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Math Help - Area of A triangle

  1. #1
    Junior Member
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    Area of A triangle

    Thank you so much to everyone that is helping me
    Find the area of the triangle with vertices , , and
    so for this one i found 2 vectors
    4i-j+0k
    -6i+10j+0k

    took the cross product of the two to get 40k - (6k) which equals 34k, so
    where am i going wrong?
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  2. #2
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    Hello, Allie!

    You did nothing wrong . . . leep going!


    Find the area of the triangle with vertices (0,-3), (4,-4) and (-2,6)


    So for this one i found 2 vectors: . 4\vec i - \vec j + 0\vec k\:\text{ and }\;-6\vec i + 10\vec j + 0\vec k . . . . Right!

    took the cross product of the two to get: . 40\vec k - 6\vec k \:=\:34\vec k . . . . Yes!

    Where am i going wrong?

    Recall that the magnitude of the cross product gives the area of the parallelogram.


    Area of parallelogram: . |34\vec k| \:=\:34

    Therefore, area of triangle: . \frac{1}{2}\times 34 \;=\;17

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  3. #3
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    ohh yess i see now thank you so much
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