Question

If $\theta$ is the angle between two vectors $\hat{i}-2\hat{j}+3\hat{k}$ and $3\hat{i}-2\hat{j}+\hat{k}$, find $\theta$.

Answer 1

Given vectors are $\overrightarrow{a}=\hat{i}-2\hat{j}+3\hat{k}$ and $\overrightarrow{b}=3\hat{i}-2\hat{j}+\hat{k}$,

If $\theta$ is the angle between the given vectors, then

$\cos \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}$

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

$=3+4+3=10$

Also, $|\overrightarrow{a}|=\sqrt{1^2+(-2{)}^2+3^2}=\sqrt{1+4+9}=\sqrt{14}$ and

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

$\therefore \cos \theta =\frac{10}{14}=\frac{5}{7}$

$\Rightarrow \theta ={\cos }^{-1}\left(\frac{5}{7}\right)$

Hence, $\theta ={\cos }^{-1}\left(\frac{5}{7}\right)$ is the angle between the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$.