Question
Find the angle between the vectors $\vec{\text{a}} $ and $\vec{\text{b}},$ where$\vec{\text{a}}=\hat {\text{i}}-\hat{\text{j}}$ and $\vec{\text{b}} = \hat{\text{j}}+\hat{\text{k}}$

Answer

Let $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$.
$\big|\vec{\text{a}}\big|=\sqrt{(1)^2+(-1)^2}=\sqrt{2}$
$\big|\vec{\text{b}}\big|=\sqrt{(1)^2+(1)^2}=\sqrt{2}$
$\vec{\text{a}}.\vec{\text{b}}=0-1+0=-1$
$\cos\theta=\frac{\vec{\text{a }}.\vec{\text{ b}}}{\big|\vec{a}\big|\big|\vec{b}\big|}=\frac{-1}{\sqrt{2}\sqrt{2}}=\frac{-1}{2}$
$\Rightarrow\theta=\cos^{-1}\big(\frac{-1}{2}\big)=\frac{2\pi}{3}$

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