Class 12-NCERT Solutions-Chapter-10-Vector Algebra-Ex 10.4

Solution:

\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -7 & 7 \\ 3 & -2 & 2 \end{vmatrix} \]

\[ = \hat{i}(-14 - (-14)) - \hat{j}(2 - 21) + \hat{k}(-2 - (-21)) \]

\[ = \hat{i}(0) - \hat{j}(-19) + \hat{k}(19) = 19\hat{j} + 19\hat{k} \]

Magnitude \( |\vec{a} \times \vec{b}| = \sqrt{0^2 + 19^2 + 19^2} = \sqrt{2 \times 19^2} = 19\sqrt{2} \).

Solution:

Let \( \vec{c} = \vec{a} + \vec{b} = 4\hat{i} + 4\hat{j} + 0\hat{k} \).

Let \( \vec{d} = \vec{a} - \vec{b} = 2\hat{i} + 0\hat{j} + 4\hat{k} \).

Vector perpendicular to both is \( \vec{n} = \vec{c} \times \vec{d} \).

\[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{vmatrix} = \hat{i}(16) - \hat{j}(16) + \hat{k}(-8) = 16\hat{i} - 16\hat{j} - 8\hat{k} \]

Magnitude \( |\vec{n}| = \sqrt{16^2 + (-16)^2 + (-8)^2} = \sqrt{256 + 256 + 64} = \sqrt{576} = 24 \).

Unit vector \( \hat{n} = \pm \frac{16\hat{i} - 16\hat{j} - 8\hat{k}}{24} = \pm \left( \frac{2}{3}\hat{i} - \frac{2}{3}\hat{j} - \frac{1}{3}\hat{k} \right) \).

Solution:

Let direction cosines be \( l, m, n \). Given \( l = \cos \frac{\pi}{3} = \frac{1}{2} \), \( m = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} \), \( n = \cos \theta \).

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 \( (l, m, n) = (\frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2}) \).

Solution:

\[ \begin{aligned} LHS &= (\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) \\ &= \vec{a} \times \vec{a} + \vec{a} \times \vec{b} - \vec{b} \times \vec{a} - \vec{b} \times \vec{b} \end{aligned} \]

We know \( \vec{a} \times \vec{a} = \vec{0} \) and \( \vec{b} \times \vec{b} = \vec{0} \).

Also, \( \vec{b} \times \vec{a} = -(\vec{a} \times \vec{b}) \).

\[ = \vec{0} + \vec{a} \times \vec{b} - (-(\vec{a} \times \vec{b})) - \vec{0} \]

\( = 2(\vec{a} \times \vec{b}) = RHS \).

Solution:

Cross product equals zero implies the vectors are parallel. Ratios of components must be equal.

\( \frac{2}{1} = \frac{6}{\lambda} = \frac{27}{\mu} \).

\( \frac{6}{\lambda} = 2 \Rightarrow 2\lambda = 6 \Rightarrow \lambda = 3 \).

\( \frac{27}{\mu} = 2 \Rightarrow 2\mu = 27 \Rightarrow \mu = \frac{27}{2} \).

Solution:

\( \vec{a} \cdot \vec{b} = 0 \) implies \( \vec{a} \perp \vec{b} \) (if non-zero).

\( \vec{a} \times \vec{b} = \vec{0} \) implies \( \vec{a} \parallel \vec{b} \) (if non-zero).

Two non-zero vectors cannot be both perpendicular and parallel at the same time.

Therefore, either \( \vec{a} = \vec{0} \) or \( \vec{b} = \vec{0} \).

Solution:

This is the distributive property of vector product over addition.

It can be proved by expanding the determinants for \( \vec{a} \times (\vec{b} + \vec{c}) \) and showing it equals sum of determinants of \( \vec{a} \times \vec{b} \) and \( \vec{a} \times \vec{c} \).

LHS involves components \( (b_i + c_i) \). Determinant properties allow splitting rows/columns with sums into sum of two determinants.

Solution:

The converse is: "If \( \vec{a} \times \vec{b} = \vec{0} \), then either \( \vec{a} = \vec{0} \) or \( \vec{b} = \vec{0} \)".

This is false. The cross product is zero if the vectors are parallel (collinear), even if they are non-zero.

Example: Let \( \vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k} \) and \( \vec{b} = 4\hat{i} + 6\hat{j} + 8\hat{k} \).

Here \( \vec{b} = 2\vec{a} \), so they are parallel. \( \vec{a} \times \vec{b} = \vec{0} \), but neither is a zero vector.

Solution:

Area = \( \frac{1}{2} |\vec{AB} \times \vec{AC}| \).

\( \vec{AB} = (2-1)\hat{i} + (3-1)\hat{j} + (5-2)\hat{k} = \hat{i} + 2\hat{j} + 3\hat{k} \).

\( \vec{AC} = (1-1)\hat{i} + (5-1)\hat{j} + (5-2)\hat{k} = 0\hat{i} + 4\hat{j} + 3\hat{k} \).

\[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 0 & 4 & 3 \end{vmatrix} = \hat{i}(6-12) - \hat{j}(3-0) + \hat{k}(4-0) = -6\hat{i} - 3\hat{j} + 4\hat{k} \]

Magnitude = \( \sqrt{(-6)^2 + (-3)^2 + 4^2} = \sqrt{36 + 9 + 16} = \sqrt{61} \).

Area = \( \frac{\sqrt{61}}{2} \) sq. units.

Solution:

Area = \( |\vec{a} \times \vec{b}| \).

\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 3 \\ 2 & -7 & 1 \end{vmatrix} = \hat{i}(-1+21) - \hat{j}(1-6) + \hat{k}(-7+2) = 20\hat{i} + 5\hat{j} - 5\hat{k} \]

Magnitude = \( \sqrt{20^2 + 5^2 + (-5)^2} = \sqrt{400 + 25 + 25} = \sqrt{450} = \sqrt{225 \times 2} = 15\sqrt{2} \).

Area = \( 15\sqrt{2} \) sq. units.

Answer: (B)

Solution:

\( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta = 1 \).

\( 3 \times \frac{\sqrt{2}}{3} \times \sin \theta = 1 \).

\( \sqrt{2} \sin \theta = 1 \Rightarrow \sin \theta = \frac{1}{\sqrt{2}} \).

\( \theta = \frac{\pi}{4} \).

Answer: (C)

Solution:

Adjacent sides \( \vec{AB} \) and \( \vec{AD} \).

\( \vec{AB} = \vec{B} - \vec{A} = 2\hat{i} \). Magnitude = 2.

\( \vec{AD} = \vec{D} - \vec{A} = -\hat{j} \). Magnitude = 1.

Area = \( |\vec{AB}| \times |\vec{AD}| = 2 \times 1 = 2 \) sq. units.