Notes:Reflection of ray given normal

Attempt 1

Best diagram I have to hand

We are given two lines:

• $\ell_1$ by $y:\eq mx+c$ and
• $\ell_2$ by $y:\eq m'x+c'$

and

• let $t> 0$ for $t\in\mathbb{R}_{>0}$

We will find $\theta$ - the angle between $\ell_1$ and $\ell_2$

We make the following definitions:

• $I$ - the intersection point of the lines $\ell_1$ and $\ell_2$
• $p$ - the point which is $t$ $x$-units behind $I$ on $\ell_1$
• $N$ - the normal line to $\ell_1$ through $p$
• $q$ - the intersection of $N$ and $\ell_2$
• $b$ - the vector $I-p$
• $c$ - the vector $I-q$
• $a$ - the vector $q-p$

Then we can use any one of the following (for $\Vert\cdot\Vert$ the Euclidean norm):

1. $\theta\eq\arccos\left(\frac{\Vert b\Vert}{\Vert c\Vert}\right)$
2. $\theta\eq\arctan\left(\frac{\Vert a\Vert}{\Vert b\Vert}\right)$
3. $\theta\eq\arcsin\left(\frac{\Vert a\Vert}{\Vert c\Vert}\right)$

Solutions

To ease many expressions we make the following definitions:

1. $$\alpha:\eq\frac{c-c'}{m'-m}$$ ,
2. $$\beta:\eq m+\frac{1}{m}$$ and
3. $$\gamma:\eq m'+\frac{1}{m}$$

To obtain:

• $I\eq\big(\alpha,\ell_1(\alpha)\big)\eq\big(\alpha,\ell_2(\alpha)\big)$
• $p\eq\big(\alpha-t,\ell_1(\alpha-t)\big)$
• $q\eq\big(q_x,\ell_2(q_x)\big)$
  • For: $$q_x:\eq\frac{c-c+(\alpha-t)\beta}{\gamma}$$
• $\Vert a\Vert\eq\sqrt{a_x^2+a_y^2}$, for:
  • $$a_x:\eq \frac{c-c'+(\alpha-t)(\beta-\gamma)}{\gamma}$$ , and
  • $$a_y:\eq \frac{c'-c+(\alpha-t)(\beta m m'-m^2m'-m)}{mm'+1}$$
• $\Vert b\Vert\eq \vert t\vert\sqrt{1+m^2} \eq t\sqrt{1+m^2}$ as in our model $t\ge 0$ by definition

$\theta$ is found

So:

• $$\theta\eq\arctan\left(\frac{\Vert a\Vert}{\Vert b\Vert} \right)$$

Work required

Then we just find a line with angle $2\theta$ from the incoming ray, or $-\theta$ from the normal.

Attempt 2

I'm going to put 15 minutes into the form of line:

• $$\left(\begin{array}{c}x\\y\end{array}\right)\eq\left(\begin{array}{c}u\\v\end{array}\right)t+\left(\begin{array}{c}a\\b\end{array}\right)$$ and perhaps if we make it unit speed parametrisation (see regular curve and arc length) things might simplify a bit more. Recall
  • Dot product: $A\cdot B\eq \Vert a\Vert\Vert b\Vert \cos(\theta)$