# What is the angle between <7,-3,1>  and < -1,3,2 >?

Angl between above vctors ${119.2}^{0}$
Let  vec u =[7, -3, 1] ; vec v=[-1,3,2] and $\theta$ be the angle between them. We know $\cos \theta = \frac{\vec{u} . \vec{v}}{| | \vec{u} | | . | | \vec{v} | |}$ Now $\vec{u} . \vec{v} = \left(7 \cdot - 1\right) + \left(- 3 \cdot 3\right) + \left(1 \cdot 2\right) = - 14$ ; $| | \vec{u} | | = \sqrt{{7}^{2} + {\left(- 3\right)}^{2} + {1}^{2}} = \sqrt{59}$ and $| | \vec{v} | | = \sqrt{{\left(- 1\right)}^{2} + {\left(3\right)}^{2} + {2}^{2}} = \sqrt{14}$ $\therefore \cos \theta = - \frac{14}{\sqrt{59} \cdot \sqrt{14}} = - .487 \mathmr{and} \theta = {\cos}^{-} 1 \left(- .487\right) = {119.2}^{0}$[Ans]