If $\vec{a}=-4 \hat{i}+2 \hat{j}+4 \hat{k}, \vec{b}=\sqrt{2} \hat{i}-\sqrt{2} \hat{j}$ are two vectors then angle between the vectors $2 \vec{a}$ and $\frac{\vec{b}}{2}$ is

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Aicharya · Accepted Answer

Angle between $2 \vec{a}$ and $\frac{\vec{b}}{2}$ $\begin{aligned} & \cos 	heta=\frac{(2 \vec{a}) \cdot\left(\frac{\vec{b}}{2}ight)}{|2 \vec{a}|\left|\frac{\vec{b}}{2}ight|}=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \ & =\frac{-4 \sqrt{2}-2 \sqrt{2}}{\sqrt{16+4+16} \sqrt{2+2}}=\frac{-6 \sqrt{2}}{6 	imes 2}=-\frac{1}{\sqrt{2}} \ & \cos 	heta=-\frac{1}{\sqrt{2}}=\cos \left(180^{\circ}-45^{\circ}ight)=\cos 135^{\circ} \Rightarrow 	heta=135^{\circ}\end{aligned}$

If $\vec{a}=-4 \hat{i}+2 \hat{j}+4 \hat{k}, \vec{b}=\sqrt{2} \hat{i}-\sqrt{2} \hat{j}$ are two vectors then…

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