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Math Help - The Cross Product

  1. #1
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    The Cross Product

    Thanks!!
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  2. #2
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    Using the normal, n, that you found and the point P
    n = \left[ {\begin{array}{*{20}c}<br />
   3  \\<br />
   3  \\<br />
   { - 3}  \\<br />
\end{array}} \right]\quad P = \left[ {\begin{array}{*{20}c}<br />
   1  \\<br />
   0  \\<br />
   { - 2}  \\<br />
\end{array}} \right]

    write the equation ot the plane
    \Pi :\;n \cdot (R - P) = 0,\quad R = \left[ {\begin{array}{*{20}c}<br />
   x  \\<br />
   y  \\<br />
   z  \\<br />
\end{array}} \right].
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  3. #3
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    Re:

    I'm still a bit unsure...

    Equation of the plane for n(3,3,-3)

    3x+3y-3x=0

    ???
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  4. #4
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    It should be written as 3\left( {x - 1} \right) + 3y - 3\left( {z + 3} \right) = 0.
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  5. #5
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    Re:

    So,

    3x-3y-3z=12

    Thanks Again,
    qbkr21
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