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Math Help - Area of a Parallelogram

  1. #1
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    Area of a Parallelogram

    I'm given 4 points and asked to find the area of the parallelogram.

    K(1,2,3) , L(1,3,6), M(3,8,6), and N(3,7,3)

    I want to use the cross product to do this. Don't I just have to take the cross product of KN and KL? I did this and got an answer of 11 cubic units which is not the answer. It's 16.28 and I'm not seeing what I did wrong =/
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  2. #2
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    \vec{KL} = \langle 0, 1, 3 \rangle
    \vec{KN} = \langle 2, 5, 0 \rangle

    \vec{KL} \times \vec{KN} = \langle -15, 6, -2 \rangle

    |\vec{KL} \times \vec{KN}| = \sqrt{(-15)^2 + 6^2 + (-2)^2} \approx 16.28
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