The vector $\frac{1}{3}(2 \hat{i}-2 \hat{j}+\hat{k})$ is\

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Professor PC Thomas Classes · Answer

(a, c, d) : $|\vec{a}|^{2}=\frac{1}{9}(4+4+1) 1 \Rightarrow|\vec{a}|=1$ Let $\vec{b}=2 \hat{i}-4 \hat{j}+3 \hat{k}$ then\ $\cos 	heta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\frac{5}{\sqrt{29}} \Rightarrow 	heta 
eq \pi / 3$\ Let $\vec{c}=-\hat{i}+\hat{j}-\frac{1}{2} \hat{k}=\frac{-3}{2} \vec{a} \Rightarrow \vec{c} \| \vec{a}$\ Let $\vec{d}=3 \hat{i}+2 \hat{j}+2 \hat{k}$ then $\vec{a} \cdot \vec{d}=0 \Rightarrow \vec{a} \perp \vec{d}$\