What is the projection of #<4,-6,8># onto #<4,-6,3 >#?

1 Answer
Jan 11, 2018

Answer:

The projection is #=76/61<4, -6,3>#

Explanation:

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

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

#veca=<4,-6,3>#

#vecb= <4, -6,8>#

The dot product is

#veca.vecb =<4,-6,3>. <4,-6,8> #

#= 4*4+(-6)*(-6)+3*8=16+36+24=76#

The modulus of #veca# is

#=||veca||=||<4,-6,3>|| =sqrt(4^2+(-6)^2+3^2)=sqrt61#

Therefore,

#proj_(veca)vecb=76/61<4, -6,3>#