If vec a = 2hat i -hat j + 3hat k and vec b = 2hat i+2hat j+hat k, th…

The correct answer is Option (3) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)

List-I	List-II
(A) Projection of $\vec a$ on $\vec b$ is	(III) $\frac{5}{3}$
(B) $\vec a×\vec b$ is	(I) $-7\hat i+4\hat j+6\hat k$
(C) unit vector along $\vec a +\vec b$ is	(IV) $\frac{1}{\sqrt{33}}(4\hat i+\hat j+4\hat k)$
(D) Unit vector perpendicular to both $\vec a$ & $\vec b$ is	(II) $\frac{1}{\sqrt{101}}(-7\hat i +4\hat j+6\hat k)$

Given:

$\vec{a} = 2\hat{i} - \hat{j} + 3\hat{k}$, $\vec{b} = 2\hat{i} + 2\hat{j} + \hat{k}$

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(A) Projection of $\vec{a}$ on $\vec{b}$

Formula: $\text{proj}_{\vec{b}} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \cdot \vec{b}$

$\vec{a} \cdot \vec{b} = 2(2) + (-1)(2) + 3(1) = 4 - 2 + 3 = 5$
$|\vec{b}|^2 = 2^2 + 2^2 + 1^2 = 4 + 4 + 1 = 9$

$\Rightarrow \text{proj}_{\vec{b}} \vec{a} = \frac{5}{9} \cdot \vec{b} = \frac{5}{9}(2\hat{i} + 2\hat{j} + \hat{k})$

= $\frac{10}{9} \hat{i} + \frac{10}{9} \hat{j} + \frac{5}{9} \hat{k}$

Projection =$\frac{5}{3}$

(A) → (III)

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(B) $\vec{a} \times \vec{b}$

$\vec{a} \times \vec{b} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 3 \\ 2 & 2 & 1 \\ \end{array} \right|$

$= \hat{i}((-1)(1) - (3)(2)) - \hat{j}((2)(1) - (3)(2)) + \hat{k}((2)(2) - (-1)(2))$
$= \hat{i}(-1 - 6) - \hat{j}(2 - 6) + \hat{k}(4 + 2)$
$= -7\hat{i} + 4\hat{j} + 6\hat{k}$

(B) → (I)

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(C) Unit vector along $\vec{a} + \vec{b}$

$\vec{a} + \vec{b} = (2+2)\hat{i} + (-1+2)\hat{j} + (3+1)\hat{k} = 4\hat{i} + \hat{j} + 4\hat{k}$

Magnitude = $\sqrt{4^2 + 1^2 + 4^2} = \sqrt{16 + 1 + 16} = \sqrt{33}$

Unit vector = $\frac{1}{\sqrt{33}}(4\hat{i} + \hat{j} + 4\hat{k})$

(C) → (IV)

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(D) Unit vector perpendicular to both $\vec{a}, \vec{b}$

$\vec{a} \times \vec{b} = -7\hat{i} + 4\hat{j} + 6\hat{k}$, magnitude = $\sqrt{49 + 16 + 36} = \sqrt{101}$

Unit vector = $\frac{1}{\sqrt{101}}(-7\hat{i} + 4\hat{j} + 6\hat{k})$

(D) → (II)