Reflection of an Incident Ray Across a Mirror

The given problem is related to the past weekly problem .

For the ray to project from the point of collision to the target, the line passing through point $(10,10)$ must be parallel to the line of equation $2x - 3y + 15 = 0$ , so that the lines of equation $\beta_1$ and $\beta_2$ are equidistant from the bolded lines. Using the standard form of the linear equation $Ax + By + C = 0$ , $A$ and $B$ are same for both equations $\beta_1$ and $\beta_2$ . The only values that are different involve the $C$ variable, which shifts the equation vertically with respect to $B$ . With that being said,

• Equation $\beta_1$ passing through $(10,10)$ is $2x - 3y + 10 = 0$ .
• Because of $A,B,C$ -linearity, Equation $\beta_2$ is $2x - 3y + 15 + (15 - 10) = 2x - 3y + 20 = 0$ .

Mirroring point $(10,10)$ past the bolded line shows that the line of equation $\alpha_2$ is perpendicular to these 3 lines. So equation $\alpha_2$ is $3x + 2y + 50 = 0$ , which passes through point $(10,10)$ . With equations $\alpha_2$ and $\beta_2$ , we can determine their point of arrival, which is the intersection point $\left(\frac{110}{13}, \frac{160}{13}\right)$ .

So the answer is $\boxed{y = -8(x - 10)}$ , which is easily found by computing the equation passing through $(10,0)$ and the image point.

Let $P = (10, 0)$ and $Q = (10, 10)$ . Since $Q$ lies on the reflected ray, then its reflection across the mirror lies on the extension of the original ray (behind the mirror). So, the only thing to this problem is to find the image $Q'$ under reflection across the given line. There is a standard formula for this computation, and it is as follows. Let the mirror line equation be $M \cdot (r - r_0 ) = 0$ where $r , r_0$ , and $M$ are two-dimensional vectors, with $r = (x, y)^T$ , $r_0 = (x_0, y_0)^T$ is any point that lies on the mirror line, and $M$ is the normal to the line. For the given equation, $M = (2, -3)^T$ , and we can take $r_0 = (0, 5)^T$ . Now the formula for finding the image of point $Q$ is:

$Q' = r_0 + (I - 2 \dfrac{M M^T}{ M^T M } ) (Q - r_0 )$

where $I$ is the $2 \times 2$ identity matrix. Using the given value of the vector $M$ , we obtain

$M^T M = (2, -3) (2, -3)^T = 13$ , and

$M M^T = \begin{bmatrix} 2 \\ -3 \end{bmatrix} \begin{bmatrix} 2 && -3 \end{bmatrix} = \begin{bmatrix} 4 && -6 \\ -6 && 9 \end{bmatrix}$

Therefore,

$I - 2 \dfrac{M M^T}{ M^T M } = \begin{bmatrix} 1 && 0 \\ 0 && 1 \end{bmatrix} - \dfrac{2}{13}\begin{bmatrix} 4 && -6 \\ -6 && 9 \end{bmatrix} = \dfrac{1}{13} \begin{bmatrix} 5 && 12 \\ 12 && -5 \end{bmatrix}$

Therefore,

$Q' = (0, 5)^T + \dfrac{1}{13} \begin{bmatrix} 5 && 12 \\ 12 && -5 \end{bmatrix} ( (10, 10) - (0, 5) )^T$

Performing the above computation, results in,

$Q' = ( \dfrac{110}{13}, \dfrac{160}{13} )$

What is left is to find an equation of the line passing through $P$ and $Q'$ . The slope is given by

$m = \dfrac{ ( \dfrac{160}{13} - 0 ) } { \dfrac{110}{13} - 10 } = \dfrac{ 160 }{-20}= -8$

Therefore, the equation of the line (using point-slope formulation) is $y = -8 (x - 10)$ .

Yet another way to do this problem is to write the incident ray line in parametric form as follows

$r = P + t d$

where $d$ is a direction vector, and $t$ is an arbitrary real parameter, then the reflected ray will have the equation

$r' = r_0 + (I - 2 \dfrac{M M^T}{ M^T M } ) (r - r_0)$

Substituting $r = P + t d$

$r' = r_0 + (I - 2 \dfrac{M M^T}{ M^T M } ) ( P - r_0 + t d)$

Since Q lies on the reflected ray, then

$Q = r_0 + (I - 2 \dfrac{M M^T}{ M^T M } ) (P - r_0 + t d)$

for some t. Noting that the inverse of $(I - 2 \dfrac{M M^T}{ M^T M } )$ is itself, it follows that

$(I - 2 \dfrac{M M^T}{ M^T M } ) (Q - r_0) - (P - r_0) = t d$

From which the direction of d can be determined..

Evaluating the expression above, results in

$\dfrac{1}{13} \begin{bmatrix} 5 && 12 \\ 12 && -5 \end{bmatrix} ( (10,10) - (0, 5) ) - ( (10, 0) - (0, 5) ) = t d$

$( \dfrac{-20}{13} , \dfrac{160}{13} ) = t d$

Therefore, vector $d$ can be taken as $(-1, 8)$ . And the equation of the incident ray becomes,

$r = (10, 0) + t (-1, 8)$

Pre-multiplying by an orthogonal vector to $(-1,8)$ , for example , $(8, 1 )$ , results in the algebraic equation of the line,

$(8, 1)^T r = 8 x + y = 80$ , hence $y = -8 x + 80 = -8 (x - 10)$ .