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

Answer

We have
$\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$
Let $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
$|\vec{\text{a}}|=\sqrt{(1)^2+(-1)^2+(1)^2}=\sqrt{3}$
$\big|\vec{\text{b}}\big|=\sqrt{(1)^2+(1)^2+(-1)^2}=\sqrt{3}$
and
$\vec{\text{a}}.\vec{\text{b}}=1-1-1=-1$
Now,
$\cos\theta=\frac{\vec{\text{a}}.\vec{\text{b}}}{|\vec{\text{a}}|\big|\vec{\text{b}}\big|}=\frac{-1}{\sqrt{3}\sqrt{3}}=\frac{-1}{3}$
$\therefore\theta=\cos^-1\big(\frac{-1}{3}\big)$

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