If the vertices $A,B,C$ of a triangle $ABC$ are$(1,2,3),( - 1,0,0),(0,1,2)$ , respectively, then find $\angle ABC.[\angle ABC$ is the angle between the vectors $\overrightarrow {BA}$ and $\overrightarrow {BC}$].

If $O$ be the origin, then
$\overrightarrow {OA} = \hat i + 2\hat j + 3\hat k,\overrightarrow {OB} = - \hat i$ and $\overrightarrow {OC} = \hat j + 2\hat k$

$\therefore$ $\overrightarrow {BC} = \overrightarrow {OC} - \overrightarrow {OB} = (\hat j + 2\hat k) - ( - \hat i) = \hat i + \hat j + 2\hat k$

and $\overrightarrow {BA} = \overrightarrow {OA} - \overrightarrow {OB} = (\hat i + 2j + 3\hat k) - ( - \hat i) = 2\hat i + 2j + 3\hat k$

$\therefore$ $\cos \angle ABC = \cfrac{{\overrightarrow {BC} \cdot \overrightarrow {BA} }}{{|\overrightarrow {BC} ||\overrightarrow {B \cdot A} |}} = \cfrac{{(\hat i + \hat j + 2\hat k) \cdot (2\hat i + 2\hat j + 3\hat k)}}{{\sqrt {{1^2} + {1^2} + {2^2}} \sqrt {{2^2} + {2^2} + {3^2}} }}$

$= \cfrac{{(1)(2) + (1)(2) + (2)(3)}}{{\sqrt {1 + 1 + 4} \sqrt {4 + 4 + 9} }} = \cfrac{{2 + 2 + 6}}{{\sqrt 6 \sqrt {17} }} = \cfrac{{10}}{{\sqrt {102} }}$

Hence, $\angle ABC = {\cos ^{ - 1}}\left( {\cfrac{{10}}{{\sqrt {102} }}} \right)$