 1 – \(\left(\frac{(\sqrt{3}-1)^2}{8}+\frac{1}{4}\right)\)
= 1 – \(\frac{1}{4}\) – \(\left(\frac{3+1-2 \sqrt{3}}{8}\right)\)
= \(\frac{3}{4}-\frac{4-2 \sqrt{3}}{8}\)
= \(\frac{6-(4-2 \sqrt{3})}{8}\)
= \(\frac{2(1+\sqrt{3})}{2 \times 4}\) = \(\frac{1+\sqrt{3}}{4}\)
∴ cos²α = \(\frac{1+\sqrt{3}}{4}\)

Question 5.
If a line has the direction ratios -18, 12, -4 then what arc its direction cosines ?
Solution:
D.r’s of the given line = (a, b, c)= (-18, 12, -4)
∴ \(\sqrt{a^2+b^2+c^2}\)
= \(\sqrt{(-18)^2+(12)^2+(-4)^2}\)
= \(\sqrt{484}\) = 22
DC’s = \(\left(\frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}}\right)\)
= \(\left(\frac{-18}{22}, \frac{12}{22}, \frac{-4}{22}\right)\)
= \(\left(\frac{-9}{11}, \frac{6}{11}, \frac{-2}{11}\right)\)

Question 6.
Find the direction cosines of the joining the points (-2, 6, 7) and (1, 2, -5).
Solution:
Let A(x₁, y₁, z₁) = (-2, 6, 7), B(x₂, y₂, z₂) = (1, 2, -5)
Dr’s of AB = (a, b, c) = (x₂ – x₁, y₂ – y₁, z₂ – z₁) = (1+2, 2-6, -5-7) = (3, -4, -12)
∴ \(\sqrt{a^2+b^2+c^2}\)
= \(\sqrt{9+16+144}\)
= \(\sqrt{169}\) = 22
DC’s of AB = \(\left(\frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}}\right)\)
= \(\left(\frac{3}{13}, \frac{-4}{13}, \frac{-12}{13}\right)\)

Question 7.
Show that the points (2, 3, 4), (-1, -2, 1), (5, 8, 7) are collinear.
Solution:
Given points are A(2, 3, 4), B(-1, -2, 1), C(5, 8, 7).
D.r’s of line joining A and B are (-1-2, -2-3, 1-4) is (-3, -5, -3) ⇒ (3, 5, 3)
D.r’s of line joining A and C are (5-2, 8-3, 7-4) ⇒ 3, 5, 3
Hence D.r’s of AB and AC are proportional.
∴ A,B,C are collinear.

Question 8.
If the points (2, -1, -3), (4, a, 1) and (3, 1, b) are collinear, then find the ratio between a and A?
Solution:
Let A(x₁, y₁, z₁) = (2, -1, 3), B (x₂, y₂, z₂) = (4, a, 1), C(x₃, y₃, z₃) = (3, 1, b)
If A,B,C are collinear then \(\frac{x_1-x_2}{x_2-x_3}=\frac{y_1-y_2}{y_2-y_3}=\frac{z_1-z_2}{z_2-z_3}\)
⇒ \(\frac{2-4}{4-3}=\frac{-1-a}{a-1}=\frac{3-1}{1-b}\)
⇒ \(\frac{-2}{1}=\frac{-1-a}{a-1}=\frac{2}{1-b}\)
⇒ \(\frac{-2}{1}=\frac{-1-a}{a-1}\)
⇒ 2a – 2 = 1 + a ⇒ 2a – a = 1 + 2 ⇒ a = 3
⇒ \(\frac{-2}{1}=\frac{2}{1-b}\) ⇒ \(\frac{-1}{1}=\frac{1}{1-b}\)
⇒ -1 + b = 1
⇒ b = 2
∴ Required ratio \(\frac{a}{b}=\frac{3}{2}\)
⇒ 3 : 2

II.

Question 1.
Find the direction cosines of the sides of the triangle whose vertices are (3, 5, -4),(-1, 1, 2) and (-5, -5, -2).
Solution:
Let A(x₁, y₁, z₁) = (3, 5, -4), B (x₂, y₂, z₂) = (-1, 1, 2), C(x₃, y₃, z₃) = (-5, -5, -2)

Question 2.
Find the direction cosines of the medians of the triangles whose vertices are (1, 0, 2), (4. 3, 2) and (0, 7, 6).
Solution:
Vertices of the triangle are A = (1, 0, 2), B (4, 3, 2) and C(0, 7, 6).

Referring to the AP Inter 2nd Year Maths Study Material Chapter 10 Vector Algebra Exercise 10b Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Vector Algebra Solutions Exercise 10b

I.

Question 1.
Compute the magnitude of the vector \(\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}\)
Solution:
Given that \(\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}\)
∴ \(|\vec{a}|=\sqrt{x^2+y^2+z^2}=\sqrt{1+1+1}=\sqrt{3}\)

Question 2.
Compute the magnitude of the vector \(\overrightarrow{\mathbf{b}}=2 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}\)
Solution:
Given that \(\overrightarrow{\mathbf{b}}=2 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}\)
\(|\overrightarrow{\mathrm{b}}|=\sqrt{(2)^2+(-7)^2+(-3)^2}=\sqrt{4+49+9}=\sqrt{62} .\)

Question 3.
Compute the magnitude of the vector \(\vec{c}=\frac{1}{\sqrt{3}} \hat{\mathbf{i}}+\frac{1}{\sqrt{3}} \hat{\mathbf{j}}-\frac{1}{\sqrt{3}} \hat{\mathbf{k}}\)
Solution:
Given that \(\vec{c}=\frac{1}{\sqrt{3}} \hat{\mathbf{i}}+\frac{1}{\sqrt{3}} \hat{\mathbf{j}}-\frac{1}{\sqrt{3}} \hat{\mathbf{k}}\)
\(|\overrightarrow{\mathrm{c}}|=\sqrt{\left(\frac{1}{\sqrt{3}}\right)^2+\left(\frac{1}{\sqrt{3}}\right)^2+\left(\frac{-1}{\sqrt{3}}\right)^2}=\sqrt{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}=\sqrt{\frac{3}{3}}=\sqrt{1}=1 .\)

Question 4.
Write two different vectors having same direction.
Solution:
Let \(\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}\) and \(\vec{b}=2(\hat{i}+2 \hat{j}+3 \hat{k})=2 \vec{a}\)
Then \(\overrightarrow{\mathrm{b}}=\mathrm{m} \overrightarrow{\mathrm{a}}\) where m = 2 > 0.
∴ Vectors \(\vec{a} \text { and } \vec{b}\) have the same direction. But \(\overrightarrow{\mathrm{b}} \neq \overrightarrow{\mathrm{a}}\) as the corresponding components are distinct. We can write an infinite number of such vectors.

Question 5.
Write two different vectors having same magnitude.
Solution:
Let \(\vec{a}=(\hat{i}-2 \hat{j}+3 \hat{k})\) and \(\overrightarrow{\mathrm{b}}=(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-3 \hat{\mathrm{k}})\)
\(|\vec{a}|=\sqrt{1^2+(-2)^2+3^2}=\sqrt{1+4+9}=\sqrt{14}\)
\(|\vec{b}|=\sqrt{2^2+1^2+(-3)^2}=\sqrt{4+1+9}=\sqrt{14}\)
But \(\overrightarrow{\mathrm{a}} \neq \overrightarrow{\mathrm{b}}\) as the corresponding components are distinct.
We can write an infinite number of such vectors.

Question 6.
Find the values of x and y so that the vectors \(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}} \text { and } x \hat{\mathbf{i}}+y \hat{\mathbf{j}}\) are equal.
Solution:
Given \(2 \hat{i}+3 \hat{j}=x \hat{i}+y \hat{j}\)
Comparing coefficients of \(\hat{i} \text { and } \hat{j}\) on both sides, we have x = 2 and y = 3.

Question 7.
Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (- 5, 7).
Solution:
Let \(\overrightarrow{\mathrm{AB}}\) be the vector with initial point A(2, 1) and terminal point B(-5, 7)
⇒ PV (Position Vector)of point A(2, 1)is \(\overrightarrow{\mathrm{OA}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}\) and P.V. of point B (-5, 7) is \(\overrightarrow{\mathrm{OB}}=-5 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}\)
∴ \(\overrightarrow{\mathrm{AB}}\) = PV of point B – PV of point A = \((-5 \hat{\mathrm{i}}+7 \hat{\mathrm{j}})-(2 \hat{\mathrm{i}}+\hat{\mathrm{j}})=-5 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}-2 \hat{\mathrm{i}}-\hat{\mathrm{j}}\)
⇒ \(\overrightarrow{\mathrm{AB}}=-7 \hat{\mathrm{i}}+6 \hat{\mathrm{j}} .\)
∴ By definition, scalar components of the vectors \(\overrightarrow{\mathrm{AB}}\) are -7 and 6 and vector components of the vector \(\overrightarrow{\mathrm{AB}}\) are \(-7 \hat{\mathrm{i}} \text { and } 6 \hat{\mathrm{j}} .\)

Question 8.
Find the sum of the vectors \(\vec{a}=\hat{i}-2 \hat{j}+\hat{k}, b=-2 \hat{i}+4 \hat{j}+5 \hat{k}\) and \(\vec{c}=\hat{\mathrm{i}}-6 \hat{\mathrm{j}}-7 \hat{\mathrm{k}} .\)
Solution:
Given vectors \(\vec{a}=\hat{i}-2 \hat{j}+\hat{k}, \vec{b}=-2 \hat{i}+4 \hat{j}+5 \hat{k} \text { and } \vec{c}=\hat{i}-6 \hat{j}-7 \hat{k} .\)
∴ \(\vec{a}+\vec{b}+\vec{c}=(1-2+1) \hat{i}+(-2+4-6) \hat{j}+(1+5-7) \hat{k}=0 \hat{i}-4 \hat{j}-\hat{k}=-4 \hat{j}-\hat{k}\)

Question 9.
Find the unit vector in the direction of the vector \(\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}} .\)
Solution:
We know that a unit vector in the direction of the vector
\(\vec{a}=\hat{i}+\hat{j}+2 \hat{k} \text { is } \hat{a}=\frac{\vec{a}}{|\vec{a}|}=\frac{\hat{i}+\hat{j}+2 \hat{k}}{\sqrt{1+1+4}} \Rightarrow \hat{a}=\frac{\hat{i}+\hat{j}+2 \hat{k}}{\sqrt{6}}=\frac{1}{\sqrt{6}} \hat{i}+\frac{1}{\sqrt{6}} \hat{j}+\frac{2}{\sqrt{6}} \hat{k} .\)

Question 10.
Find the unit vector in the direction of vector \(\overrightarrow{\mathrm{PQ}}\), where P and Q are the points (1, 2, 3) and (4, 5, 6), respectively.
Solution:
Given points are P(1, 2, 3) and Q(4, 5, 6)
∴ Position vector of the point P(1, 2, 3) is \(\overrightarrow{\mathrm{OP}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\)
and position vector of point Q(4, 5, 6) is \(\overrightarrow{\mathrm{OQ}}=4 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}\) where O is the origin.
∴ \(\overrightarrow{P Q}=\overrightarrow{O Q}-\overrightarrow{O P}=(4 \hat{i}+5 \hat{j}+6 \hat{k})-(\hat{i}+2 \hat{j}+3 \hat{k})=3 \hat{i}+3 \hat{j}+3 \hat{k}\)
∴ a unit vector in the direction of vector \(\overline{\mathrm{PQ}}=\frac{\overline{\mathrm{PQ}}}{|\overline{\mathrm{PQ}}|}=\frac{3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}}{\sqrt{9+9+9=27=9 \times 3}}\)
\(\frac{3(\hat{i}+\hat{j}+\hat{k})}{3 \sqrt{3}}=\frac{(\hat{i}+\hat{j}+\hat{k})}{\sqrt{3}}=\frac{1}{\sqrt{3}} \hat{i}+\frac{1}{\sqrt{3}} \hat{j}+\frac{1}{\sqrt{3}} \hat{k}\)

Question 11.
For given vectors, \(\overrightarrow{\mathbf{a}}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}} \text { and } \overrightarrow{\mathbf{b}}=-\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}\). Find the unit vector in the direction of \(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}} .\)
Solution:
Given vectors \(\overrightarrow{\mathbf{a}}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}} \text { and } \overrightarrow{\mathbf{b}}=-\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}\)
∴ \(\vec{a}+\vec{b}=2 \hat{i}-\hat{j}+2 \hat{k}-\hat{i}+\hat{j}-\hat{k}=\hat{i}+0 \hat{j}+\hat{k}\)
∴ \(|\vec{a}+\vec{b}|=\sqrt{(1)^2+(0)^2+(1)^2}=\sqrt{2}\)
∴ A unit vector in the direction of \(\vec{a}+\vec{b} \text { is } \frac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|}=\frac{\hat{i}+0 \hat{j}+\hat{k}}{\sqrt{2}}=\frac{\hat{i}+\hat{k}}{\sqrt{2}}=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{k} .\)

Question 12.
Find a vector in the direction of vector \(5 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}\) which has magnitude 8 units.
Solution:
Let \(\vec{a}=5 \hat{i}-\hat{j}+2 \hat{k} .\)
∴ A vector in the direction of vector \(\vec{a}\) which has magnitude 8 units = \(8 \hat{a}=8 \frac{\vec{a}}{|\vec{a}|}=\frac{8(5 \hat{i}-\hat{j}+2 \hat{k})}{\sqrt{25+1+4}}\)
= \(\frac{8}{\sqrt{30}}(5 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})=\frac{40}{\sqrt{30}} \hat{\mathrm{i}}-\frac{8}{\sqrt{30}} \hat{\mathrm{j}}+\frac{16}{\sqrt{30}} \hat{\mathrm{k}}\)

Question 13.
Show that the vectors \(2 \hat{i}-3 \hat{j}+4 \hat{k} \text { and }-4 \hat{i}+6 \hat{j}-8 \hat{k}\) are coilinear.
Solution:
Let \(\vec{a}=2 \hat{i}-3 \hat{j}+4 \hat{k} \text { and } \vec{b}=-4 \hat{i}+6 \hat{j}-8 \hat{k}=-2(2 \hat{i}-3 \hat{j}+4 \hat{k})=-2 \vec{a}\)
Here, \(\overrightarrow{\mathrm{b}}=-2 \overrightarrow{\mathrm{a}}=\mathrm{m} \overrightarrow{\mathrm{a}}\) where m = -2 < 0
∴ Vectors \(\vec{a} \text { and } \vec{b}\) are coilinear (unlike because m = -2 < 0)

Question 14.
Find the direction cosines of the vector \(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\)
Solution:
The given vector is \(\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k} \Rightarrow|\vec{a}|=\sqrt{i^2+2^2+3^2}=\sqrt{24}\)
\(\hat{a}=\frac{\vec{a}}{|\vec{a}|}=\frac{\hat{i}+2 \hat{j}+3 \hat{k}}{\sqrt{14}}=\frac{1}{\sqrt{14}} \hat{i}+\frac{2}{\sqrt{14}} \hat{j}+\frac{3}{\sqrt{14}} \hat{k}\)
We know that direction consines of a vectors \(\vec{a}\) are coefficients of \(\hat{\mathrm{i}}, \hat{\mathrm{j}}, \hat{\mathrm{k}} \text { in } \hat{\mathrm{a}} \frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}\)
∴ Dc’s of the given vector = \(\left(\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}\right)\)

Question 15.
Find the direction c„osines of the vector joining the points A (1, 2, -3) and B (-1, -2, 1), directed from A to B.
Solution:
Given points A (1, 2, -3) andB (-1, -2, 1) ⇒ \(\overrightarrow{O A}=\hat{i}+2 \hat{j}-3 \hat{k}, \overrightarrow{O B}=-\hat{i}-2 \hat{j}+\hat{k}\)
\(\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=(-1-1) \hat{i}+(-2-2) \hat{j}+[1-(-3)] \hat{k}=-2 \hat{i}-4 \hat{j}+4 k\)
\(|\overrightarrow{\mathrm{AB}}|=\sqrt{(-2)^2+(-4)^2+4^2}=\sqrt{4+16+16}=\sqrt{36}=6\)
∴ A unit vector along AB = \(\frac{\overline{A B}}{|\overrightarrow{A B}|}=\frac{-2 \hat{i}-4 \hat{j}+4 \hat{k}}{6}=-\frac{2}{6} \hat{i}-\frac{4}{6} \hat{j}+\frac{4}{6} \hat{k}=\frac{-1}{3} \hat{i}-\frac{2}{3} \hat{j}+\frac{2}{3} \hat{k} .\)
∴ Direction Cosines of the vector \(\frac{-1}{3}, \frac{-2}{3}, \frac{2}{3}\)

Question 16.
Show that the vector \(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}\) is equally inclined to the axes OX, OY and OZ.
Solution:
Let \(\vec{a}=\hat{i}+\hat{j}+\hat{k}\)
\(|\overrightarrow{\mathbf{a}}|=\sqrt{1^2+1^2+1^2}=\sqrt{3}\)
Thus, the DCs of \(\overrightarrow{\mathrm{a}} \text { are }\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)\)
Now, α β and γ be the angles formed by \(\vec{a}\) with the positive directions of x, y and z axes respectively.
Then, cosα = \(\frac{1}{\sqrt{3}}\), cosβ = \(\frac{1}{\sqrt{3}}\), cosγ = \(\frac{1}{\sqrt{3}}\)
Hence, the vector is equally inclined to OX, OY and OZ.

Question 17.
Find the position vector of the mid point of the vector joining the points P(2, 3, 4) and Q(4, 1, -2).
Solution:
The position vector of the midpoint R is \(\overrightarrow{\mathrm{OR}}=\frac{\overrightarrow{\mathrm{OP}}+\overrightarrow{\mathrm{OQ}}}{2}=\frac{(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})+(4 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}})}{2}\)
\(=\frac{(2+4) \hat{i}+(3+1) \hat{j}+(4-2) \hat{k}}{2}=\frac{6 \hat{i}+4 \hat{j}+2 \hat{k}}{2}=3 \hat{i}+2 \hat{j}+\hat{k}\)

II.

Question 1.
Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are \(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}} \text { and }-\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}\) respectively in the ratio 2 : 1.
(i) internally
(ii) externally
Solution:
Position vectors of P and Q are given as \(\overrightarrow{\mathrm{OP}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}} \text { and } \overrightarrow{\mathrm{OQ}}=-\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
i) The position vector of R which divides the line joining two points P and Q internally in the ratio
2 : 1 is \(\overline{O R}=\frac{2(-\hat{i}+\hat{j}+\hat{k})+1(\hat{i}+2 \hat{j}-\hat{k})}{2+1}=\frac{-2 \hat{i}+2 \hat{j}+2 \hat{k}+\hat{i}+2 \hat{j}-\hat{k}}{3}\)
= \(\frac{-\hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{3}=\frac{-1}{3} \hat{\mathrm{i}}+\frac{4}{3} \hat{\mathrm{j}}+\frac{1}{3} \hat{\mathrm{k}}\)
ii) The position vector of R which divides the line joining two points P and Q externally in the 2 : 1 is \(\overrightarrow{\mathrm{OR}}=\frac{2(-\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})-1(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}})}{2-1}=\frac{-2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}-\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{1}=-3 \hat{\mathrm{i}}+3 \hat{\mathrm{k}}\)

Question 2.
Show that the points A, B and C with position vectors, \(\vec{a}=3 \hat{i}-4 \hat{j}-4 \hat{k}, \quad \vec{b}=2 \hat{i}-\hat{j}+\hat{k}\) and \(\vec{c}=\hat{\mathbf{i}}-3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}\) respectively form the vertices of a right angled triangle.
Solution:
Let \(\overline{\mathrm{OA}}=3 \overline{\mathrm{i}}-4 \overline{\mathrm{j}}-4 \overline{\mathrm{k}}, \overline{\mathrm{OB}}=2 \overline{\mathrm{i}}-\overline{\mathrm{j}}+\overline{\mathrm{k}}, \overline{\mathrm{OC}}=\overline{\mathrm{i}}-3 \overline{\mathrm{j}}-5 \overline{\mathrm{k}}\)
\(\overline{A B}=\overline{O B}-\overline{O A}=(2 \bar{i}-\bar{j}+\bar{k})-(3 \bar{i}-4 \bar{j}-4 \bar{k})=-\bar{i}+3 \bar{j}+5 \bar{k} \Rightarrow|\overline{A B}|=\sqrt{1+9+25}=\sqrt{35}\)
\(\overline{\mathrm{BC}}=\overline{\mathrm{OC}}-\overline{\mathrm{OB}}=(\overline{\mathrm{i}}-3 \overline{\mathrm{j}}-5 \overline{\mathrm{k}})-(2 \overline{\mathrm{i}}-\overline{\mathrm{j}}+\overline{\mathrm{k}})=-\overline{\mathrm{i}}-2 \overline{\mathrm{j}}-6 \overline{\mathrm{k}} \Rightarrow|\overline{\mathrm{BC}}|=\sqrt{1+4+36}=\sqrt{41}\)
\(\overline{C A}=\overline{O A}-\overline{O C}=(3 \bar{i}-4 \bar{j}-4 \bar{k})-(\bar{i}-3 \bar{j}-5 \bar{k})=2 \bar{i}-\bar{j}+\bar{k} \Rightarrow|\overline{C A}|=\sqrt{2+1+1}=\sqrt{6}\)
Here \(|\overrightarrow{\mathrm{OC}}|=(\sqrt{41})^2=41=35+6=|\overrightarrow{\mathrm{AB}}|^2+|\overrightarrow{\mathrm{CA}}|^2\)
∴ Points A, B, C are the vertices of a right angled triangle.