# What is the projection of < -3 , 1 ,3 > onto < 8 , 8 , -4 >?

Oct 19, 2016

The answer is $\left(- \frac{14}{9} , - \frac{14}{9} , \frac{7}{9}\right)$

#### Explanation:

The vectorprojection of $\vec{u}$ onto $\vec{v}$ is
=(vecu.vecv.) /(vecv*vecv)*vecv
here $\vec{u} = \left(- 3 , 1 , 3\right)$ and $\vec{v} = \left(8 , 8 , - 4\right)$
Applying this we get
$= \frac{- 24 + 8 - 12}{64 + 64 + 16} \cdot \vec{v} = - \frac{28}{144} \cdot \vec{v} = - \frac{7}{36} \cdot \vec{v}$
so the answer is $\left(\frac{- 7 \cdot 8}{36} , \frac{- 7 \cdot 8}{36} , \frac{4 \cdot 7}{36}\right) = \left(- \frac{14}{9} , - \frac{14}{9} , \frac{7}{9}\right)$
or $\frac{7}{9} \cdot \left(- 2 , - 2 , 1\right)$