What is the projection of #<5,0,-2 ># onto #<1,-1,0 >#?

1 Answer
Feb 14, 2016

Answer:

#vec c=<5/2,-5/2,0>#

Explanation:

#vec a=<5,0,-2>#
#vec b=<1,-1,0>#
#vec a*vec b=5*1+0*(-1)+0*(-2)=5 #
#"dot product between vec a and vec b;"#
#|b|=sqrt(1^2+1^2+0^2)=sqrt 2#
#vec c=(vec a*vec b)/(|b|*|b|)*vec b#
#vec c=5/(sqrt 2*sqrt 2)*<1,-1,0>#
#vec c=5/2<1,-1,0>#
#vec c=<5/2,-5/2,0>#