What is the projection of #<2,-4,3 ># onto #<1,2,2 >#? Physics 2D Motion Vector Operations 1 Answer Cosmic Defect Feb 28, 2016 #\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. Answer link Related questions What are vectors used for? Why vectors cannot be added algebraically? How do we represent the magnitude of a vector in physics? How do you find the equation of a vector orthogonal to a plane? Why are vectors important? How does a vector quantity differ from a scalar quantity? How can I calculate the magnitude of vectors? How do vectors subtract graphically? How do force vectors affect an object in motion? How can vectors be represented? See all questions in Vector Operations Impact of this question 1100 views around the world You can reuse this answer Creative Commons License