What is the projection of # <-3,0,1># onto #<1,2,-4>#?

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Answer 1 · 2017-10-28T12:25:30

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

The projection of #vecb# onto #veca# is

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

Here,

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

#vecb= <-3,0,1>#

The dot product is

#veca.vecb=<1,2,-4>.<-3,0,1>=(1)xx(-3)+(2)xx(0)+(-4)xx(1)=-3-4=-7#

The modulus of #veca# is

#=||veca|| = ||<1,2,-4>||= sqrt((1)^2+(2)^2+(-4)^2)=sqrt(1+4+16)=sqrt21#

Therefore,

The projection is

#proj_(veca)vecb = (-7)/(21)<1,2,-4> =-1/3<1,2,-4> #

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