Question
Find the angle between the vectors $\vec{\text{a}} $ and $\vec{\text{b}},$ where$\vec{\text{a}}=\hat {\text{i}}+2\hat{\text{j}}-\hat{\text{k}},$ and $\vec{\text{b}} =\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$

Answer

Let $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$.$\big|\vec{\text{a}}\big|=\sqrt{(1)^2+(2)^2+(-1)^{2}}=\sqrt{6}$
$\big|\vec{\text{b}}\big|=\sqrt{(1)^2+(-1)^2+(1)^{2}}=\sqrt{3}$
$\vec{\text{a}}.\vec{\text{b}}=1-2-1=-2$
$\cos\theta=\frac{\vec{\text{a }}.\vec{\text{ b}}}{\big|\vec{a}\big|\big|\vec{b}\big|}=\frac{-2}{\sqrt{6}\sqrt{3}}=\frac{-2}{\sqrt{18}}=\frac{-\sqrt{2}\times\sqrt{2}}{\sqrt{2}\times\sqrt{9}}=\frac{-\sqrt{2}}{3}$
$\Rightarrow\theta=\cos^{-1}\Big(\frac{-\sqrt{2}}{{3}}\Big)$

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