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Math Help - eqn of line from a plane and another lineFind the

  1. #1
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    eqn of line from a plane and another lineFind the

    find the parametric eqn for the line lying on the plane x+y+z=3 and intersecting the line x=2+t , y=1-t, z=-3t perpendicularly.

    I found the point of intersection to be (2,1,0).

    how do i get the other point? I was thinking that if the plane would intersect the line perpendicularly, wouldnt any other point on that plane, that satisfies x+y+z = 3 be enough to form the parametric eqn that i need?

    eg. B (1,1,1)? or (0,0,3). is there a unique solution?
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    Quote Originally Posted by adamantine View Post
    find the parametric eqn for the line lying on the plane x+y+z=3 and intersecting the line x=2+t , y=1-t, z=-3t perpendicularly.

    I found the point of intersection to be P(2,1,0). <<<<< OK

    how do i get the other point? I was thinking that if the plane would intersect the line perpendicularly, wouldnt any other point on that plane, that satisfies x+y+z = 3 be enough to form the parametric eqn that i need?

    eg. B (1,1,1)? or (0,0,3). is there a unique solution?
    1. The given line has the direction vector \vec u = (1,-1,-3)

    2. Let Q(a,b,c) denote the point in the plane who is situated on the line perpendicular to the given line. Then you know:

    \overrightarrow{PQ} \cdot \vec u = 0~\wedge~a+b+c=3

    3. Plug in the known components:

    (2-a, 1-b, -c) \cdot (1,-1,-3)=0~\implies~-a+b+3c=-1 and a+b+c=3 yields (2+c, 1-2c, c)

    4. Choose a suitable value for c. I used c = 1. That means the Point Q is Q(3, -1, 1).

    The equation of the line is
    (x,y,z) = \overrightarrow{OP}+r \cdot \overrightarrow{PQ}

    5. I attached a sketch: The given line in green, the point P in red, the point Q in yellow and the line PQ in orange.
    Attached Thumbnails Attached Thumbnails eqn of line from a plane and another lineFind the-eben_gerad_senkrecht.png  
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