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

1 Answer
Feb 28, 2016

Answer:

#\vec{A_{}}# id perpendicular to #\vec{B_{}#. So one's projection on the other must be a null vector ( #<0,0,0>#)

Explanation:

The projection of a #\vec{A_{}}# onto another vector #\vec{B_{}}# is:

#\vec{A_B} = \frac{\vec{A_{}}.\vec{B_{}}}{B}\hat{B}=\frac{\vec{A_{}}.\vec{B_{}}}{B^2}\vec{B_{}}#

Solution: #\vec{A_{}}=<2,-4,3>; \qquad \vec{B_{}}=<1,2,2>;#

#\vec{A_{}}.\vec{B_{}}=(2\times1-4\times2+3\times2)=0#

Since #\vec{A_{}}.\vec{B_{}} = 0#, it is clear that #\vec{A_{}}# is perpendicular to #\vec{B_{}}#. So its projection on #\vec{B_{}}# is a null vector.