Notes:Reflection of ray given normal

Attempt 1

Best diagram I have to hand

We are given two lines:

• [ilmath]\ell_1[/ilmath] by [ilmath]y:\eq mx+c[/ilmath] and
• [ilmath]\ell_2[/ilmath] by [ilmath]y:\eq m'x+c'[/ilmath]

and

• let [ilmath]t> 0[/ilmath] for [ilmath]t\in\mathbb{R}_{>0} [/ilmath]

We will find [ilmath]\theta[/ilmath] - the angle between [ilmath]\ell_1[/ilmath] and [ilmath]\ell_2[/ilmath]

We make the following definitions:

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

Then we can use any one of the following (for [ilmath]\Vert\cdot\Vert[/ilmath] the Euclidean norm):

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

Solutions

To ease many expressions we make the following definitions:

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

To obtain:

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

[ilmath]\theta[/ilmath] is found

So:

• [math]\theta\eq\arctan\left(\frac{\Vert a\Vert}{\Vert b\Vert} \right)[/math]

Work required

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

Attempt 2

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

• [math]\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)[/math] and perhaps if we make it unit speed parametrisation (see regular curve and arc length) things might simplify a bit more. Recall
  • Dot product: [ilmath]A\cdot B\eq \Vert a\Vert\Vert b\Vert \cos(\theta)[/ilmath]