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

Let$\vec a = \hat i - 2\hat j + 3\hat k$and$\vec b = 3\hat i - 2\hat j + \hat k$

Then, $|\vec a| = \sqrt {{1^2} + {{( - 2)}^2} + {3^2}} = \sqrt {1 + 4 + 9} = \sqrt {14}$
$|\vec b| = \sqrt {{3^2} + {{( - 2)}^2} + {1^2}} = \sqrt {9 + 4 + 1} = \sqrt {14}$

and $\vec a \cdot \vec b = \left( {\hat i - 2\hat j + 3\hat k} \right) \cdot \left( {3\hat i - 2\hat j + \hat k} \right)$

$= \left( 1 \right)\left( 3 \right) + \left( { - 2} \right)\left( { - 2} \right) + \left( 3 \right)\left( 1 \right) = 3 + 4 + 3 = 10$.

If $\theta$ be the angle between $\vec a$ and $\vec b$ then

$\cos \theta = \cfrac{{\vec a \cdot \vec b}}{{|\vec a||\vec b|}} = \cfrac{{10}}{{\sqrt {14} \sqrt {14} }} = \cfrac{{10}}{{14}} = \cfrac{5}{7} \Rightarrow \theta = {\cos ^{ - 1}}\cfrac{5}{7}$