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Math Help - Show that a line is parallel to a plane

  1. #1
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    Show that a line is parallel to a plane

    The question:
    Show that the line x = t(2 1 3) is parallel to the plane 3x - 3y - z = 2.

    I'm not sure how to check if it's parallel. I converted the plane to parametric form, but the line doesn't seem to be a scalar multiple of either directional vectors of the plane. How do I solve this?

    Thanks.
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  2. #2
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    Hello, Glitch!

    Show that the line x \,=\, t\langle2\;1\;3\rangle is parallel to the plane 3x - 3y - z \:=\: 2.

    The direction of the line is: . \vec v \:=\:\langle 2,\:1,\:3\rangle

    The normal of the plane is: . \vec n \:=\:\langle 3,\:\text{-}3,\:\text{-}1\rangle


    If the line is parallel to the plane, then: . \vec v \:\perp\:\vec n


    Got it?

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  3. #3
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    Thank you! I didn't know the normal was just the coefficients of the plane equation. Not sure why my textbook doesn't mention that...
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  4. #4
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    Here's another way to do it:

    A line is parallel to a plane if and only if it does not intersect that plane. So what happens if you try to find where that line intersects the plane?

    The line is x= 2t, y= t, z= 3t and the plane is 3x- 3y- z= 2. Putting those expressions for x, y, and z into the equation of the plane, 3(2t)- 3(t)- (3t)= 0t= 2. Obviously, that cannot be solved: the line and plane do not intersect. Of course the left hand side is essentially the dot product of the two vectors <2, 1, 3> and <3, 3, -1>, the one in the direction of the line, the other perpendicular to the plane. Since the two vectors have 0 dot product, they are perpendicular which means that the line is parallel to the plane.
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  5. #5
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by Glitch View Post
    The question:
    Show that the line x = t(2 1 3) is parallel to the plane 3x - 3y - z = 2.

    I'm not sure how to check if it's parallel. I converted the plane to parametric form, but the line doesn't seem to be a scalar multiple of either directional vectors of the plane. How do I solve this?

    Thanks.

    if you have line  l :

     \displaystyle \frac {x-x_1}{m} =  \frac {y-y_1}{n} = \frac {z-z_1}{p}

    and plane  \alpha :

     Ax+By+Cz+D=0

    so you have vectors :

     \vec{q} =(m,n,p) and  \vec {n} = (A,B,C)

    so to line and plane to be parallel it must be that :

     (\vec{n},\vec{q}) =0

    or

     Am+Bn+Cp =0
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