# What is the projection of <5,8,-2 > onto <4,-5,2 >?

##### 1 Answer
Jul 13, 2017

$p r o {j}_{\vec{B}} \vec{A} = < - \frac{32}{15} , \frac{8}{3} , - \frac{16}{15} >$

#### Explanation:

We're asked to find the projection of a vector onto another.

If we call the first vector $\vec{A}$ and the second one $\vec{B}$, the projection of $\vec{A}$ onto $\vec{B}$ (notated $p r o {j}_{\vec{B}} \vec{A}$) is found using the formula

proj_(vecB)vecA = (vecA·vecB)/(B^2)vecB

where

• vecA · vecB is the dot product of the two vectors

This is found using the equation

vecA·vecB = A_xB_x + A_yB_y + A_zB_z

So

vecA·vecB = (5)(4) + (8)(-5) + (-2)(2) = color(red)(-24

• $B$ is the magnitude of $\vec{B}$, which is

$B = \sqrt{{\left({B}_{x}\right)}^{2} + {\left({B}_{y}\right)}^{2} + {\left({B}_{z}\right)}^{2}}$

B = sqrt((4)^2 + (-5)^2 + (-2)^2) = color(green)(sqrt45

We therefore have

$p r o {j}_{\vec{B}} \vec{A} = \frac{\textcolor{red}{- 24}}{\left({\textcolor{g r e e n}{\sqrt{45}}}^{2}\right)} \vec{B}$

$= - \frac{8}{15} \vec{B}$

Now all we have to do is multiply the scalar through all components of $\vec{B}$ to find our vector projection:

$= < \left(- \frac{8}{15}\right) 4 , \left(- \frac{8}{15}\right) - 5 , \left(- \frac{8}{15}\right) 2 >$

= color(red)( < -32/15, 8/3, -16/15 >