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Math Help - Area of Triangle using cross product

  1. #1
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    Area of Triangle using cross product

    A triangle has vertices A(-2,1,3), B(7,8,-4), and C(5,0,2). Determine the area of triangle ABC.

    Vector AB = B-A = [9,7,-7]
    Vector AC = C-A = [7, -1, -1]

    Area of triangle = (ACxAB)/2

    I end up getting vector [-14, -40, -40] and the magnitude is not the correct answer of 35.9 square units. Can someone help?
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  2. #2
    Senior Member Spec's Avatar
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    AB \times AC=(-14,-40,-58)
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  3. #3
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    Hello, millerst!

    A small error . . .


    A triangle has vertices A(-2,1,3), B(7,8,-4), and C(5,0,2).
    Determine the area of triangle ABC.

    \overrightarrow{AB} \:=\:\langle9,7,\text{-}7\rangle
    \overrightarrow{AC} \:=\: \langle7, \text{-}1, \text{-}1\rangle

    Area of triangle = \:\frac{\overrightarrow{AC} \times \overrightarrow{AB}}{2} . . . . Correct so far!

    I end up getting vector \langle \text{-}14, \text{-}40, \text{-}40\rangle . ← Here!

    The correct answer is 35.9 unitsē.
    Can someone help?

    The cross product is: . \langle\text{-}14, \text{-}40, {\color{blue}\text{-}58}\rangle



    Edit: Too fast for me, spec!
    .
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