Question
Find the angle between the vectors $\vec{\text{a}} $ and $\vec{\text{b}},$ where$\vec{\text{a}}=2\hat {\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$ and $\vec{\text{b}} =4\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}$

Answer

Let $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$.
$\big|\vec{\text{a}}\big|=\sqrt{(2)^2+(-1)^2+(2)^{2}}=\sqrt{9}=3$
$\big|\vec{\text{b}}\big|=\sqrt{(4)^2+(4)^2+(-2)^{2}}=\sqrt{36}=6$
$\vec{\text{a}}.\vec{\text{b}}=8-4-4=0$
$\cos\theta=\frac{\vec{\text{a }}.\vec{\text{ b}}}{\big|\vec{a}\big|\big|\vec{b}\big|}=\frac{0}{(3)(6)}=0$
$\Rightarrow\theta=\cos^{-1}(0)=\frac{\pi}{2}$

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