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Prove a vector is perpendicular to a plane?

  1. Jan 16, 2008 #1
    1. The problem statement, all variables and given/known data
    Vectors A, B and C are vectors from the origin to the points a, b, c respectively, and the Vector D is defined as

    D= (AxB)+(BxC)+(CxA)

    Show that D is Perpendicular to the plane in which the points A, B, and C lie

    2. Relevant equations
    Cross Product

    3. The attempt at a solution
    I tried writing everything out doing the Det of each Cross product to find anything to cancel with out any luck.

    I also understand that if a b and c are points lying on a plane, then crossing each of these vectors AxB, BxC, and CxA should give me a vector perpendicular to that plane...how do I show this mathmatically?

    Thanks for the help =)
  2. jcsd
  3. Jan 16, 2008 #2


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    Couple of things I suppose. The dot product of perpendicular vectors is zero. so D dotted into the vector from A to B or B to C or C to A etc. should all be zero. Do you know the vector equation of a plane?
    Last edited: Jan 16, 2008
  4. Jan 16, 2008 #3

    n . (r->p) = 0

    if n is the normal vector ( D in this case I think )
    r and p are the points in the plane ( A->B, B->C, C->A)

    I'm still not understanding how it relates =(
  5. Jan 17, 2008 #4

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    Hi dangsy, a shortcut (if it's allowed by your teacher)!

    Three points always lie in a plane -- that we know. Chose the point 'a' as the origin.

    Then A = 0, which gives,

    D = BXC, which we know is perp to the plane.
  6. Jan 17, 2008 #5
    Let [tex]\overrightarrow{n}[/tex]= Normal vector of the plane where points a, b and c lie on.

    \overrightarrow{n} &= \overrightarrow{ab}\ x\ \overrightarrow{bc}
    &= (\overrightarrow{B} - \overrightarrow{A})\ x\ (\overrightarrow{C} - \overrightarrow{B})
    &= [(\overrightarrow{B} - \overrightarrow{A})\ x\ \overrightarrow{C}] - [(\overrightarrow{B} - \overrightarrow{A})\ x\ \vec{B}]
    &= [(\overrightarrow{B} \ x\ \overrightarrow{C})\ - (\overrightarrow{A} \ x\ \overrightarrow{C})] \ -\ [(\overrightarrow{B} \ x\ \overrightarrow{B})\ - (\overrightarrow{A} \ x\ \overrightarrow{B})]
    &= [(\overrightarrow{B} \ x\ \overrightarrow{C})\ + (\overrightarrow{C} \ x\ \overrightarrow{A})] \ +\ (\overrightarrow{A} \ x\ \overrightarrow{B})]
    &=(\overrightarrow{A} \ x\ \overrightarrow{B})\ + (\overrightarrow{B} \ x\ \overrightarrow{C}) \ +\ (\overrightarrow{C} \ x\ \overrightarrow{A})

    \ Note\ that: -(\overrightarrow{A} \ x\ \overrightarrow{C}) = \overrightarrow{C} \ x\ \overrightarrow{A}\ and \ \overrightarrow{B} \ x\ \overrightarrow{B} = \overrightarrow{0}
    &\overrightarrow{n} = \overrightarrow{D}
    &\ which\ means\ \overrightarrow{n} \ is \ parallel \ with\ \overrightarrow{D}
    &\overrightarrow{n}\ is \ perpendicular \ to \ the \ plane.\ Therefore,\ \overrightarrow{D} \ is\ also \ perpendicular\ to \ the \ plane.

    This Latex is extremely difficult to use.
    Last edited: Jan 17, 2008
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