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Homework Help: Equation of a Plane

  1. Jun 7, 2010 #1
    1. The problem statement, all variables and given/known data

    A flashlight located at the origin (0, 0, 0) shines a beam of light towards a flat mirror. The beam reflects off of the mirror at (8, 4, 1) and then passes through (10, 8, 5). What is the equation of the plane that contains the mirror?

    2. Relevant equations

    The only assumed knowledge is what has been covered in this specific chapter; three-dimensional Cartesian coordinates, vectors, dot product, cross product, and finding the equations of lines and planes in space.

    3. The attempt at a solution

    I know that to find the equation of a plane one needs:
    * a point on the plane, and
    * a vector n, perpendicular to the plane.

    The point where the light strikes the mirror satisfies the first requirement, so our point is P(8, 4, 1). Finding a vector normal to the plane (mirror) is where I'm stumped. I found the vector created by the reflecting light, (10-8, 8-4, 5-1) = (2, 4, 4). I initially tried to use the cross product with vector (8, 4, 1) and vector (2, 4, 4). I know that gives me a vector perpendicular to both, but I quickly realized that I don't have any way (or don't know how) to find the orientation of that vector to the plane I'm trying to find. So it doesn't really help.

    Someone suggested that the normal vector to the mirror is the bisector vector of (-8, -4, -1) and (2, 4, 4). While I imagine this would work, I'm almost certain that is not the approach we are intended to use. We haven't covered that, and I'm pretty sure it's not assumed knowledge at this point.

    I also tried all types of other stuff, but the truth is that I was just getting desperate and trying everything I could think of (without success).

    Any help would be much appreciated.
  2. jcsd
  3. Jun 7, 2010 #2
    I think the easiest way to do this is to use the bisector of the two vectors. "Finding the bisector" might sound scary especially because you said you haven't done anything like that before. However, it is actually very simple:

    There isn't a "formula" for it although you can systematically express the answer by using a simple equation. I won't give you the equation; it's best that you derive it yourself. Here is how you do it: (Do this on a paper for 2D vectors) Draw a mirror, an incident ray and the reflected ray. Now draw the bisector of the two. Try to do something with two of those vectors to come up with the third. (maybe add them?..)

    I hope this is useful.
  4. Jun 7, 2010 #3
    Thank you very much for the reply.

    I'll certainly attempt to apply the bisector of the two vectors since you're the second person to suggest that approach to me. I'm not reluctant to give it a try because it sounds intimidating or because I'm not familiar with it, but because I imagine my teacher expects us to solve the problem another way.

    I'll see what I can do with what you've said, but I still have to ask... does anyone see another method of solving this problem that would fall within the material covered that I previously mentioned?
  5. Jun 8, 2010 #4
    Ah, got it. Thanks for the help. I popped in to see my professor and mentioned the fact that the bisector vector approach had been suggested. He said that was, indeed, what he was hoping we'd see. As soon as he started to draw the diagram in two-dimensions (as you suggested) it clicked.
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