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Math Help - urgent question!!! intersection of planes

  1. #1
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    urgent question!!! intersection of planes

    Okay I'm freaking out. I'm reviewing for my final tomorrow, and I realized I don't get the plane portion of the final. I can't find my notes on it.



    So for intersection of a plane. I'm gonna make up a problem and show how far I can get into it.

    Intersection between 2x + 3y - 4z = 12 and 5x + 2y + 3z = 7.

    So you take the cross product of [2, 3, 4] and [5, 2, 3], right? How do you find this cross product? is it [10, 6, 12] or something else?

    Then after you find that...say it is [1,2,3] even though it isn't.

    So set x to 0 for the two planes. Then solve for y and z. Say it turns out to be [0, 6, 7].

    then it is [0 6, 7] + t(1,2,3). is that right?

    Okay urgent how do you find a cross product?? I have no idea how.
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  2. #2
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    Quote Originally Posted by view360 View Post
    So for intersection of a plane. I'm gonna make up a problem and show how far I can get into it.

    Intersection between 2x + 3y - 4z = 12 and 5x + 2y + 3z = 7.

    So you take the cross product of [2, 3, 4] and [5, 2, 3], right? How do you find this cross product? is it [10, 6, 12] or something else?

    Then after you find that...say it is [1,2,3] even though it isn't.

    So set x to 0 for the two planes. Then solve for y and z. Say it turns out to be [0, 6, 7].

    then it is [0 6, 7] + t(1,2,3). is that right?

    Okay urgent how do you find a cross product?? I have no idea how.
    I hope that you are more careful on your exam than you are here.
    The normal is \left\langle {2,3, - 4} \right\rangle not [MATh]\left\langle {2,3, 4} \right\rangle [/tex]
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  3. #3
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    Sorry, was trying to figure out the problem on my paper, and I was trying it both ways. I must have mixed them up.

    Can anyone help me with this? I can't find my notes on how to cross product, and I tried googling it but I did not learn anything even close to what it is pulling up?
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  4. #4
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    Hello, view360!

    I have an entire different approach.
    I'll use my own example . . .


    Intersection between: . 2x - y +z \:=\: 3\:\text{ and }\:3x + 2y - 3z \:=\: 9

    We know that the intersection of two planes is a line.


    We try to solve the system of equations: . \begin{array}{cccc}2x - y + z &=& 3 & [1] \\ 3x + 2y - 3z &=& 9 & [2] \end{array}

    Eliminate one of the variables.

    . . \begin{array}{cccc}\text{Multiply [1] by 2:} & 4x - 2y + 2z &=& 6 \\<br />
\text{Add [2]:} & 3x + 2y - 3z &=& 9 \end{array}

    And we have: . 7x - z \:=\:15 \quad\Rightarrow\quad{\color{blue} z \:=\:7x - 15}

    Substitute into [1]: . 2x - y + (7x - 15) \:=\:3 \quad\Rightarrow\quad 9x - y \:=\:18 \quad\Rightarrow\quad{\color{blue} y \:=\:9x - 18}


    We have these equations: . \begin{array}{ccc}x &=&x \\ y &=& 9x - 18 \\ z &=& 7x-15\end{array}


    On the right side, replace x with a parameter t\!:\;\;\begin{Bmatrix}x &=& t \\ y &=& 9t - 18 \\ z&=& 7t-15 \end{Bmatrix}

    There are the parametric equations of the line of intersection.

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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, view360!

    I have an entire different approach.
    I'll use my own example . . .



    We know that the intersection of two planes is a line.


    We try to solve the system of equations: . \begin{array}{cccc}2x - y + z &=& 3 & [1] \\ 3x + 2y - 3z &=& 9 & [2] \end{array}

    Eliminate one of the variables.

    . . \begin{array}{cccc}\text{Multiply [1] by 2:} & 4x - 2y + 2z &=& 6 \\<br />
\text{Add [2]:} & 3x + 2y - 3z &=& 9 \end{array}

    And we have: . 7x - z \:=\:15 \quad\Rightarrow\quad{\color{blue} z \:=\:7x - 15}

    Substitute into [1]: . 2x - y + (7x - 15) \:=\:3 \quad\Rightarrow\quad 9x - y \:=\:18 \quad\Rightarrow\quad{\color{blue} y \:=\:9x - 18}


    We have these equations: . \begin{array}{ccc}x &=&x \\ y &=& 9x - 18 \\ z &=& 7x-15\end{array}


    On the right side, replace x with a parameter t\!:\;\;\begin{Bmatrix}x &=& t \\ y &=& 9t - 18 \\ z&=& 7t-15 \end{Bmatrix}

    There are the parametric equations of the line of intersection.

    That makes so much more sense! I like your method a lot, finding x, y, and z just is so much more logical, lol.

    one more question -- is this the type of thing you can put into [x,y,z] + t(a,b,c) form? Cuz for most of our plane things I remember we used that.
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  6. #6
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    Hello again, view360!

    Is this the type of thing you can put into (x_1,y_1,z_1) + t(a,b,c) form? . Yes!

    We have: . \begin{Bmatrix}x &=& t \\ y &=& 9t-18 \\ z &=&7t - 15 \end{Bmatrix}

    I like to re-write them: . \begin{Bmatrix} x &=& 0 + 1t \\ y &=& \text{-}18 + 9t \\ z &=& \text{-}15 + 7y \end{Bmatrix}


    Therefore: . (x,y,z) \;=\;(0,\text{-}18,\text{-}15) + t(1,9,7)

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