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

##### 1 Answer

#### Explanation:

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

If we call the first vector

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#vecB# , which is

#B = sqrt((B_x)^2 + (B_y)^2 + (B_z)^2)#

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

We therefore have

Now all we have to do is multiply the scalar through all components of