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Thread: Symmetric equation of the line of intersection of the two planes

  1. #1
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    Symmetric equation of the line of intersection of the two planes

    given the two planes

    3x+4y-5z=12
    4x+y+z=15

    how can I find the symmetric equation of the line of intersection of the two planes.

    i got the cross product=9i-23j-13k
    but why is the point (3,2,1) a point on both the planes?
    Last edited by mr fantastic; September 21st 2010 at 02:48 PM. Reason: Re-titled.
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  2. #2
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    You have to get any point which is found on both planes. The point you take can be any point, as long as it satisfies both equations.

    I could take for example,

    If I take x = 1, I get z = 17/3 and y = 28/3

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  3. #3
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    Hello, statman101!

    The cross-product approach is the best way.

    Here's another (primitive and longer) approach.
    (Use it when you're desperate.)


    \text{Given the two planes: }\;\begin{Bmatrix}3x+4y-5z& =& 12 & [1] \\ 4x+y+z&=& 15 & [2] \end{Bmatrix}

    \text{Find the symmetric equation of the line of intersection of the two planes.}

    \begin{array}{cccccccc}<br />
\text{Multiply [2] by -4:} & \text{-}16x - 4y - 4z &=& \text{-}60 \\<br />
\text{Add [1]:} & 3x + 4y - 5x &=& 12 \end{array}

    And we have: . -13x - 9z \:=\:-48 \quad\Rightarrow\quad x \:=\:\dfrac{48-9z}{13}

    Substitute into [2]: . 4\left(\frac{48-9z}{13}\right) + y + z \:=\:15 \quad\Rightarrow\quad y \:=\:\dfrac{3+23z}{13}

    So we have: . \begin{Bmatrix}x &=& \frac{48-9z}{13} &=& \frac{48}{13} - \frac{9}{13}z\\ \\[-3mm] <br />
y &=& \frac{3 + 23z}{13} &=& \frac{3}{13} + \frac{23}{13}z \\ \\[-3mm]  z &=& z \end{Bmatrix}


    On the right side, replace \,z with a parameter \,t.

    . . . . . \begin{Bmatrix} x &=& \frac{48}{13} - \frac{9}{13}t \\ \\[-3mm]<br />
y &=& \frac{3}{13} + \frac{23}{13}t \\ \\[-3mm]<br />
z &=& 0 + t \end{Bmatrix}


    These are the parametric equations of the line through \left(\frac{48}{13},\:\frac{3}{13},\:0\right)

    . . with direction vector: . \vec v \:=\:\langle \text{-}\frac{9}{13},\:\frac{23}{13},\:1\rangle \;=\;\langle \text{-}9,\:23,\:13\rangle


    The symmetric equation is: . \displaystyle \frac{x-\frac{48}{13}}{-9} \;=\;\frac{y-\frac{3}{13}}{23} \;=\;\frac{z-0}{1}



    Note: There are fifteen brizillion different forms of the symmetric equation.
    . . . . .This is just one of them.
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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, statman101!

    The cross-product approach is the best way.

    Here's another (primitive and longer) approach.
    (Use it when you're desperate.)



    \begin{array}{cccccccc}<br />
\text{Multiply [2] by -4:} & \text{-}16x - 4y - 4z &=& \text{-}60 \\<br />
\text{Add [1]:} & 3x + 4y - 5x &=& 12 \end{array}

    And we have: . -13x - 9z \:=\:-48 \quad\Rightarrow\quad x \:=\:\dfrac{48-9z}{13}

    Substitute into [2]: . 4\left(\frac{48-9z}{13}\right) + y + z \:=\:15 \quad\Rightarrow\quad y \:=\:\dfrac{3+23z}{13}

    So we have: . \begin{Bmatrix}x &=& \frac{48-9z}{13} &=& \frac{48}{13} - \frac{9}{13}z\\ \\[-3mm] <br />
y &=& \frac{3 + 23z}{13} &=& \frac{3}{13} + \frac{23}{13}z \\ \\[-3mm]  z &=& z \end{Bmatrix}


    On the right side, replace \,z with a parameter \,t.

    . . . . . \begin{Bmatrix} x &=& \frac{48}{13} - \frac{9}{13}t \\ \\[-3mm]<br />
y &=& \frac{3}{13} + \frac{23}{13}t \\ \\[-3mm]<br />
z &=& 0 + t \end{Bmatrix}


    These are the parametric equations of the line through \left(\frac{48}{13},\:\frac{3}{13},\:0\right)

    . . with direction vector: . \vec v \:=\:\langle \text{-}\frac{9}{13},\:\frac{23}{13},\:1\rangle \;=\;\langle \text{-}9,\:23,\:13\rangle


    The symmetric equation is: . \displaystyle \frac{x-\frac{48}{13}}{-9} \;=\;\frac{y-\frac{3}{13}}{23} \;=\;\frac{z-0}{1}



    Note: There are fifteen brizillion different forms of the symmetric equation.
    . . . . .This is just one of them.
    Only fifteen brizillion? I could have sworn there were more!
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