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Equation of reflecting plane when incoming and reflected rays are known

  1. Nov 16, 2009 #1
    1. The problem statement, all variables and given/known data

    A light ray with direction (1, 2, -2) passes through the point (-1, -3, 5) and is reflected farther away on a plane. The reflected ray has the direction (3, 0, 4) and passes through the point (4, 1, 5). Determine the equation for the plane which reflects the ray. The vector space is orthonormal.

    2. Relevant equations

    3. The attempt at a solution

    I think I've solved this in a reasonably protracted way, so I'll name the general steps of it. If you can think of a simpler solution or some shortcuts, or if it seems like a decent or even viable approach, I'll be happy to know!

    1: Assign v=(1, 2, -2), u=(3, 0, 4) as the two vectors and determine the equations of the incoming and outgoing lines (Lv and Lu) in parameterized form.

    2: Determine the point of intersection Ps=(1, 1, 1) between the lines.

    3: Use Ps and the two given vectors to determine the equation of the plane which they're in, ([tex]4x-5y-3z+4=0[/tex]) (e.g. by assuming that the relevant determinant is 0 so that the three vectors are linearly dependent). The normal vector of this plane, [tex]\vec{Nv}[/tex]=(4, -5, -3) is then parallel to the sought plane.

    4: Note that the dot product u[tex]*[/tex]v = |u||v|[tex]\cos{(\pi - 2 \alpha)} = 1*3 + 2*0 + (-2)*4 = -5 => \cos{\alpha} = \sqrt{2/3}[/tex] (Solutions outside the interval [tex]0 < \alpha <\pi/2[/tex] should be irrelevant, I think. Can anyone verify this step?)

    5: Choose two points on the lines Lv and Lu that are equidistant from Ps, say, a distance D away. Two such points exist on each line. Determine their coordinates. Choose one of these points on each line. The point Pm with the arithmetic mean of their coordinates will reside either on the sought plane or on a vector parallel to the normal vector of the sought plane that passes through Ps.

    6: Noticing that [tex]\cos{\alpha} = \sqrt{2/3} > \sqrt{1/2} = 1/ \sqrt{2} = \cos{\pi / 4}[/tex], it is realized that [tex]0 < \alpha < \pi / 4[/tex]. Thus, if [tex]|\vec{PsPm}| < D \cdot 1/ \sqrt{2}[/tex], Pm is on the sought plane. If not, choose another pair of points in (5).

    7: Use [tex]\vec{PsPm}[/tex], [tex]\vec{Nv}[/tex] and Ps, determine the equation of the sought plane. My result is:

    [tex]-\frac{10}{3}x - \frac{34}{15}y - \frac{2}{3}z + \frac{94}{15} = 0[/tex]

    I'm not sure if this is correct, but I haven't found any particular problems with my approach, other than it being somewhat ugly. I suppose I'll recheck my arithmetic.
  2. jcsd
  3. Nov 16, 2009 #2
    I ran through the arithmetic of my solution again, and realized that I forgot to divide one of the coordinates of one of the points on Pm by two. So, in step (5), I've chosen (2/3, 1/3, 5/3) on Lv and (2/5, 1, 1/5) on Lu, both being a distance of 1 away from the the intersection point Ps of the two lines. This gives Pm = (8/15, 2/3, 14/15). Subsequently, [tex]\vec{PsPm}[/tex] = (8/15 - 1, 2/3 - 1, 14/15 - 1) = (-7/15, -1/3, -1/15).

    And so, with the point Ps = (1, 1, 1) and the two vectors [tex]\vec{Nv}[/tex] = (4, -5, -3) and [tex]\vec{PsPm}[/tex] = (-7/15, -1/3, -1/15) in the plane, we get

    [tex]\begin{vmatrix}x-1& -7/15 & 4\\
    y-1& -1/3 & -5\\
    z-1& -1/15 & -3\end{vmatrix} = \frac{1}{3}(2x -5y + 11z -8) = 0[/tex]

    Looks better?

    For some reason, every problem I've attempted since this one has seemed a lot easier than usual...
  4. Nov 16, 2009 #3


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    Well, you are obviously pretty good at this stuff. You got the correct answer. About the only shortcut I can recommend is that you can find the direction vector of the bisector without finding any points. If d1 is the normalized direction vector heading into the reflection point and d2 is the normalized direction vector heading out, then (d1+(-d2))/2 is a direction vector of the angle bisector, hence a normal vector to the reflection plane. Do you see why? Once you've got the normal and the intersection point you've got the plane. As for the dot product stuff, I don't think there is any constraint on the angle between the incoming and outgoing vectors until you know something about the reflection plane. Once again, well done. I can see why the other problems don't seem so difficult. You have a good intuitive grasp of the concepts.
  5. Nov 23, 2009 #4
    Wow, I should reply to these things more quickly. Thank you for looking over my work. Also, thank you for providing me with that shortcut. Realizing that sooner would have made things a lot easier, but hopefully I learnt something along the way...
  6. Dec 1, 2009 #5
    Using the fact that the incident ray and reflected ray make equal angles to the Normal to the plane of the reflecting surface, we can compute the normal to that plane with the formula,

    Unit Normal vector N, <N> = (-(Unit Incident vector, I) + (Unit Reflected Vector, R))/2
    which is,

    N = (-(1/3,2/3,-2/3)+(3/5,0,4/5))/2 = 1/15(2,-5,11)

    This is the equation of the normal to the plane.

    Please comment on whether this is right.
  7. Dec 2, 2009 #6


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    It's a normal alright. It's not a unit normal.
  8. Dec 2, 2009 #7
    Hi Dick,
    Thanks, yes, it is a normal to the plane and not a unit normal.
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