Question

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

Answer 1

Let $\overrightarrow{a}=2\hat{i}+\hat{j}+3\hat{k},\overrightarrow{b}=3\hat{i}-2\hat{j}+\hat{k}$

We know that $\overrightarrow{a}\cdot \overrightarrow{b}=|\overrightarrow{a}||\overrightarrow{b}|cos\theta$, where $\theta$ is the angle between $a$ and $b$.

$\overrightarrow{a}\cdot \overrightarrow{b}=(2\hat{i}+\hat{j}+3\hat{k})\cdot (3\hat{i}-2\hat{j}+\hat{k})$

$=6-2+3$

$\overrightarrow{a}\cdot \overrightarrow{b}=7$

$|\overrightarrow{a}|=\sqrt{2^2+1^2+3^2}=\sqrt{4+1+9}=\sqrt{14}$

$|\overrightarrow{b}|=\sqrt{3^2+(-2{)}^2+1^2}=\sqrt{9+4+1}=\sqrt{14}$

Substitute the values in $\overrightarrow{a}\cdot \overrightarrow{b}=|\overrightarrow{a}||\overrightarrow{b}|cos\theta$ we get

$7=\sqrt{14}\cdot \sqrt{14}cos\theta$

$\Rightarrow 7=14cos\theta$

$\Rightarrow cos\theta =\frac{7}{14}$

$\Rightarrow cos\theta =\frac{1}{2}$

$\Rightarrow \theta =co{s}^{-1}\left(\frac{1}{2}\right)$

$\Rightarrow \theta =\frac{\pi }{3}$