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Math Help - area of triangle using 3d vectors

  1. #1
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    area of triangle using 3d vectors

    can someone explain how to do this problem, i got the first part correct but i dont understand how to get the area exactly. THanks!
    Consider the points below. P(1, 0, 0), Q(0, -2, 0), R(0, 0, -3)

    (a) Find a nonzero vector orthogonal to the plane through the points P, Q, and R.
    Enter a number.
    6i + Enter a number.
    -3j + Enter a number.
    -2k

    (b) Find the area of the triangle PQR.
    Enter an exact number as an integer fraction or decimal.
    3







    NVM I FIGURED IT OUT THANKS FOR LOOKING AT IT!!
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  2. #2
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    Quote Originally Posted by ahawk1 View Post
    can someone explain how to do this problem, i got the first part correct but i dont understand how to get the area exactly. THanks!
    Consider the points below. P(1, 0, 0), Q(0, -2, 0), R(0, 0, -3)

    (a) Find a nonzero vector orthogonal to the plane through the points P, Q, and R.
    Enter a number.
    6i + Enter a number.
    -3j + Enter a number.
    -2k

    (b) Find the area of the triangle PQR.
    Enter an exact number as an integer fraction or decimal.
    3
    It will be half of the magnitude of the cross product of PQ and RQ
    PQ=\vec i +2 \vec j and RQ =2 \vec j - 3\vec k

    \begin{vmatrix}<br />
i & j & k \\<br />
1 & 2 & 0 \\<br />
0 & 2 & -3 \\<br />
\end{vmatrix} = (6-0)\vec i-(-3-0)\vec j+(2-0)\vec k=6 \vec i +3 \vec j +2\vec k

    Now we take half of the magnitude to get

    \frac{1}{2}\sqrt{(6)^2+(3)^2+(2)^2}=\frac{1}{2}\sq  rt{49}=\frac{7}{2}
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