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Math Help - Parametic equation of the straight line

  1. #1
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    Parametic equation of the straight line

    Find the parametic equation of the line given by x+y-z=2,2x+3y+z=4

    Attempt to solution:

    Is the line given by this equation the intersection of these two planes ?
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  2. #2
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    Dear nyasha,

    Yes it is. I will give you a hint; the first step is to take z=0. If you need further assistance please don't hesitate to reply me.
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  3. #3
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    Quote Originally Posted by Sudharaka View Post
    Dear nyasha,

    Yes it is. I will give you a hint; the first step is to take z=0. If you need further assistance please don't hesitate to reply me.

    Can l use matrix row reduced echeloen form ?
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  4. #4
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    Dear nyasha,

    I don't know what row reduced echeloen form is. But you could easily find the line in the following method.

    x+y-z=2------(1)

    2x+3y+z=4---------(2)

    (1)+(2)\Rightarrow{3x+4y=6}

    This is the equation of the line. Since you need to find the parametric form,

    x=\frac{6-4y}{3}=t

    Therefore, x=t

    y=\frac{6-3t}{4}

    Hope this helps.
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  5. #5
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    Hello, nyasha!

    Here's my approach . . .


    Find the parametic equation of the line given by: . \begin{array}{cccc} x+y-z&=&2 & [1] \\ 2x+3y+z&=&4 & [2] \end{array}

    Add [1] and [2]: . 3x + 4y \:=\:6

    . . Solve for x\!:\;\;x \:=\:\frac{6-4y}{3}


    Substitute into [1]: . \frac{6-4y}{3} + y - z \:=\:2

    . . Solve for z\!:\;\;z \:=\:-\frac{y}{3}


    We have: . \begin{Bmatrix}x &=& 2 - \frac{4}{3}y \\ \\[-4mm] y &=& y \\ \\[-3mm] z &=& -\frac{1}{3}y \end{Bmatrix}



    On the right, replace y with 3t

    . . and we have: . \begin{Bmatrix}x &=& 2-4t \\ y &=& 3t \\ z &=& -t \end{Bmatrix}

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