1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Equation of a plane through 2 points, parallel to a line.

  1. Nov 1, 2009 #1
    1. The problem statement, all variables and given/known data
    I'm trying to understand my professor's solution to a quiz problem. I can follow it up to a point, but then I'm getting confused as to how he continues. The question is:

    Find an equation of the plane [tex]p[/tex] through [tex]A(2,1,-1)[/tex] and [tex]B(1,5,-4)[/tex] parallel to the line [tex]L: x=1 + 2t, y=-1+t, z=-2-t[/tex].


    2. Relevant equations
    n/a


    3. The attempt at a solution
    This is the solution that was given:

    [tex]
    p: A(x-2)+B(y-1)+C(z+1)=0
    [/tex]

    [tex]
    B \in p: A(1-2)+B(5-1)+C(-4+1)=0
    [/tex]

    [tex]
    L \|p \Leftrightarrow \vec{d}\bot \vec{n}\Leftrightarrow 2A+B-C=0
    [/tex]

    He then solves for [tex]C[/tex], getting [tex]C=2A+B[/tex].

    The next step is where I get lost:

    [tex]
    9B=7C \Rightarrow B=\frac{7}{9}C
    [/tex]

    [tex]
    A=4B-3C=4 \cdot \frac{7}{9}C - 3C=\frac{1}{9}C
    [/tex]

    [tex]
    A=1, B=7, C=9
    [/tex]

    [tex]
    \vec{n}=(1,7,9)
    [/tex]

    [tex]
    x-2+7(y-1)+9(z+1)=0 \Rightarrow x+7y+9z=0
    [/tex]

    It is the [tex]9B=7C[/tex] part that I don't get. Where are those coefficients coming from? Thanks for any assistance anyone can provide.
     
    Last edited: Nov 1, 2009
  2. jcsd
  3. Nov 1, 2009 #2

    Mark44

    Staff: Mentor

    I think what your prof is doing is using the idea that the cross product of a vector in the plane with the direction of the line will be parallel to the normal to the plane.

    I.e., <-1, 4, -3> X <2, 1, -1> = k<A, B, C>

    The first vector in the equation above is AB, the second is the direction of the line.

    It is unfortunate that A and B are used both for points in the plane and for the coordinates of the plane's normal. This duplication can lead to confusion.
     
  4. Nov 1, 2009 #3
    Thank you so much. Taking the cross product does lead to the normal vector he eventually arrived at. I still don't understand exactly what algebraic manipulations he was performing to get there his way, but as long as I have some way of arriving at the correct answer, I'll be happy.

    Thanks again.
     
  5. Nov 1, 2009 #4

    Mark44

    Staff: Mentor

    The cross product works in this problem because the vector from A to B is not parallel to the direction vector of the line.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook