NCERT Solutions Class-12-Chapter-10-Vector Algebra
Miscellaneous Exercise on Chapter 10
Note:
Vector (Cross) Product: \( \vec{a} \times \vec{b} = |\vec{a}| |\vec{b}| \sin \theta \hat{n} \), where \( \hat{n} \) is a unit vector perpendicular to both \( \vec{a} \) and \( \vec{b} \).
Magnitude: \( |\vec{a} \times \vec{b}| = \sqrt{(y_1 z_2 - z_1 y_2)^2 + (z_1 x_2 - x_1 z_2)^2 + (x_1 y_2 - y_1 x_2)^2} \).
Area of Triangle: \( \frac{1}{2} |\vec{a} \times \vec{b}| \).
Area of Parallelogram: \( |\vec{a} \times \vec{b}| \) (adjacent sides).
Q1
Find \( |\vec{a} \times \vec{b}| \), if \( \vec{a} = \hat{i} - 7\hat{j} + 7\hat{k} \) and \( \vec{b} = 3\hat{i} - 2\hat{j} + 2\hat{k} \).
Solution:
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -7 & 7 \\ 3 & -2 & 2 \end{vmatrix} \]
\[ = \hat{i}(-14 - (-14)) - \hat{j}(2 - 21) + \hat{k}(-2 - (-21)) \]
\[ = 0\hat{i} + 19\hat{j} + 19\hat{k} \]
\( |\vec{a} \times \vec{b}| = \sqrt{0^2 + 19^2 + 19^2} = \sqrt{2(19)^2} = 19\sqrt{2} \).
Q2
Find a unit vector perpendicular to each of the vector \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b} \), where \( \vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k} \) and \( \vec{b} = \hat{i} + 2\hat{j} - 2\hat{k} \).
Solution:
Let \( \vec{c} = \vec{a} + \vec{b} = 4\hat{i} + 4\hat{j} \).
Let \( \vec{d} = \vec{a} - \vec{b} = 2\hat{i} + 4\hat{k} \).
Vector perpendicular to both is \( \vec{n} = \vec{c} \times \vec{d} \).
\[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{vmatrix} = \hat{i}(16) - \hat{j}(16) + \hat{k}(-8) \]
\( \vec{n} = 16\hat{i} - 16\hat{j} - 8\hat{k} \).
Magnitude \( |\vec{n}| = \sqrt{256 + 256 + 64} = \sqrt{576} = 24 \).
Unit vector \( \hat{n} = \pm \frac{16\hat{i} - 16\hat{j} - 8\hat{k}}{24} = \pm (\frac{2}{3}\hat{i} - \frac{2}{3}\hat{j} - \frac{1}{3}\hat{k}) \).
Q3
If a unit vector \( \vec{a} \) makes angles \( \frac{\pi}{3} \) with \( \hat{i} \), \( \frac{\pi}{4} \) with \( \hat{j} \) and an acute angle \( \theta \) with \( \hat{k} \), then find \( \theta \) and hence, the components of \( \vec{a} \).
Solution:
Direction cosines are \( l = \cos \frac{\pi}{3} = \frac{1}{2} \), \( m = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} \), \( n = \cos \theta \).
Since \( \vec{a} \) is a unit vector, \( l^2 + m^2 + n^2 = 1 \).
\( \frac{1}{4} + \frac{1}{2} + \cos^2 \theta = 1 \Rightarrow \frac{3}{4} + \cos^2 \theta = 1 \Rightarrow \cos^2 \theta = \frac{1}{4} \).
\( \cos \theta = \pm \frac{1}{2} \). Since \( \theta \) is acute, \( \cos \theta = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{3} \).
Components of \( \vec{a} \) are \( (\frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2}) \).
Q4
Show that \( (\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) = 2(\vec{a} \times \vec{b}) \).
Solution:
\[ LHS = \vec{a} \times \vec{a} + \vec{a} \times \vec{b} - \vec{b} \times \vec{a} - \vec{b} \times \vec{b} \]
Since \( \vec{a} \times \vec{a} = \vec{0} \) and \( \vec{b} \times \vec{b} = \vec{0} \):
\[ = \vec{0} + \vec{a} \times \vec{b} - (-\vec{a} \times \vec{b}) - \vec{0} \]
\( = 2(\vec{a} \times \vec{b}) = RHS \).
Q5
Find \( \lambda \) and \( \mu \) if \( (2\hat{i} + 6\hat{j} + 27\hat{k}) \times (\hat{i} + \lambda\hat{j} + \mu\hat{k}) = \vec{0} \).
Solution:
Cross product is zero means vectors are parallel. Components are proportional.
\( \frac{2}{1} = \frac{6}{\lambda} = \frac{27}{\mu} \).
\( \lambda = \frac{6}{2} = 3 \).
\( \mu = \frac{27}{2} \).
Q6
Given that \( \vec{a} \cdot \vec{b} = 0 \) and \( \vec{a} \times \vec{b} = \vec{0} \). What can you conclude about the vectors \( \vec{a} \) and \( \vec{b} \)?
Solution:
\( \vec{a} \cdot \vec{b} = 0 \implies \vec{a} \perp \vec{b} \).
\( \vec{a} \times \vec{b} = \vec{0} \implies \vec{a} \parallel \vec{b} \).
Vectors cannot be both parallel and perpendicular unless one of them is a zero vector.
Thus, \( \vec{a} = \vec{0} \) or \( \vec{b} = \vec{0} \).
Q7
Let the vectors \( \vec{a}, \vec{b}, \vec{c} \) be given as components. Show that \( \vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c} \).
Solution:
Using determinant expansion:
LHS uses components \( (b_1+c_1, b_2+c_2, b_3+c_3) \).
Determinant property allows splitting rows: \( |A + B| = |A| + |B| \).
Thus, \( \vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c} \).
Q8
If either \( \vec{a} = \vec{0} \) or \( \vec{b} = \vec{0} \), then \( \vec{a} \times \vec{b} = \vec{0} \). Is the converse true? Justify your answer.
Solution:
No, the converse is not true.
\( \vec{a} \times \vec{b} = \vec{0} \) implies vectors are parallel. They can be non-zero.
Example: \( \vec{a} = \hat{i} \) and \( \vec{b} = 2\hat{i} \). Cross product is zero, but vectors are non-zero.
Q9
Find the area of the triangle with vertices \( A(1, 1, 2) \), \( B(2, 3, 5) \) and \( C(1, 5, 5) \).
Solution:
\( \vec{AB} = \hat{i} + 2\hat{j} + 3\hat{k} \).
\( \vec{AC} = 0\hat{i} + 4\hat{j} + 3\hat{k} \).
Area = \( \frac{1}{2} |\vec{AB} \times \vec{AC}| \).
\( \vec{AB} \times \vec{AC} = -6\hat{i} - 3\hat{j} + 4\hat{k} \).
Magnitude \( = \sqrt{36 + 9 + 16} = \sqrt{61} \).
Area = \( \frac{\sqrt{61}}{2} \) sq. units.
Q10
Find the area of the parallelogram whose adjacent sides are determined by the vectors \( \vec{a} = \hat{i} - \hat{j} + 3\hat{k} \) and \( \vec{b} = 2\hat{i} - 7\hat{j} + \hat{k} \).
Solution:
Area = \( |\vec{a} \times \vec{b}| \).
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 3 \\ 2 & -7 & 1 \end{vmatrix} = 20\hat{i} + 5\hat{j} - 5\hat{k} \]
Magnitude = \( \sqrt{400 + 25 + 25} = \sqrt{450} = 15\sqrt{2} \).
Area = \( 15\sqrt{2} \) sq. units.
Q11
Let vectors \( \vec{a} \) and \( \vec{b} \) be such that \( |\vec{a}| = 3 \) and \( |\vec{b}| = \frac{\sqrt{2}}{3} \), then \( \vec{a} \times \vec{b} \) is a unit vector if the angle is
(A) \( \pi/6 \) (B) \( \pi/4 \) (C) \( \pi/3 \) (D) \( \pi/2 \)
(A) \( \pi/6 \) (B) \( \pi/4 \) (C) \( \pi/3 \) (D) \( \pi/2 \)
Answer: (B)
Solution:
\( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta = 1 \).
\( 3 \times \frac{\sqrt{2}}{3} \sin \theta = 1 \Rightarrow \sqrt{2} \sin \theta = 1 \Rightarrow \sin \theta = \frac{1}{\sqrt{2}} \).
\( \theta = \frac{\pi}{4} \).
Q12
Area of a rectangle having vertices A, B, C, D with position vectors ... is
(A) \( \frac{1}{2} \) (B) 1 (C) 2 (D) 4
(A) \( \frac{1}{2} \) (B) 1 (C) 2 (D) 4
Answer: (C)
Solution:
Calculate adjacent sides \( \vec{AB} = \vec{B} - \vec{A} \) and \( \vec{AD} = \vec{D} - \vec{A} \).
\( \vec{AB} = 2\hat{i} \), magnitude 2.
\( \vec{AD} = -\hat{j} \), magnitude 1.
Area = \( |\vec{AB} \times \vec{AD}| = |2\hat{i} \times (-\hat{j})| = |-2\hat{k}| = 2 \).