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

1 Answer
Nathan L.
Jul 13, 2017

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 >

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