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Proving the vector OA is normal to the plane OBC

  1. Aug 19, 2012 #1
    1. The problem statement, all variables and given/known data

    Given position vectors: OA ( i + 2j - k) ; OB ( -i + 2j +3k); OC ( 2i + j + 4k)

    Given that OA is perpendicular to OB.

    The Question : Show that OA is normal to the plane OBC.

    2. Relevant equations
    r . n = d

    To find the normal of the plane OBC, I used n = BO x BC

    d = OB . n

    3. The attempt at a solution

    Equation of plane OBC:

    BO = -OB = i - 2j - 3k

    BC = OC - OB = (2,1,4) - (-1,2,3) = (3,-1,1)

    n = BO x BC = (5,10,-5)

    d = OB . n = (-1,2,3) . (5,10,-5) = 0

    ∴ Equation of plane OBC = r. (5,10,-5) = 0


    How do I show that OA is normal to the plane? I know that it is related to the statement - 'OA is perpendicular to OB'. However, I do not know how can it be applied into the question.
    I hope someone can help me out.

    Thank you for your time! :smile:
     
  2. jcsd
  3. Aug 19, 2012 #2

    CAF123

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    I agree with the normal vector to the plane OBC, that is [tex] \vec{n} = 5\vec{i} +10\vec{j} -5\vec{k}.[/tex]
    You are given that OA is perpendicular to OB, which you can easily verify.
    To show that OA is normal to the plane, show that the cross product of OA with [itex] \vec{n} [/itex] is 0.
     
  4. Aug 19, 2012 #3
    Thank you very much for your help! :smile:
     
  5. Aug 19, 2012 #4

    LCKurtz

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    Or even easier, observe that OA is a scalar multiple of ##\vec n##.
     
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