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Thread: Calculating the unit normal of a plane

  1. #1
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    Calculating the unit normal of a plane

    Hi,
    I've tried to solve a problem and I was just wondering if my answer is correct or not:

    Assume three points $\displaystyle p_1, p_2$ and $\displaystyle p_3$:
    $\displaystyle p_1 = (1,2,3)^T$
    $\displaystyle p_2 = (2,2,1)^T$
    $\displaystyle p_3 = (0,1,3)^T$

    The problem is to find the unit normal of this plane.

    I take the cross product of $\displaystyle p_1p_2$ and $\displaystyle p_1p_3$, I get:

    $\displaystyle N = -2i + 2j - k$
    I find that the length of N is $\displaystyle sqrt((-2)^2 + 2^2 + (-1)^2) = 3$.

    I.e. the answer is: $\displaystyle (-2i + 2j - k) / 3$

    Is this correct?
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    Correct.
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