Equation of one of the two lines whose angle bisector is given.

Question

A ray of light falling along the line $ lx+my+n=0 $ strikes a plane mirror at point $P$. Find the equation of reflected ray if $px+qy+r=0$ is the equation of normal to the plane at point $P$.

Answer 1

Let's normalize things so that $l^2 + m^2 =1$ and $p^2 + q^2 = 1$. Then $\mathbf{L} = (-m,l)$ is a unit vector parallel to the incoming ray, and $\mathbf{N}=(p,q)$ is a unit vector perpendicular to the mirror. The incoming ray is along the line $(\mathbf{X} - \mathbf{P}) \times \mathbf{L} = 0$, and the mirror plane has equation $(\mathbf{X} - \mathbf{P}) \cdot \mathbf{N} = 0$. Let's use $\mathbf{M}$ to denote a unit vector parallel to the the out-going ray. Draw a little picture to convince yourself that $\mathbf{M} = \mathbf{L} + 2[(\mathbf{L} \cdot \mathbf{N})\mathbf{N} - \mathbf{L} ]$ or $\mathbf{M} = 2(\mathbf{L} \cdot \mathbf{N})\mathbf{N} - \mathbf{L}$. Writing out all the coordinates: $$ \mathbf{M} = (2lpq - 2mp^2 + m, 2lq^2 -2mpq -l) $$ Now you have $\mathbf{M}$, you're done, because the out-going ray has equation $(\mathbf{X} - \mathbf{P}) \times \mathbf{M} = 0$.