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Finding a plane based off 2 points and the z axis

  1. Jul 13, 2014 #1
    1. The problem statement, all variables and given/known data

    Find an equation of the plane that passes through the points (4,2,1) and (-2, 9, 6), and is parallel to the z axis.

    2. Relevant equations

    a(X-Xo) + b(Y-Yo) + c(Z-Zo) = 0



    3. The attempt at a solution

    Okay so for this one, I first tried to make a vector out of the two given points.
    If P(4,2,1) and Q(-2,9,6),

    then PQ = <-6,7,5>.

    Next I fumbled around a bit trying to figure out how to either a) find another vector to form a cross product with, or b) somehow figure out what the <a,b,c> numbers should look like in order to be parallel to z axis.

    When I tried approach a) I came up with another point on the z axis, (0,0,1), used it as a vector of its own, crossed it with PQ, picked up the resultant, and plugged in the direction numbers to the plane equation.

    For part b) I tried making a vector out of (0,0,1) and P, crossing it with PQ and using the resultant direction numbers.

    However, both approaches seem to be rather wrong.
     

    Attached Files:

  2. jcsd
  3. Jul 13, 2014 #2

    SammyS

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    Why pick a point on the z-axis?

    Instead, use a vector which is parallel to the z-axis .
     
  4. Jul 14, 2014 #3
    Does the unit vector z count? <0,0,1>
     
  5. Jul 14, 2014 #4

    SammyS

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    Why not?



    Beginning or ending at either of the given points, where does this vector take you ?
     
  6. Jul 14, 2014 #5
    Well, since the unit vector always points up by 1 unit, for point P(4,2,1), the vector says to go up one to (4, 2, 2).

    For point Q(-2,9,6), it says go up one unit in the z axis, to (-2,9,8).

    With these new points, the cross product of the resulting vectors should yield the direction numbers <a,b,c> of the normal vector to the plane that I am trying to describe, no?
     
  7. Jul 14, 2014 #6
    Maybe this is still wrong, but I have a resultant of 7i + 6j + 0k from crossing <0,0,1> with <-6,7,5>, which I then used to produce 7(x-4) + 6(y-2) = 0, which turns out to

    7x + 6y - 40 = 0.

    However, upon asking Wolfram Alpha to plot it, I got a line!
     
  8. Jul 14, 2014 #7
    Well actually, after some meddling, got this:
     

    Attached Files:

  9. Jul 14, 2014 #8

    SammyS

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    That plane does contain the two points, and is parallel to (or contains) <0, 0, 1>, right? Since the plane doesn't pass through the origin, it's parallel to the z-axis.
     
  10. Jul 15, 2014 #9
    Excellent. Thank you for the assistance :D
     
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