imported>Paul Wormer |
imported>Paul Wormer |
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| ===Point-normal representation=== | | ==Parabolic mirror== |
| {{Image|Param1 plane.png|right|350px|<small>Fig. 1. Equation for plane. ''P'' is arbitary point in plane; <font style<nowiki>=</nowiki> "vertical-align: top;"> <math>\scriptstyle\vec{d}</math></font> and <font style<nowiki>=</nowiki> "vertical-align: text-top;"> <math>\scriptstyle \vec{n}_0 </math></font> are collinear.</small>}} | | {{Image|Refl parab.png|right|350px|Fig. 2. Reflection in a parabolic mirror}} |
| In [[analytic geometry]] several closely related algebraic equations are known for a plane in three-dimensional Euclidean space. One such equation is illustrated in figure 1. Point ''P'' is an arbitrary point in the plane and ''O'' (the origin) is outside the plane. The point ''A'' in the plane is chosen such that vector
| | Parabolic mirrors concentrate incoming vertical light beams in their focus. We show this. |
| :<math>
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| \vec{d} \equiv \overrightarrow{OA}
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| </math>
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| is orthogonal to the plane. The collinear vector
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| :<math>
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| \vec{n}_0 \equiv \frac{1}{d} \vec{d} \quad \hbox{with}\quad d \equiv \left|\vec{d}\,\right|
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| </math>
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| is a unit (length 1) vector normal (perpendicular) to the plane which is known as the ''normal of the plane in point A''. Note that ''d'' is the distance of ''O'' to the plane. The following relation holds for an arbitrary point ''P'' in the plane
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| :<math>
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| \left(\vec{r}-\vec{d}\;\right)\cdot \vec{n}_0 = 0 \quad\hbox{with}\quad \vec{r} \equiv\overrightarrow{OP}\quad\hbox{and}\quad \vec{r}-\vec{d} = \overrightarrow{AP} .
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| </math>
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| This equation for the plane can be rewritten in terms of coordinates with respect to a Cartesian frame with origin in ''O''. Dropping arrows for component vectors (real triplets) that are written bold, we find
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| :<math>
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| \left( \mathbf{r} - \mathbf{d}\right)\cdot \mathbf{n}_0 = 0
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| \Longleftrightarrow
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| x a_0 +y b_0+z c_0 = d
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| </math>
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| with
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| :<math>
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| \mathbf{d} = (a,\;b,\; c), \quad
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| \mathbf{n}_0 = (a_0,\;b_0,\; c_0), \quad
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| \mathbf{r} = (x,\;y,\; z),
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| </math>
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| and
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| :<math>
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| \mathbf{d}\cdot \mathbf{n}_0 = \frac{1}{d} \mathbf{d}\cdot \mathbf{d} = d = \sqrt{a^2+b^2+c^2}.
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| </math>
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|
| | Consider in figure 2 the arbitrary vertical light beam (blue, parallel to the ''y''-axis) that enters the parabola and hits it at point ''P'' = (''x''<sub>1</sub>, ''y''<sub>1</sub>). The parabola (red) has focus in point ''F''. The incoming beam is reflected at ''P'' obeying the well-known law: incidence angle is angle of reflection. The angles involved are with the line ''APT'' which is tangent to the parabola at point ''P''. It will be shown that the reflected beam passes through ''F''. |
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| Conversely, given the following equation for a plane
| | Clearly ∠''BPT'' = ∠''QPA'' (they are vertically opposite angles). Further ∠''APQ'' = ∠''FPA'' because the triangles ''FPA'' and ''QPA'' are congruent and hence ∠''FPA'' = ∠''BPT''. |
| :<math>
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| ax+by+cz = e, \,
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| </math>
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| it is easy to derive the same equation.
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| Write
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| :<math>
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| \mathbf{r} = (x,\;y,\; z), \quad\mathbf{f} = (a,\;b,\; c), \quad\hbox{and}\quad
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| \mathbf{d} \equiv \left(\frac{e}{a^2+b^2+c^2}\right) \mathbf{f}.
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| </math>
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| It follows that
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| :<math>
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| \mathbf{f}\cdot\mathbf{r} = e = \mathbf{f}\cdot \mathbf{d}.
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| </math>
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| Hence we find the same equation,
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| :<math>
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| \mathbf{f}\cdot(\mathbf{r}-\mathbf{d}) = 0 \;\Longrightarrow\; (\mathbf{r}-\mathbf{d})\cdot\mathbf{n}_0 = 0 \quad\hbox{with}\quad \mathbf{n}_0 = \frac{1}{\sqrt{a^2+b^2+c^2}}\mathbf{f}
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| </math>
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| where '''f''' , '''d''', and '''n'''<sub>0</sub> are collinear. The equation may also be written in the following mnemonically convenient form
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| :<math>
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| \mathbf{d}\cdot(\mathbf{r}-\mathbf{d}) = 0,
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| </math>
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| which is the equation for a plane through a point ''A'' perpendicular to <font style="vertical-align: super;"><math>\overrightarrow{OA}</math></font>.
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| ===Three-point representation===
| | We prove the congruence of the triangles: By the definition of the parabola the line segments ''FP'' and ''QP'' are of equal length, because the length of the latter segment is the distance of ''P'' to the directrix and the length of ''FP'' is the distance of ''P'' to the focus. The point ''F'' has the coordinates (0,''f'') and the point ''Q'' has the coordinates (''x''<sub>1</sub>, −''f''). The line segment ''FQ'' has the equation |
| {{Image|Param2 plane.png|right|350px|<small> Fig. 2. Plane through points ''A'', ''B'', and ''C''.</small>}}
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| Figure 2 shows a plane that by definition passes through non-coinciding points ''A'', ''B'', and ''C'' that moreover are not on one line. The point ''P'' is an arbitrary point in the plane and the reference point ''O'' is outside the plane. Referring to figure 2 we introduce the following definitions
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| :<math>
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| \vec{a} = \overrightarrow{OA},\quad \vec{b} = \overrightarrow{OB},\quad\vec{c} = \overrightarrow{OC},\quad \vec{r} = \overrightarrow{OP}.
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| </math> | |
| Clearly the following two non-collinear vectors belong to the plane
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| :<math> | | :<math> |
| \vec{u} = \overrightarrow{AB}= \vec{b}-\vec{a} ,\quad \vec{v} = \overrightarrow{AC}= \vec{c}-\vec{a}. | | \lambda\begin{pmatrix}0\\ f\end{pmatrix} + (1-\lambda)\begin{pmatrix}x_1\\ -f\end{pmatrix}, \quad 0\le\lambda\le 1. |
| </math> | | </math> |
| Because a plane (an [[affine space]]), with a given fixed point as origin is a 2-dimensional [[linear space]] and two non-collinear vectors with "tails" in the origin are [[linearly independent]], it follows that any vector in the plane can be written as a linear combination of these two non-collinear vectors (this is also expressed as: any vector in the plane can be decomposed into components along the two non-collinear vectors). In particular, taking ''A'' as origin in the plane,
| | The midpoint ''A'' of ''FQ'' has coordinates (λ = ½): |
| :<math> | | :<math> |
| \overrightarrow{AP}= \vec{r}-\vec{a} = \lambda \vec{u} + \mu\vec{v},\qquad \lambda,\mu \in \mathbb{R}. | | \frac{1}{2}\begin{pmatrix}0\\ f\end{pmatrix} + \frac{1}{2}\begin{pmatrix}x_1\\ -f\end{pmatrix} = |
| | \begin{pmatrix}\frac{1}{2} x_1\\ 0\end{pmatrix}. |
| </math> | | </math> |
| The real numbers λ and μ specify the direction of <font style="vertical-align: super;"><math>\overrightarrow{AP}</math></font>. Hence the following equation for the position vector <math>\vec{r}</math> of the arbitrary point ''P'' in the plane:
| | Hence ''A'' lies on the ''x''-axis. |
| | The parabola has equation, |
| :<math> | | :<math> |
| \vec{r} = \vec{a} + \lambda \vec{u} + \mu\vec{v}
| | y = \frac{1}{4f} x^2. |
| </math> | | </math> |
| is known as the ''point-direction representation'' of the plane. This representation is equal to the ''three-point representation''
| | The equation of the tangent at ''P'' is |
| :<math>
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| \vec{r} = \vec{a}+ \lambda (\vec{b}-\vec{a}) + \mu(\vec{c}-\vec{a}),
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| </math>
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| where <math>\vec{a}</math>, <font style="vertical-align: top;"><math>\vec{b}</math></font>, and <math>\vec{c}</math> are the position vectors of the three points that define the plane.
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| Writing for the position vector of the arbitrary point ''P'' in the plane
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| :<math> | | :<math> |
| \vec{r} = (1-\lambda-\mu)\, \vec{a}+ \lambda\, \vec{b} + \mu\,\vec{c} \;\equiv\; \xi_1\, \vec{a} +\xi_2\,\vec{b} + \xi_3\, \vec{c}\; ,
| | y = y_1 + \frac{x_1}{2f} (x-x_1)\quad \hbox{with}\quad y_1 = \frac{x_1^2}{4f}. |
| </math> | | </math> |
| we find that the real triplet (ξ<sub>1</sub>, ξ<sub>2</sub>, ξ<sub>3</sub>) with ξ<sub>1</sub> + ξ<sub>1</sub> + ξ<sub>1</sub> = 1 forms a set of coordinates for ''P''. The numbers {ξ<sub>1</sub>, ξ<sub>2</sub>, ξ<sub>3</sub> | ξ<sub>1</sub>+ ξ<sub>2</sub>+ ξ<sub>3</sub> = 1 } are known as the ''[[barycentric coordinates]]'' of ''P''. It is trivial to go from barycentric coordinates to the "three-point representation",
| | This line intersects the ''x''-axis at ''y'' = 0, |
| :<math> | | :<math> |
| \vec{r} = \xi_1 \vec{a} + \xi_2\vec{b} + \xi_3 \vec{c}\quad\hbox{with}\quad \xi_1 = 1- \xi_2-\xi_3 | | 0 = \frac{x_1^2}{4f} - \frac{x_1^2}{2f} + \frac{x_1}{2f} x |
| \;\Longleftrightarrow\; | | \Longrightarrow \frac{x_1}{2f} x = \frac{x_1^2}{4f} \longrightarrow x = \tfrac{1}{2}x_1. |
| \vec{r} = \vec{a} + \xi_2 (\vec{b}-\vec{a}) + \xi_3(\vec{c}-\vec{a}).
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| </math> | | </math> |
| | The intersection of the tangent with the ''x''-axis is the point ''A'' = (½''x''<sub>1</sub>, 0) that lies on the midpoint of ''FQ''. The corresponding sides of the triangles ''FPA'' and ''QPA'' are of equal length and hence the triangles are congruent. |
Parabolic mirror
PD Image Fig. 2. Reflection in a parabolic mirror
Parabolic mirrors concentrate incoming vertical light beams in their focus. We show this.
Consider in figure 2 the arbitrary vertical light beam (blue, parallel to the y-axis) that enters the parabola and hits it at point P = (x1, y1). The parabola (red) has focus in point F. The incoming beam is reflected at P obeying the well-known law: incidence angle is angle of reflection. The angles involved are with the line APT which is tangent to the parabola at point P. It will be shown that the reflected beam passes through F.
Clearly ∠BPT = ∠QPA (they are vertically opposite angles). Further ∠APQ = ∠FPA because the triangles FPA and QPA are congruent and hence ∠FPA = ∠BPT.
We prove the congruence of the triangles: By the definition of the parabola the line segments FP and QP are of equal length, because the length of the latter segment is the distance of P to the directrix and the length of FP is the distance of P to the focus. The point F has the coordinates (0,f) and the point Q has the coordinates (x1, −f). The line segment FQ has the equation
The midpoint A of FQ has coordinates (λ = ½):
Hence A lies on the x-axis.
The parabola has equation,
The equation of the tangent at P is
This line intersects the x-axis at y = 0,
The intersection of the tangent with the x-axis is the point A = (½x1, 0) that lies on the midpoint of FQ. The corresponding sides of the triangles FPA and QPA are of equal length and hence the triangles are congruent.
