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The vertical plane through a point and containing a vector.

  1. Oct 24, 2011 #1
    1. The problem statement, all variables and given/known data
    f(x,y) = x2 + y2
    P(9,6)
    v=<2,4,0>
    1)Find a vertical plane that passes through the point P(9,6) and has the vector v=<2,4,0>
    2)What is the vector function of the curve of intersection of the vertical plane and z=f(x,y)

    2. Relevant equations
    A(x - x0) + B(y - y0) + C(z - z0)

    normal vector
    n= <0,0,1>

    given point and vector:
    P(9,6)
    v = < 2, 4 0>

    3. The attempt at a solution

    I tried plugging the give point and and the normal vector into A(x - x0) + B(y - y0) + C(z - z0)
    and I got 3x+2y but that's not a vertical plane.

    I also tried plugging in the point and the given vector and got x+2y-21 and that doesn't look right either.

    I found the unit vector to be <1/√3 , 2/√3 , 0>. I'm not sure if I need to use that find the vertical plane.
     
  2. jcsd
  3. Oct 24, 2011 #2

    Mark44

    Staff: Mentor

    A vertical plane implies three dimensions, so any point on it should have three coordinates. Is there a typo in P(9, 6)?
     
  4. Oct 24, 2011 #3
    okay, so the point would then be (9,6,0).
    But where do I go from there?
     
  5. Oct 25, 2011 #4

    Mark44

    Staff: Mentor

    I hope you're drawing a sketch of this situation.

    Draw the point P(9, 6, 0).
    Put the tail of the vector u = <2, 4, 0> at P.
    Since the plane is vertical, the vector v = <0, 0, 1> is in the plane (provided that we put the tail of this vector at P.

    What operation can be applied to two vectors to get another vector that is perpendicular to both of the other vectors? That will be a normal to the plane. Once you know a normal to the plane and a point in the plane, you can get the equation of the plane pretty easily.
     
  6. Oct 25, 2011 #5
    Oh, I see. I was drawing the vector from the origin so I couldn't see how the point and the vector touched.

    I take the cross product of <2,4,0> and <0,0,1> and get 4i - 2j to be a vector that is perpendicular.

    And then to find the equation of this plane I can use the equation I had above.
    I get 2x-y=2 or y=2x+2.

    If I graph this, the plane will go through the origin, and be on the z-axis correct?
    I'm using maple to graph it so I might be doing that wrong...
     
    Last edited: Oct 25, 2011
  7. Oct 25, 2011 #6
    And for question two, I plugged in y = 2x+2 into f(x,y).
    so f(x,y)=z=5x2 + 8x + 4 and y = 2x+2.

    So my vector function is <x, 2x+2, 5x2+8x+4>
    can it also be written r(t) = <t, 2t+2, 5t2+8t+4> ?
     
  8. Oct 25, 2011 #7

    Mark44

    Staff: Mentor

    No, this isn't right. The point (9, 6, 0) is supposed to be on the plane, and so should be a solution of this equation, but it isn't.


    No, the plane doesn't go through the origin, and it doesn't include the z-axis.
     
  9. Oct 25, 2011 #8
    I need a vertical plane, so I'm trying to find a plane normal to, for example, z=1 right?
     
  10. Oct 25, 2011 #9

    Mark44

    Staff: Mentor

    Yes, the plane you're looking for will be normal to the plane z = 1 (or for that matter, the plane z = 0), but so what?

    Fix the problems in post #6 and you should get a vector that is perpendicular to your plane. Once you have a normal to the plane and a point in the plane, it's straightforward to get the plane's equation.
     
  11. Oct 25, 2011 #10
    I think I got it now. If looking straight down on the xy-plane you get a straight line with a slope of 2. The equation of the straight line is 2x-y=12 which is also the equation of the plane.

    And for question 2, I got the vector function to be r(t) = <t, 2t-12, 5t2 - 48t + 144>
     
  12. Oct 25, 2011 #11

    Mark44

    Staff: Mentor

    Looks good to me!
     
  13. Oct 25, 2011 #12
    Thank you!!!
     
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