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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