Question

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

Answer 1

`proj_ (vecB)vecA = < -32/15, 8/3, -16/15 >`

Explanation:

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

If we call the first vector `vecA` and the second one `vecB`, the projection of `vecA` onto `vecB` (notated `proj_(vecB)vecA`) 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 `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

`proj_(vecB)vecA = (color(red)(-24))/((color(green)(sqrt45)^2))vecB`

`= -8/15vecB`

Now all we have to do is multiply the scalar through all components of `vecB` to find our vector projection:

`= < (-8/15)4, (-8/15)-5, (-8/15)2 >`

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