Question

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

Answer 1

Let the given vectors are $\overrightarrow{a}=\hat{i}-2\hat{j}+3\hat{k}$ and $\overrightarrow{b}=3\hat{i}-2\hat{j}+\hat{k}$

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

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

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

$=1.3+(-2)(-2)+3.1$

$=3+4+3=10$

Also, we know that $\overrightarrow{a}\cdot \overrightarrow{b}=|\overrightarrow{a}||\overrightarrow{b}|\cos \theta$

$\therefore 10=\sqrt{14}\sqrt{14}\cos \theta$

$\Rightarrow \cos \theta =\frac{10}{14}$

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