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Q.
Angle between vectors $\left(\hat{i}+\hat{j}\right)$ and $\left(\hat{j}+\hat{k}\right)$ is
UP CPMTUP CPMT 2007
Solution:
Let $\vec{A}=\hat{i}+\hat{j}$ and $\vec{B}=\hat{j}+\hat{k}$ and $\theta$ is the angle between them
Also, $\left|\vec{A}\right|=\sqrt{\left(1\right)^{2}+\left(1\right)^{2}}=\sqrt{2}$;
$\left|\vec{B}\right|=\sqrt{\left(1\right)^{2}+\left(1\right)^{2}}=\sqrt{2}$;
$\left|\vec{B}\right|=\sqrt{\left(1\right)^{2}+\left(1\right)^{2}}=\sqrt{2}$
$\therefore \vec{A}\cdot\vec{B} =\left|\vec{A}\right|\left|\vec{B}\right| cos\, \theta$
$\left(\hat{i}+\hat{j}\right)\cdot\left(\hat{j}+\hat{k}\right)=\sqrt{2}\sqrt{2}cos\,\theta$
$\left[0+0+1+0\right]=2cos \theta$
or $cos\,\theta=\frac{1}{2}$ or $ \theta=cos^{-1}\left(\frac{1}{2}\right)$
or $\theta=60^{\circ}$