Let vec{a} be a vector which is perpendicular | vedclass

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(C) Given $\vec{a} \perp (3 \hat{i} + \frac{1}{2} \hat{j} + 2 \hat{k})$ and $\vec{a} 	imes (2 \hat{i} + \hat{k}) = 2 \hat{i} - 13 \hat{j} - 4 \hat{k}

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$\frac{5}{3}$

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(C) Given $\vec{a} \perp (3 \hat{i} + \frac{1}{2} \hat{j} + 2 \hat{k})$ and $\vec{a} 	imes (2 \hat{i} + \hat{k}) = 2 \hat{i} - 13 \hat{j} - 4 \hat{k}$.Using the vector triple product identity $(\vec{a} 	imes \vec{b}) 	imes \vec{c} = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{b} \cdot \vec{c}) \vec{a}$,let $\vec{b} = 2 \hat{i} + \hat{k}$ and $\vec{c} = 3 \hat{i} + \frac{1}{2} \hat{j} + 2 \hat{k}$.Since $\vec{a} \cdot \vec{c} = 0$,we have $(\vec{a} 	imes \vec{b}) 	imes \vec{c} = -(\vec{b} \cdot \vec{c}) \vec{a}$.Calculate $\vec{b} \cdot \vec{c} = (2)(3) + (0)(\frac{1}{2}) + (1)(2) = 6 + 0 + 2 = 8$.Now,compute the cross product $(2 \hat{i} - 13 \hat{j} - 4 \hat{k}) 	imes (3 \hat{i} + \frac{1}{2} \hat{j} + 2 \hat{k})$:$= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -13 & -4 \ 3 & 0.5 & 2 \end{vmatrix} = \hat{i}(-26 - (-2)) - \hat{j}(4 - (-12)) + \hat{k}(1 - (-39)) = -24 \hat{i} - 16 \hat{j} + 40 \hat{k}$.So,$-8 \vec{a} = -24 \hat{i} - 16 \hat{j} + 40 \hat{k} \implies \vec{a} = 3 \hat{i} + 2 \hat{j} - 5 \hat{k}$.The projection of $\vec{a}$ on $\vec{v} = 2 \hat{i} + 2 \hat{j} + \hat{k}$ is $\frac{\vec{a} \cdot \vec{v}}{|\vec{v}|}$.$\vec{a} \cdot \vec{v} = (3)(2) + (2)(2) + (-5)(1) = 6 + 4 - 5 = 5$.$|\vec{v}| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{9} = 3$.Projection $= \frac{5}{3}$.

Let $\vec{a}$ be a vector which is perpendicular to the vector $3 \hat{i} + \frac{1}{2} \hat{j} + 2 \hat{k}$. If $\vec{a} \times (2 \hat{i} + \hat{k}) = 2 \hat{i} - 13 \hat{j} - 4 \hat{k}$,then the projection of the vector $\vec{a}$ on the vector $2 \hat{i} + 2 \hat{j} + \hat{k}$ is

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