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Math Help - [SOLVED] Area of right triangle with 3 vertices

  1. #1
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    [SOLVED] Area of right triangle with 3 vertices

    I want to find the area of a right triangle with vertices P,Q,R.

    P (1,-1,2)
    Q (0,3,4)
    R (6,1,8)

    Area of right triangle is |PQ x PR| / 2

    vector PQ (-1,4,2)
    vector PR (5,2,6)
    cross product vector PQ and vector PR (20,-16,-22)

    The area should be |20^2 + -16^2 + -22^2| / 2

    = square root (20^2 + -16^2 + -22^2) / 2

    = 570

    I think I got this, I'm just doubting the formula. I have trouble knowing when |...| means what if it's absolute value or the square root.
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  2. #2
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    "| |" never means "square root". It means absolute value if it is around a number, length if it is around a vector. Of course, you then use the square root to find the length of the vector (and to find an absolute value: |x|= \sqrt{x^2}) but it is incorrect to say that it means square root!

    By the way, this is NOT a right triangle but, fortunately, the area of any triangle, "right" or not, with two sides formed by vectors \vec{u} and \vec{v} is \frac{1}{2}|\vec{u}\times\vec{v}|. There the | | clearly means "length" (or "magnitude") because \vec{u}\times\vec{v} is a vector.
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  3. #3
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    I see, so the abs value is put there since an area really isn't negative. How are my calculations?
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