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Homework Statement
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.
Homework Equations
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.