Difference between revisions of "Notes:Reflection of ray given normal"
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** {{MM|a_x:\eq \frac{c-c'+(\alpha-t)(\beta-\gamma)}{\gamma} }}, and | ** {{MM|a_x:\eq \frac{c-c'+(\alpha-t)(\beta-\gamma)}{\gamma} }}, and | ||
** {{MM|a_y:\eq \frac{c'-c+(\alpha-t)(\beta m m'-m^2m'-m)}{mm'+1} }} | ** {{MM|a_y:\eq \frac{c'-c+(\alpha-t)(\beta m m'-m^2m'-m)}{mm'+1} }} | ||
| + | * {{M|\Vert b\Vert\eq \vert t\vert\sqrt{1+m^2} \eq t\sqrt{1+m^2} }} as in our model {{M|t\ge 0}} by definition | ||
| + | ===={{M|\theta}} is found==== | ||
| + | So: | ||
| + | * {{MM|\theta\eq\arctan\left(\frac{\Vert a\Vert}{\Vert b\Vert} \right)}} | ||
===Work required=== | ===Work required=== | ||
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Then we just find a line with angle {{M|2\theta}} from the incoming ray, or {{M|-\theta}} from the normal. | Then we just find a line with angle {{M|2\theta}} from the incoming ray, or {{M|-\theta}} from the normal. | ||
==Attempt 2== | ==Attempt 2== | ||
Latest revision as of 21:02, 2 January 2018
Contents
Attempt 1
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):
- [ilmath]\theta\eq\arccos\left(\frac{\Vert b\Vert}{\Vert c\Vert}\right)[/ilmath]
- [ilmath]\theta\eq\arctan\left(\frac{\Vert a\Vert}{\Vert b\Vert}\right)[/ilmath]
- [ilmath]\theta\eq\arcsin\left(\frac{\Vert a\Vert}{\Vert c\Vert}\right)[/ilmath]
Solutions
To ease many expressions we make the following definitions:
- [math]\alpha:\eq\frac{c-c'}{m'-m} [/math],
- [math]\beta:\eq m+\frac{1}{m} [/math] and
- [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]
