The coordinates of a line in two dimensions may be given in this form:
If [MATH]a[/MATH] and [MATH]b[/MATH] take some vector that is perpendicular to the line in question, then it is a simple matter to solve for [MATH]c[/MATH] using a point on the line, giving the line's coordinates.
Say you take two such lines, [MATH]ax + by + c = 0[/MATH] and [MATH]dx + ey + f = 0[/MATH], and find their point of intersection:
[MATH]x = \frac{bf - ce}{ae - bd}, y = \frac{cd - af}{ae - bd}[/MATH]
If the lines are parallel, then [MATH]ae = bd[/MATH] and a division by zero results.
Now pretend we live in an alternate universe where the coordinates of these lines in two dimensions are actually the components of vectors in three dimensions and take their cross product:
This winds up as the same operation as above, giving the position of the point of intersection in homogeneous coordinates such that [MATH]x' = \frac{x}{w}, y' = \frac{y}{w}[/MATH]. If the lines are parallel, then [MATH]w = 0[/MATH] and a division by zero results.
It's clear by following the algebra from the first example that it ultimately becomes the same operations as the cross product, but... why? Is there some relationship between two-dimensional lines and three-dimensional vectors? Or is it just a coincidence?
[MATH]ax + by + c = 0[/MATH]
If [MATH]a[/MATH] and [MATH]b[/MATH] take some vector that is perpendicular to the line in question, then it is a simple matter to solve for [MATH]c[/MATH] using a point on the line, giving the line's coordinates.
Say you take two such lines, [MATH]ax + by + c = 0[/MATH] and [MATH]dx + ey + f = 0[/MATH], and find their point of intersection:
Solve for [MATH]x[/MATH]:
Solve for [MATH]y[/MATH]:
[MATH]aex + bey + ec = 0[/MATH][MATH]bdx + bey + bf = 0[/MATH][MATH]aex - bdx + ec - bf = 0[/MATH][MATH](ae - bd)x = bf - ec[/MATH][MATH]x = \frac{bf - ec}{ae - bd}[/MATH]
Solve for [MATH]y[/MATH]:
[MATH]adx + aey + af = 0[/MATH][MATH]adx + bdy + cd = 0[/MATH][MATH]aey - bdy + af - cd = 0[/MATH][MATH](ae - bd)y = cd - af[/MATH][MATH]y = \frac{cd - af}{ae - bd}[/MATH]
[MATH]x = \frac{bf - ce}{ae - bd}, y = \frac{cd - af}{ae - bd}[/MATH]
If the lines are parallel, then [MATH]ae = bd[/MATH] and a division by zero results.
Now pretend we live in an alternate universe where the coordinates of these lines in two dimensions are actually the components of vectors in three dimensions and take their cross product:
[MATH]\begin{bmatrix}x\\y\\w\end{bmatrix} = \begin{bmatrix}a\\b\\c\end{bmatrix} \times \begin{bmatrix}d\\e\\f\end{bmatrix}= \begin{bmatrix}\textit{bf - ce}\\\textit{cd - af}\\\textit{ae - bd}\end{bmatrix}[/MATH]
This winds up as the same operation as above, giving the position of the point of intersection in homogeneous coordinates such that [MATH]x' = \frac{x}{w}, y' = \frac{y}{w}[/MATH]. If the lines are parallel, then [MATH]w = 0[/MATH] and a division by zero results.
It's clear by following the algebra from the first example that it ultimately becomes the same operations as the cross product, but... why? Is there some relationship between two-dimensional lines and three-dimensional vectors? Or is it just a coincidence?