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

1 Answer
Oct 28, 2017

Answer:

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

Explanation:

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>#