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How do I prove that A^B is orthogonal to A?

  1. Nov 9, 2011 #1
    If A = (2,-2,1) and B = (2, 0, -1)

    show by explicit calculation that;

    i) A^B is orthogonal to A

    ii) (A^B)^B lies in the same plane as A and B by expressing it as a linear combination of A and B

    I'm using;
    A^B = |A||B|sin θ

    I know that when you do the cross product of two vectors the result will be a vector that is perpendicular to both and I can draw a diagram to demonstrate. However I can't show it by explicit calculation.

    I thought maybe if I could prove that the angle between them was ∏/2...
     
  2. jcsd
  3. Nov 9, 2011 #2

    phyzguy

    User Avatar
    Science Advisor

    When you write A^B = |A||B|sin θ, that only gives the magnitude of A^B. You need to include the direction information as well. In three dimensions the dual of A^B is a vector, which is given by the cross product. So, for part A, try forming the dot product of (AxB) with A and see what this gives.
     
  4. Nov 9, 2011 #3

    Mark44

    Staff: Mentor

    Since you're talking about the cross product, why don't you write it as A x B?
     
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