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Math Help - Line of intersection of two planes

  1. #1
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    Line of intersection of two planes

    How would you find the line of intersection of the planes:

    \mathbf{x}\cdot\begin{pmatrix}1\\-1\\3\end{pmatrix}=0 and \mathbf{x}=\lambda\begin{pmatrix}2\\1\\2\end{pmatr  ix}+\mu\begin{pmatrix}3\\1\\-3\end{pmatrix}
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  2. #2
    Member Haven's Avatar
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    We have:

    \mathbf{x}\cdot\begin{pmatrix}1\\-1\\3\end{pmatrix}=0 (1)

    \mathbf{x}=\lambda\begin{pmatrix}2\\1\\2\end{pmatr  ix}+\mu\begin{pmatrix}3\\1\\-3\end{pmatrix} (2)

    If x lies on the intersection, then we must have that x satisfies both (1) and (2)

    So.
    \left(\lambda\begin{pmatrix}2\\1\\2\end{pmatrix}+\  mu\begin{pmatrix}3\\1\\-3\end{pmatrix}\right) \cdot \begin{pmatrix}1\\-1\\3\end{pmatrix} = 0
    \Rightarrow \lambda\begin{pmatrix}2\\1\\2\end{pmatrix} \cdot \begin{pmatrix}1\\-1\\3\end{pmatrix} +\mu\begin{pmatrix}3\\1\\-3\end{pmatrix} \cdot \begin{pmatrix}1\\-1\\3\end{pmatrix} = 0
    \Rightarrow 7\lambda + -7\mu = 0<br />
\Rightarrow \lambda = \mu
    Which determines a line in \mathbb{R}^2
    Last edited by Haven; June 3rd 2010 at 10:41 PM.
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  3. #3
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    Ref attachment, you will know how to do
    Traditional method is complex.

    To introduce new ***

    Cross product could be replaced sometimes, im(A)^{\perp}=ker(A^T)

    For example:

    Plane 2 ( V_2) equation \mathbf{x}=\lambda\begin{pmatrix}2\\1\\2\end{pmatr  ix}+\mu\begin{pmatrix}3\\1\\-3\end{pmatrix}

    It means V_{2}=im(\begin{bmatrix}2\\1\\2\end{bmatrix},\begi  n{bmatrix}3\\1\\-3\end{bmatrix})

    Normal vector of Plane 2 is \vec{v}_2=V_{2}^{\perp}=im(A)^{\perp}=ker(A^T)

    By this method , we could skip cross product especially when dimension is more than 3

    Finally, line cross original point, equation

    <br />
\frac{x}{5}=\frac{y}{2}=\frac{z}{-1}<br />

    ***
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    Last edited by math2009; June 7th 2010 at 03:35 AM.
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