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Homework Help: A sphere between two non-paralell planes - finding the centre

  1. Dec 6, 2011 #1
    Allright, I've got mock-exams coming up this winter and I'm struggling with the damn vectors. I made a thread about this before but it turned out terrible due to my poor English skills and because I wrote it in a rush. Anyway:

    The problem: Plane α: x-2y+z=2. Plane β: 2x+y+z=5

    Point P= (0,1,4) is on the line of intersection, which has the directional vector L, between the two planes.

    So L=directional vector of the line of intersection between the two planes.

    A sphere is touching plane alpha at point A and plane beta at point B. Show that PB is perpendicular to L and the directional vector of plane Beta, and that PA is perpendicular to L and the directional vector of plane Alpha. Find the coordinates for the centre of the sphere, point S.

    point A=(11,9,9) point B= (4,-14,-1).

    My work: I found L by finding the cross product between the 2 directional vectors of the planes. I also confirmed that vectors PB & PA are both perpendicular to L, and PB is perpendicular to the directional vector of plane alpha while PA is perpendicular to the directional vector of plane beta.

    Where I'm stuck: I set the centre coordinates as: x,y,z. Using some geometry I set up a parametric function where x= 7t y=21t+1 z=4.

    I then figured that the radius of the sphere equals= (11-x)^2 + (9-y)^2 + (9-z)^2 = |SA| = |SB| = (-4-x)^2 + (14-y)^2 + (-1-z)^2. This is derived from the formula of a sphere.

    Well, I tried to do the algebra, substituting x with 7t, y with 21t+1 and z with 4 but I just ended up with 0t=0.

    I tried finding the radius as well, but these attempts too failed.

    I spent almost 2 hours on this but I simply couldn't do it, it was (and still is) too hard for me. Can somebody here please help?
  2. jcsd
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