What is the projection of #(32i-38j-12k)# onto # (18i -30j -12k)#?

1 Answer
Feb 14, 2016

Answer:

#vec c= <24,47i,-40,79j,-16,32k>#

Explanation:

#vec a= <32i,-38j,-12k>#
#vec b= <18i,-30j,-12k>#
#vec a * vec b =18*32+38*30+12*12=#
#vec a * vec b=576+1140+144=1860#
#|b|=sqrt(18^2+30^2+12^2)#
#|b|=sqrt(324+900+144)#
#|b|=sqrt1368#
#vec c=(vec a*vec b)/(|b|*|b|)*vec b#
#vec c=1860/(sqrt 1368 *sqrt 1368)<18i,-30j,-12k>#
#vec c=1860/1368<18i,-30j,-12k>#
#vec c= <(1860*18i) /1368, (-1860*30j)/1368,(-1860*12k)/1368>#
#vec c= <24,47i,-40,79j,-16,32k>#