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Q.
If $\vec{A}=\hat{i}+2 \hat{j}-\hat{k}, \vec{B}=-\hat{i}+\hat{j}-2 \hat{k}$, then angle between $\vec{A}$ and $\vec{B}$ is
Motion in a Plane
Solution:
$\vec{A}=\hat{i}+2 \hat{j}-\hat{k} ; \vec{B}=-\hat{i}+\hat{j}-2 \hat{k}$
$A=\sqrt{1^2+2^2+(-1)^2}=\sqrt{6} ; B=\sqrt{(-1)^2+1^2+(-2)^2}=\sqrt{6} $
$\vec{A} \cdot \vec{B}=(\hat{i}+2 \hat{j}-\hat{k}) \cdot(-\hat{i}+\hat{j}-2 \hat{k})=-1+2+2=3$
Let $\theta$ be the angle between $\vec{A}$ and $\vec{B}$. Then
$\vec{A} \cdot \vec{B}=A B \cos \theta$
or $3=\sqrt{6} \sqrt{6} \cos \theta$
or $\cos \theta=\frac{3}{6}=\frac{1}{2}$
or $\theta=\frac{\pi}{3}$