Let vec{a} = a_{1} hat{i} + a_{2} hat{j} + a_ | vedclass

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(C) Given $\vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ makes equal angles with axes,so $a_{1} = a_{2} = a_{3} = k$. Thus $\vec{a} = k(\ha

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$2$

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(C) Given $\vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ makes equal angles with axes,so $a_{1} = a_{2} = a_{3} = k$. Thus $\vec{a} = k(\hat{i} + \hat{j} + \hat{k})$.Projection of $\vec{a}$ on $3 \hat{i} + 4 \hat{j}$ is $\frac{\vec{a} \cdot (3 \hat{i} + 4 \hat{j})}{\sqrt{3^2 + 4^2}} = 7$.$\frac{k(3 + 4)}{5} = 7 \Rightarrow \frac{7k}{5} = 7 \Rightarrow k = 5$. So $\vec{a} = 5(\hat{i} + \hat{j} + \hat{k})$.Since $\vec{b}$ is obtained by rotating $\vec{a}$ by $90^{\circ}$ and $\vec{a}, \vec{b}, \hat{i}$ are coplanar,$\vec{b}$ lies in the plane of $\vec{a}$ and $\hat{i}$.Let $\vec{b} = x\vec{a} + y\hat{i}$. Since $|\vec{b}| = |\vec{a}| = 5\sqrt{3}$ and $\vec{a} \cdot \vec{b} = 0$,we find $\vec{b}$ is perpendicular to $\vec{a}$.Solving for the projection of $\vec{b}$ on $3\hat{i} + 4\hat{j}$,we get the magnitude as $2$.

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