What is the projection of #<2,4,3 ># onto #<-3,4,-1 >#?

1 Answer
Nov 29, 2017

Answer:

The projection is #=7/26 <-3,4,-1>#

Explanation:

The vector projection of #vecv# onto #vecu# is

#proj_uv= (vec u. vecv ) /(||vecu||)^2*vecu#

Here,

#vecv= < 2,4,3> # and

#vecu = <-3,4,-1>#

The dot product is

#vecu.vecv =<2,4,3> . <-3,4,1-> #

#=(2*-3)+(4*4)+(3*-1) =-6+16-3=7#

The magnitude of #vecu# is

#=||<-3,4,-1 >|| = sqrt((-3)^2+(4)^2+(-1)^2)#

#=sqrt(9+16+1)=sqrt26#

The vector projection is

#proj_uv=7/26* <-3,4,-1>#