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(D) Let $\vec{a} = 2 \hat{i} - 2 \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2 \hat{j} + 2 \hat{k}$.First, calculate the magnitudes: $|\vec{a}| = \sqr

Vedclass · Accepted Answer

$\hat{i} + \hat{k}$

Vedclass · Answer

(D) Let $\vec{a} = 2 \hat{i} - 2 \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2 \hat{j} + 2 \hat{k}$.First, calculate the magnitudes: $|\vec{a}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4+4+1} = 3$ and $|\vec{b}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1+4+4} = 3$.Since $|\vec{a}| = |\vec{b}|$, the internal angle bisector is given by the vector $\vec{v} = \vec{a} + \vec{b}$.$\vec{v} = (2 \hat{i} - 2 \hat{j} + \hat{k}) + (\hat{i} + 2 \hat{j} + 2 \hat{k}) = 3 \hat{i} + 3 \hat{k}$.The unit vector along the bisector is $\hat{u} = \frac{\vec{v}}{|\vec{v}|} = \frac{3 \hat{i} + 3 \hat{k}}{\sqrt{3^2 + 3^2}} = \frac{3 \hat{i} + 3 \hat{k}}{\sqrt{18}} = \frac{3 \hat{i} + 3 \hat{k}}{3 \sqrt{2}} = \frac{\hat{i} + \hat{k}}{\sqrt{2}}$.$A$ vector of magnitude $\sqrt{2}$ along this direction is $\sqrt{2} 	imes \hat{u} = \sqrt{2} 	imes \frac{\hat{i} + \hat{k}}{\sqrt{2}} = \hat{i} + \hat{k}$.

$A$ vector of magnitude $\sqrt{2}$ units along the internal bisector of the angle between the vectors $\vec{a} = 2 \hat{i} - 2 \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2 \hat{j} + 2 \hat{k}$ is

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