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Math Help - Lines and Planes in 3D

  1. #1
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    Lines and Planes in 3D

    Hey there. This one has me kind of stumped. Maybe I was just a bit burnt out today, as I'd already done a few hours of work before I tried to tackle this problem. Spent a good long while on it without much luck. I'm tossing it out here in the hopes that someone might be able to give me a helpful hint or nudge in the right direction.

    Question

    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?
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  2. #2
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    Quote Originally Posted by DarrenM View Post
    Hey there. This one has me kind of stumped. Maybe I was just a bit burnt out today, as I'd already done a few hours of work before I tried to tackle this problem. Spent a good long while on it without much luck. I'm tossing it out here in the hopes that someone might be able to give me a helpful hint or nudge in the right direction.

    Question

    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?
    I won't actually solve you the question, but it shouldn't be hard. the light travels along the vector (8,4,1) initially, and then along the vector (2,4,4) (subtract the coordinates). therefore the normal to the mirror vector v is precisely the bisector vector of (-8,-4,-1) and (2,4,4). Finding the bisector vector of two vectors should be easy/standard, and after that it's easy to write the equation of the plane passing through the point (8,4,1) with the normal vector v
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  3. #3
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    Quote Originally Posted by choovuck View Post
    I won't actually solve you the question, but it shouldn't be hard. the light travels along the vector (8,4,1) initially, and then along the vector (2,4,4) (subtract the coordinates). therefore the normal to the mirror vector v is precisely the bisector vector of (-8,-4,-1) and (2,4,4). Finding the bisector vector of two vectors should be easy/standard, and after that it's easy to write the equation of the plane passing through the point (8,4,1) with the normal vector v
    Thanks a lot for the help. The first thing I tried was finding the cross product of (-8, -4, -1) and (2, 4, 4), but that didn't give me what I was hoping for. It gave me an orthogonal vector, but apparently not one normal to the mirror. Now, I haven't dealt with bisectors of vectors before and I'm nearly certain that it isn't assumed knowledge (for this problem). Which leads me to the question, is there a method of approaching this problem that does not require finding the bisector of those two vectors? If so, a hint in that direction would be greatly appreciated.
    Last edited by DarrenM; June 6th 2010 at 05:33 PM.
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  4. #4
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    Quote Originally Posted by DarrenM View Post
    Thanks a lot for the help. The first thing I tried was finding the cross product of (-8, -4, -1) and (2, 4, 4), but that didn't give me what I was hoping for. It gave me an orthogonal vector, but apparently not one normal to the mirror. Now, I haven't dealt with bisectors of vectors before and I'm nearly certain that it isn't assumed knowledge (for this problem). Which leads me to the question, is there a method of approaching this problem that does not require finding the bisector of those two vectors? If so, a hint in that direction would be greatly appreciated.
    ergh, i was afraid u'll reply this way. there're two standard ways of finding the bisectors (that i know of). first is that it's a vector that lies in the same plane and creates equal angles with each of the vectors. the condition of equality of angles is easy (scalar products divided by norms are equal), but this works very nicely in 2D, not in 3D since in 3D there is the whole *plane* satisfying the condition, not a line. the second way, which is probably the correct one to use here, is that this line is precisely the set of points that satisfy the following condition: the distances from a point to both of the two lines are equal. i don't remember which way is the best to compute the distance. this could be the one: Point-Line Distance--3-Dimensional -- from Wolfram MathWorld

    hope this helps
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