Equations defining Projection in Plane/Cartesian
Theorem
Let $M$ and $N$ be distinct lines in the plane, both of which pass through the origin.
Let the angle between $M$ and the $x$-axis be $\theta$.
Let the angle between $N$ and the $x$-axis be $\phi$.
Let $\pr_{M, N}$ be the projection on $M$ along $N$:
- $\forall P \in \R^2: \map {\pr_{M, N} } P =$ the intersection of $M$ with the line through $P$ parallel to $N$.
Let $P = \tuple {x, y}$ be an arbitrary point in the plane.
Then:
- $\map {\pr_{M, N} } P = \begin {cases}
\tuple {0, y - x \tan \phi} & : \theta = \dfrac \pi 2 \\ \tuple {x, x \tan \theta} & : \phi = \dfrac \pi 2 \\ \tuple {\dfrac {x \tan \phi} {\tan \phi - \tan \theta} - \dfrac y {\tan \phi - \tan \theta}, \dfrac {y \tan \theta} {\tan \theta - \tan \phi} - \dfrac {x \tan \theta \tan \phi} {\tan \theta - \tan \phi} } & : \text {otherwise} \end {cases}$
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Proof
Let $P = \tuple {x_1, y_1} \in \R^2$ be arbitrary.
From Equation of Straight Line in Plane: Slope-Intercept Form, we can express $M$ and $N$ as follows:
| \(\ds M: \, \) | \(\ds y\) | \(=\) | \(\ds m x\) | |||||||||||
| \(\ds N: \, \) | \(\ds y\) | \(=\) | \(\ds n x\) |
where:
| \(\ds m\) | \(=\) | \(\ds \tan \theta\) | ||||||||||||
| \(\ds n\) | \(=\) | \(\ds \tan \phi\) |
Let $\pr_{M, N}$ map $P = \tuple {x_1, y_1}$ onto $P' = \tuple {x_2, y_2}$.
From Equation of Straight Line in Plane: Point-Slope Form, $PP'$ can be expressed as:
- $y - y_1 = n \paren {x - x_1}$
This holds also for $\tuple {x_2, y_2}$, and so:
| \(\ds y_2 - y_1\) | \(=\) | \(\ds n \paren {x_2 - x_1}\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds n x_2\) | \(=\) | \(\ds y_2 - y_1 + n x_1\) | |||||||||||
| \(\ds \) | \(=\) | \(\ds m x_2 - y_1 + n x_1\) | as $\tuple {x_2, y_2} \in M$ | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \paren {n - m} x_2\) | \(=\) | \(\ds n x_1 - y_1\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x_2\) | \(=\) | \(\ds \dfrac {n x_1} {n - m} - \dfrac {y_1} {n - m}\) |
Similarly:
| \(\ds y_2 - y_1\) | \(=\) | \(\ds n \paren {x_2 - x_1}\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \dfrac {y_2} n\) | \(=\) | \(\ds x_2 - x_1 + \dfrac {y_1} n\) | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {y_2} m - x_1 + \dfrac {y_1} n\) | as $\tuple {x_2, y_2} \in M$ | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \paren {\dfrac 1 n - \dfrac 1 m} y_2\) | \(=\) | \(\ds \dfrac {y_1} n - x_1\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y_2\) | \(=\) | \(\ds \dfrac {m y_1} {m - n} - \dfrac {m n x_1} {m - n}\) |
The result follows by substituting for $m$ and $n$.
$\Box$
Special Cases
When either $M$ or $N$ are coincident with the $y$-axis, then either $\tan \theta$ or $\tan \phi$ is undefined.
Hence these cases are analysed as special cases.
Let $M$ be coincident with the $y$-axis.
By definition of the projection on $M$ along $N$:
- $x_2 = 0$
while from the definition of tangent:
- $\tan \phi = \dfrac {y_1 - y_2} {x_1}$
from which it follows that:
- $y_2 = y_1 - x_1 \tan \phi$
Hence if $P = \tuple {x, y}$ is an arbitrary point in the plane, then:
- $\map {\pr_{M, N} } P = \tuple {0, y - x \tan \phi}$
Let $N$ be coincident with the $y$-axis.
By definition of the projection on $M$ along $N$:
- $x_2 = x_1$
while from the definition of tangent:
- $\tan \theta = \dfrac {y_2} {x_2}$
from which it follows that:
- $y_2 = x_1 \tan \theta$
Hence if $P = \tuple {x, y}$ is an arbitrary point in the plane, then:
- $\map {\pr_{M, N} } P = \tuple {x, x \tan \theta}$
$\blacksquare$