Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Extremely tough vectors-planes problem

  1. Sep 21, 2011 #1
    1. The problem statement, all variables and given/known data

    A plane has the equation aX + bY + cZ + d = 0. A line L goes from (0,0,0) and crosses the plane at some point. L and plane are orthogonal. express the coordinates of the crossing point P by: a, b, c and d.

    2. Relevant equations

    I know that [a,b,c] is the directional vector for the line and that

    |d|/sqrt(a^2 + b^2 + c^2) is the distance between point P and origo. I know and understand this at least.

    3. The attempt at a solution

    I'm lost lol
  2. jcsd
  3. Sep 21, 2011 #2
    ah damn, I haven't learned about directional vectors this year. but i was reminded about that stuff by a friend who helped me with this assignment.

    So, if I divide [a,b,c] by the length of the directional vector, sqrt(a^2 + b^2 + c^2) and then multiply it by the length of the OP vector, |d|/sqrt(a^2 + b^2 + c^2), I get:

    a|d|/a^2 + b^2 + c^2, b|d|/a^2 + b^2 + c^2, c|d|/a^2 + b^2 + c^2 = P.

    The answer is supposed to be (-ad/a^2 + b^2 + c^2,-bd/a^2 + b^2 + c^2,-cd/a^2 + b^2 + c^2) = P

    *sigh* does this have something to do with all the fools sometimes saying a plane's equation is ax+by+cz -d = 0 instead of ax+by+cz +d = 0 ? or that d is negative if the directional vector is posetive?

    has been a really long day at school + after 4 me and i'm sleepy as hell.. hope somebody can fill me in tomorrow
  4. Sep 21, 2011 #3


    User Avatar
    Science Advisor

    Let the line be given by x= ut, y= vt, z= wt (since (0,0,0) satifies that). If the plane and line are orthogonal, then <u, v, w> must be orthogonal to the plane and so must be parallel to <a, b, c>. That is u= ka, v=kb, w= kc. x= kat, y= kbv, z= kct. Putting that into the equation of the plane, [itex]k(a^2+ b^2+ c^2)t+ d= 0[/itex]. [itex]t= -d/(a^2+ b^2+ c^2)[/itex]. Put that into the equation of the line and solve for x, y, and z.

  5. Sep 22, 2011 #4
    hmm thanks but could you, or somebody else, answer the question in post 2?
    Last edited: Sep 22, 2011
  6. Sep 23, 2011 #5
    Can somebody explain to me the logic of why the answer should be negative [like this: (-ad/a^2 + b^2 + c^2,-bd/a^2 + b^2 + c^2,-cd/a^2 + b^2 + c^2) = P], and not in absolutes [like this (-ad/a^2 + b^2 + c^2,-bd/a^2 + b^2 + c^2,-cd/a^2 + b^2 + c^2) = P]?

    I believe the absolutes answer is plain wrong.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook