What's the geometric interpretation of this “vector cross product”?

Question

This answer on StackOverflow answers a question about intersection of two segments. Right at the beginning, it introduces a “vector cross product”.

Define the 2-dimensional vector cross product $\vec v \times \vec w$ to be $v_x w_y − v_y w_x$.

However, this doesn't seem to be a regular cross product (nor does it even produce a vector). I realize that the formula is a simple determinant of the two vectors, but I cannot understand its meaning or relation to the rest of the post.

Does it have a meaning (motivation) or is it just a “lucky-guess” operation in order to transform the equation $\vec p + t\vec r = \vec q + u \vec s$ into a solvable state? In other words, how does this operation (intuitively) relate to the described algorithm?

Answer 1

Apparently, this is known as perp product; however, the only online reference I could find after a (rather quick) search is this geomalgorithms page. It's also on Wolfram, where it mentions that Hill introduced it in 1994, in a chapter in “Graphic Gems IV”.

He firstly defines perp operator (perpendicular operator) which gives a rotation of a vector by 90 degeres:

$$ \vec v^\perp = (v_x, v_y)^\perp = (-v_y, v_x). $$

Then a perp product (perpendicular product) is defined, denoted as an infix operator $\perp$:

$$ \vec v \perp \vec w := \vec v ^ \perp \cdot \vec w = v_x w_y - v_y w_x. $$

The idea of using the perp product in the algorithm seems to come from the following property: $$\vec v \perp \vec w = 0 \Leftrightarrow \text{$\vec v$ and $\vec w$ are collinear}. $$

At the end of the algorithm, the denominators are $\vec r \perp \vec s$, which is then discussed for being zero, implying collinearity of the vectors and parallelity of the segments.

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Either way, a similar but much more intuitive discussion of the problem with the same idea of using parametric equations and perp product is given at this page called “Intersections of Lines and Planes” on geomalgorithms. It also gives more details in the algorithm at the bottom, considering cases where one or both segments are degenerated into a single point.

Answer 2

Geometrically it gives the (signed) area of the parallelogram defined by the two vectors.

If you multiply by the appropriate unit normal to the plane you get the normal three dimensional cross product. You don't get a vector in the plane though.

If you try to define a "cross product" in four dimensions, you might appreciate that the familiar situation in three dimensions is a happy coincidence which trips up people who try to generalise in the wrong way.