The angle between $\vec{A}=\hat{i}+\hat{j}$ and $\vec{B}=\hat{i}-\hat{j}$ is

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Professor PC Thomas Classes · Answer

(b) : Given: $\vec{A}=\hat{i}+\hat{j}, \vec{B}=\hat{i}-\hat{j}$ \ $	herefore \quad|\vec{A}|=\sqrt{(1)^2+(1)^2}=\sqrt{2}$ \ and $|\vec{B}|=\sqrt{(1)^2+(-1)^2}=\sqrt{2}$ \ Let $	heta$ be angle between the vectors $\vec{A}$ and $\vec{B}$. Then according to definition of scalar product (or dot product)
 $\vec{A} \cdot \vec{B}=|\vec{A}||\vec{B}| \cos 	heta$
or $\cos 	heta=\frac{\vec{A} \cdot \vec{B}}{|\vec{A}||\vec{B}|}=\frac{(\hat{i}+\hat{j}) \cdot(\hat{i}-\hat{j})}{(\sqrt{2})(\sqrt{2})}=0$
$	herefore \quad 	heta=\cos ^{-1}(0)=90^{\circ}$