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Parametrizing Planes Question

  1. Feb 12, 2015 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    Find an equation to the plane:
    1)Orthogonal to the line r = <t, 2 − 3t, 4> and passing through the origin.

    2. Relevant equations

    Equation for a plane: a(x-xi)+b(y-yi)+c(z-zi)=d

    3. The attempt at a solution

    Okay, so this is really a matter of 'slope' and understanding the values that I am giving.

    So we are giving a line:
    <0, 2, 4>+t<1,-3,0>
    and a point P on the plane at (0,0,0)

    I can automatically use the point in our equation of a plane to simplify things to
    ax+by+cz=0

    Now, I need to find the normal vector of the plane.

    So we are giving a line ON THE PLANE, this means that the slope vector of that line is NOT perpendicular to the plane, correct? So I cannot use these values as a,b,c.

    That is my first question.




    Here's what I did then:
    For some reason, I took the cross product of the lines point and the lines slope vector and got the cross product <-12, -4, 2> and the plane became
    2(-6x-2y+z)=0
    and this seems to work for point <0,2,4> and point <0,0,0>

    But I don't think I can do this... :/
     
  2. jcsd
  3. Feb 12, 2015 #2

    LCKurtz

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    If the plane is orthogonal (perpendicular) to the line, that means the direction vector (not slope) of the line is perpendicular (normal) to the plane.
     
  4. Feb 12, 2015 #3

    RJLiberator

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    So we need to find the direction vector (not slope) of the line which will be the normal vector of the plane.

    I was under the impression that the slope vector and direction vector were one in the same?
     
  5. Feb 12, 2015 #4

    LCKurtz

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    The term "slope" is used in 2d and means the rise/run for a straight line. You don't talk about the slope of a line in 3d (it isn't defined); you talk about its direction vector. And in 2d, the direction vector is not the slope.
     
  6. Feb 12, 2015 #5

    RJLiberator

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    Thank you for that clarification.

    So we already have the direction vector then at <1, -3, 0>
    We use this as abc in the plane equation and point origin (0,0,0) to net:
    x-3y=0 as the equation of this plane.

    Does this seem correct?

    Here's my thinking:
    We need a plane that is perpendicular to a giving line. This line has a direction vector that is parallel to the giving line. Since the direction vector points in the direction of the line, and we are looking for a plane that is perpendicular to this line, we can use this as the normal vector of the plane.
     
  7. Feb 12, 2015 #6

    LCKurtz

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    Do you have to ask? Can't you check for yourself whether it passes through the origin and is perpendicular to the given line?

    Apparently you understood my post #2.
     
  8. Feb 12, 2015 #7

    RJLiberator

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    Well, x-3y=0 must pass through (0,0,0).

    But from your post, it seems I have made an error.

    So a direction vector comes about from 2 points on the line, then?
    Say we use point (0, 2, 4) and then take t=2 and use point (2, -4, 4)
    We then get the line segment <2, -6, 0>
    and this can be our normal vector?
    2x-6y=0

    Hm....
     
  9. Feb 12, 2015 #8

    LCKurtz

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    It doesn't matter how long your direction vector is. Isn't ##2x-6y=0## the same line as ##x-3y=0##?
     
  10. Feb 12, 2015 #9

    RJLiberator

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    Yes, that's exactly what I observed while posting my "hm..". I was not sure how to feel about that.


    Oh, I am sorry - you were saying that I understood your post #2 and not misunderstood.

    So, indeed, my thought process was correct and my equation of the plane x-3y=0 is confirmed based on plugging in point (0,0,0) and finding another working equation of the plane with a similar direction vector.
     
  11. Feb 12, 2015 #10

    RJLiberator

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    I would like to pose one follow up question to assure my understanding:

    Equation of the plane
    Containing the line r(t) = <2 − t, 3, 4 + 2t> and point P(0, 0, 1).

    1) We have a point, we need a normal vector
    2) this time we are giving a line ON the plane, so the direction vector won't work as it is parallel to the plane.
    3) Therefore, we can find the normal vector by using the point (0,0,1) and the point from the giving line (2, 3, 4) and making their line segment which comes out to be <-2, -3, -3>.
    We then use this line segment and the direction vector <-1, 0, 2> take their cross product to net a vector that is perpendicular to the plane <6, -7, 3>.

    And the equation of this plane becomes 6x-7y+3z=3
    which satisfies both of our giving points (0, 0,1) and (2, 3, 4) on the plane.

    The thing i am unsatisfied with here is why can we use 2 points on the plane, take their difference, and then cross it with the direction vector to get the perpendicular normal vector?

    I understand that the cross product nets a perpendicular vector, but not the first part.
     
  12. Feb 12, 2015 #11

    LCKurtz

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    Forum policy is for new questions to go in new threads. It will increase your chances for an answer because I am going to be away for a while.
     
  13. Feb 12, 2015 #12

    RJLiberator

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    Thank you for all your help, LCKurtz.
     
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