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Math Help - parametric equation on line in 3d

  1. #1
    ggw
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    parametric equation on line in 3d

    find parametric equation for the line with the following properties. the lines passes through the origin, it is contained in the plane 2x+3y-z=0. and is orthogonal to the vector v=<4,1,2>.
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  2. #2
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    Hello, ggw!

    Find parametric equation for the line with the following properties:
    the line passes through the origin, it is contained in the plane 2x + 3y - z = 0.
    and is orthogonal to the vector v = <4,1,2>
    The parametric equations of a line has this form:

    . . x .= .x1 + at
    . . y .= .y1 + bt
    . . z .= .z1 + ct

    where (x1,y1,z1) is a point on the line and <a,b,c> is its direction vector.


    The line passes through the origin: (x1,y1,z1) .= .(0,0,0)
    . . We need to find its direction vector: <a,b,c>

    The line is in the plane 2x + 3y - z = 0.
    Hence, the line is orthogonal to the normal vector of the plane: <2,3,-1>
    . . and we have: .<a,b,c><2,3,-1> .= .0 . . 2a + 3b - c .= .0 . [1]

    The line is orthogonal to <4,1,2>.
    . . So we have: .<a,b,c><4,1,2> .= .0 . . 4a + b + 2c .= .0 . [2]

    Multiply [1] by 2: .4a + 6b - 2c .= .0
    . . . Subtract [2]: .4a + b + 2c .= .0

    And we have: .5b - 4c .= .0 . . b = 4c/5

    Substitute into [1]: .2a + 3(-4c/5) - c .= .0 . . a = -7c/10


    The direction vector of the line is: .<-7c/10, 4c/5, c>

    Let c = 10 and we have: .v .= .<-7, 8, 10>


    Therefore, the equations of the line are:

    . . x .= .-7t
    . . y .= . 8t
    . . z .= .10t


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