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Homework Help: Finding a plane given a point and a line help! *calculus*

  1. Apr 13, 2008 #1
    1. The problem statement, all variables and given/known data

    I know I've had alot of questions about this lately, but this will be the last one...I think I have the right idea but I'm a tad off.

    Okay, so the question is

    Find a plane containing the point (6,-3,1) and the line of intersection of the planes `-6 x -1 y + 7 z = -10` and `4 x -7 y -6 z = -25`

    So you have to find the line, find two points on that line, and then find the plane with the 3 points.

    Here are my calculations, but the numbers are funky so I doubt I'm right.

    3. The attempt at a solution

    <-6, -1, 7>
    <4, -7, -6>

    I(6 + 49) - J(36 - 28) + K(42 + 4)

    (55, -8, 46)

    -1y + 7z = -10

    -7y - 6z = -25

    -----

    -y = -10 - 7z

    7(-10 - 7z) - 6z = -25

    -70 - 49z - 6z = -25

    -55z = 45

    z = -(45/55)

    -y = -10 - 7(44/55)

    -y = -15.72

    y = 15.72

    z = -.818181

    (0, 15.72, -.818181)

    r = (0, 15.72, -.818181) + t(55, -8, 46)

    or

    x = 0 + 55t

    y = 15.72 - 8t

    z = -.818181 + 46t

    -----

    Point 1 = (6, -3, 1)
    Point 2 = (55, 7.72, 45.181818)
    Point 3 = (110, -.28, 91.181818)

    A = <51, 10.72, 44.181818>
    B = <104, 2.72, 90.181818>

    I(966.75 - 120.1745) - J(4599.2727 - 4594.91) + K(138.72 - 1114.88)

    I(846.5755) - J(4.3627) + K(-976.16)

    846.5755(x - 6) - 4.3627(y + 3) - 976.16(z - 1) = 0

    846.5755x - 4.3627y - 976.16z = 5079.453
     
  2. jcsd
  3. Apr 14, 2008 #2

    dynamicsolo

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    Homework Helper

    OK, here's the vector that is mutually perpendicular to the normal vectors of the two planes, so it must point the direction of a line lying in the desired plane.

    You chose to set x = 0, so you need to find the remaining coordinates for a point in both planes.

    Copying error: you should have z = -45/55 = -9/11 , so

    -y = -10 - 7·(-9/11) = (-110 + 63)/11 = -47/11 . (You also dropped the sign for -45/55.)

    So (0, 47/11, -9/11) is in both planes, which can be verified by substituting it back into the equation of each plane. (If you'd tried that at this point, you'd have spotted the error.)

    I believe this is how you'd proceed from here. You know the line in the desired plane which is parallel to <55, -8, 46> has to pass through this point, so set up

    (x - 0)/55 = (y - [47/11])/(-8) = (z - [-9/11])/46

    Pick any convenient point on this line and construct the vector from that point to (6,-3,1). The cross product of that vector and <55, -8, 46> will give you the normal vector to the desired plane. Finally, with those components <A,B,C> giving you the coefficients for the plane, you can set up

    A·(x - 6) + B·(y + 3) + C·(z - 1) = 0.
     
    Last edited: Apr 14, 2008
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