Question

Find the angle between two vectors $\overrightarrow{A}=2\hat{i}+\hat{j}-\hat{k}and\overrightarrow{B}=\hat{i}-\hat{k}.$

• A. $\theta ={60}^{\circ }$
• B. $\theta ={30}^{\circ }$
• C. $\theta ={150}^{\circ }$
• D. $\theta ={120}^{\circ }$

Answer 1

Answer: (B)

$\theta ={30}^{\circ }$

$A=\left|\overrightarrow{A}\right|=\sqrt{{\left(2\right)}^2+{\left(1\right)}^2+{(-1)}^2}=\sqrt{6}$
and, $B=\left|\overrightarrow{B}\right|=\sqrt{{\left(1\right)}^2+{(-1)}^2}=\sqrt{2}$
Also, $\overrightarrow{A}\cdot \overrightarrow{B}=(2\hat{i}+\hat{j}-\hat{k})\cdot (\hat{i}-\hat{k})=\left(2\right)\left(1\right)+(-1)(-1)=3$
We know that, $\cos \theta =\frac{\overrightarrow{A}\bullet \overrightarrow{B}}{AB}=\frac{3}{\sqrt{6}\cdot \sqrt{2}}=\frac{3}{\sqrt{12}}=\frac{\sqrt{3}}{2}\therefore \theta ={30}^{\circ }$