Ask Uncle Colin: The cross product form of a line
Dear Uncle Colin,
Is there any benefit to writing the equation of a line in 3D space in the form $\br{\bb{r} - \bb{r_0}}\times\bb{d} = \bb{0}$?
Doesn’t Everyone See Cartesian Are Remarkably Terse Equations, Sir?
Hi, DESCARTES, and thanks for your message((But don’t call me sir. My bank calls me sir and it makes me uneasy.))!
The answer is yes, the vector product form has several things going for it, but it might do other readers some good to talk about it a bit. It’s not immediately obvious that it defines a line, is it?
At least, until you remember that the cross product of two parallel vectors is zero. All it says is that the displacement vector of $\bb{r}$ from a reference point on the line ($\bb{r_0}$) is parallel to the direction vector $\bb{d}$.
But obviously, vector products are a pain to work out – so why not stick with one of the other forms? (I’ve nothing against the other forms, by the way).
There are (at least) a couple of neat things about this form, and for me it comes down to the parallels between this form of a line and the very similar form for a plane, $\br{\bb {r} - \bb{r_0}}\cdot \bb{n} = 0$. Look! They’re the same, except for a different kind of product and a different kind of zero.
Moreover, if the direction/normal vectors are unit vectors, we get the really lovely results for any point in space:
- $\left|\br{\bb {r} - \bb{r_0}} \cdot \bb{\hat{n}}\right| = \delta$, where $\delta$ is the distance of the point from the plane((Without the modulus, it’s the displacement in the direction of the normal vector)); and
- $\left| \br{\bb {r} - \bb{r_0}} \times \bb{\hat{d}}\right| = \Delta$, where $\Delta$ is the distance from the line.
And that’s not all!
- $-\br{\br{\bb {r} - \bb{r_0}} \cdot \bb{\hat{n}}}\bb{\hat{n}}$ is the vector that takes point $\bb{r}$ to the plane
- $\br{\br{\bb {r} - \bb{r_0}} \times \bb{\hat{d}}}\times \bb{\hat{d}}$ is the vector that takes point $\bb{r}$ to the line ((I might have a sign error here. Let me know.)).
So this form of the line and plane make it (relatively) easy to find several useful pieces of information linking any point to the object in question. (You might also be interested to think about how this relates to a quaternion product, but I’ll leave that for you to think about in your own time).
Hope that helps!
- Uncle Colin
* Edited 2022-10-17 to fix LaTeX and logic. Thanks to @RobJLow.