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Math Help - Vectors - area of parallelogram

  1. #1
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    Vectors - area of parallelogram

    Find an area of a parallelogram determined by the points P(7,-5,5), Q(-5,10,10), R(-8,8,-4) and S(-20,23,1).

    -->
    PQ = -12i + 15j + 5k
    -->
    RS = -12i + 15j + 5k
    --
    >
    PS = -27i + 28j - 4k
    -->
    RQ = 3i -2j -14k

    * This is what I have, I am lost at the moment. I have PQ and RS are parallel. I know I need to use the cross product. Any assistance on what to do next????
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  2. #2
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    I would actually compute PR and SQ. You want to use the cross product of two non-parallel outer sides (the area is the magnitude of said cross product). Because PS and RQ are not parallel or anti-parallel, you know that they are diagonals, and hence not so useful (you might be able to use them, but I'm not entirely sure how).
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  3. #3
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    The area is \left\| {\overrightarrow {PQ}  \times \overrightarrow {PS} } \right\|

    Is your question why is that the area?
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  4. #4
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    The objective is to find the area of the parallelogram determined by the folowing points:

    P(7,-5,5), Q(-5,10,10), R(-8,8,-4) and S(-20,23,1)

    Basically I just need to solve the following:

    The area is PQ x PS????
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  5. #5
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    Quote Originally Posted by sderosa518 View Post
    The objective is to find the area of the parallelogram determined by the folowing points:
    P(7,-5,5), Q(-5,10,10), R(-8,8,-4) and S(-20,23,1)
    The area is PQ x PS????
    NO It is \|\vec{PQ}\times\vec{PS}\|.

    Suppose that \vec{a}~\&~\vec{b} are adjacent sides of a parallelogram.

    Then the area of that parallelogram is \|\vec{a}\|\|\vec{b}\|\sin(\theta), where \theta is the angle between the vectors.

    Recall that \sin (\theta ) = \dfrac{{\left\| {\overrightarrow a  \times \overrightarrow b } \right\|}}{{\left\| {\overrightarrow a } \right\|\left\| {\overrightarrow b } \right\|}}.
    Last edited by Plato; March 17th 2011 at 01:32 PM.
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  6. #6
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    this is wat i got:

    -->
    PQ = -12i + 15j + 5k
    -->
    PS = -27i + 28j - 4k


    i j k
    PQ x PS = -12 15 5
    -27 28 -4

    = -200i + 357j + 69k

    = ll -200^2 + 357^2 + 69^2 ll

    = ll172210ll
    = 1312.253
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  7. #7
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    Quote Originally Posted by sderosa518 View Post
    -->
    PQ = -12i + 15j + 5k CORRECT
    -->
    PS = -27i + 28j - 4k CORRECT
    i j k
    PQ x PS = -12 15 5
    -27 28 -4
    = -200i + 357j + 69k WRONG
    \vec{PQ}\times\vec{PS}=-200i-183j+69k
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  8. #8
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    It doesnt seem right. May I ask why you choose PQ and PS???
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  9. #9
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    Quote Originally Posted by sderosa518 View Post
    It doesnt seem right. May I ask why you choose PQ and PS???
    Did you read reply #5 very carefully?
    Those two vectors are adjacent sides of a parallelogram.

    You just made an arithmetic error in the matrix of the cross product.
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  10. #10
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    Finale Answer:

    200^2 + 183^2 + 69^2 = sqrt (40000 + 33489 + 4761)


    =279.73
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  11. #11
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    That is what MathCad gives as the answer.
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