If $\bar{a}$ and $\bar{b}$ are two vectors such that $|\vec{a}|=|\vec{b}|=\sqrt{2}$ with $\vec{a} \cdot \vec{b}=-1$, then the angle between $\bar{a}$ and $\bar{b}$ is

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Aicharya · Accepted Answer

$	heta=\cos ^{-1}\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}ight)=\cos ^{-1}\left(\frac{-1}{\sqrt{2} \sqrt{2}}ight)=\cos ^{-1}\left(\frac{-1}{2}ight)=\frac{2 \pi}{3}$

If $\bar{a}$ and $\bar{b}$ are two vectors such that $|\vec{a}|=|\vec{b}|=\sqrt{2}$ with $\vec{a} \cdot…

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