Question

Find the angle between two vectors $(\overrightarrow{a}=2\hat{i}+\hat{j}-3\hat{k})$ and $(\overrightarrow{b}=3\hat{i}-2\hat{j}-\hat{k})$.

• (A) $-{160}^{\circ }$
• (B) $-{60}^{\circ }$
• (C) ${160}^{\circ }$
• (D) ${60}^{\circ }$

Answer 1

The Correct Option is D

Explanation:
Given:$\overrightarrow{\mathrm{a}}=2\hat{\mathrm{i}}+\hat{\mathrm{j}}-3\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{b}}=3\hat{\mathrm{i}}-2\hat{\mathrm{j}}-\hat{\mathrm{k}}$$\overrightarrow{a}\cdot \overrightarrow{b}=(2\hat{i}+\hat{j}-3\hat{k})(3\hat{i}-2\hat{j}-\hat{k})$$=6-2+3=7\dots .(1)$$|\overrightarrow{a}|=\sqrt{2^2+1^2+(-3{)}^2}=\sqrt{14}\dots (2)$$|\overrightarrow{b}|=\sqrt{3^2+(-2{)}^2+(-1{)}^2}=\sqrt{14}\dots .(3)$$\cos \theta =\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{|\overrightarrow{a}|\cdot |\overrightarrow{b}|}$Put the values from (1), (2) amnd (3) in above equation:$=\frac{7}{\sqrt{14}.\sqrt{14}}$$=\frac{1}{2}$$\cos \theta =\cos {60}^{\circ }$$\theta ={60}^{\circ }$Hence, the correct option is (D).