What is the projection of #(3i + 2j - 6k)# onto # (-2i- 3j + 2k)#?

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Answer 1 · 2017-11-06T07:54:18

The projection is #= <48/17,72/17,-48/17>#

Let #vecb=<3,2,-6># and #veca=<-2,-3,2>#

The projection of #vecb# onto #veca# is

#proj_(veca)vecb=(veca.vecb)/(||veca||^2)veca#

#veca.vecb = <-2,-3,2> . <3,2,-6> = (-2) * (3)+(-3) * (2)+(2) * (-6) = -6-6-12=-24#

#||veca||=||<-2,-3,2>||=sqrt((-2)^2+(-3)^2+(-2)^2)=sqrt(4+9+4) = sqrt17#

Therefore,

#proj_(veca)vecb=(veca.vecb)/(||veca||^2)veca=-24/17 <-2,-3,2>#