What is the projection of #<5,-2,3 ># onto #<1,-6,-3 >#?

1 Answer
Jan 12, 2018

Answer:

The projection is #=4/23<1, -6,-3>#

Explanation:

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

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

#veca=<1,-6,-3>#

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

The dot product is

#veca.vecb =<1,-6,-3>. <5,-2,3> #

#= 1*5+(-6)*(-2)+(-3)*3=5+12-9=8#

The modulus of #veca# is

#=||veca||=||<1,-6,-3>|| =sqrt(1^2+(-6)^2+(-3)^2)=sqrt46#

Therefore,

#proj_(veca)vecb=8/46<1, -6,-3> = 4/23<1, -6,-3>#