What is the projection of #< 5 ,- 3, 2 ># onto #< -1 , -2 , -4 >#?

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Answer 1 · 2018-03-01T17:44:44

The vector projection is #= <1/3,2/3,4/3> #

The vector projection of #vecb# onto #veca# is

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

#veca= <-1,-2,-4>#

#vecb= <5,-3,2>#

The dot product is

#veca.vecb= <-1,-2,-4> . <5, -3,2> = (-1)*(5)+(-2)*(-3)+(-4)*(2)#

#=-5+6-8=-7#

The modulus of #veca# is

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

#= sqrt(1+4+16)=sqrt(21)#

Therefore,

#proj_(veca)vecb=(-7)/(sqrt(21))^2* <-1,-2,-4>#

#=-1/3<-1,-2,-4>#

#= <1/3,2/3,4/3>#