Equation of a plane from cross product

Question

I'm working from Do Carmo, and I ran into another snag.

More specifically, 1.4.5:

Given points $p_1, p_2, p_3 \in \mathbb{R}^3$, show that the following expression gives the equation for the plane containing these points:

$((p-p_1) \wedge (p-p_2)) \bullet (p-p_3) = 0$ for an arbitrary point on the plane $p=(x,y,z)$

My issue is that Do Carmo doesn't mention any equation for planes, so I'm not sure what he wants here. Furthermore, I think I may be missing some nuance of it as direct computation was very messy and unfruitful.
One may simplify the expression to $(p \wedge p_2 + p_1 \wedge p) \bullet p_3 + (p_1 \wedge p_2) \bullet (p-p_3)$ but the usefulness of this is questionable at best.

Note that $u \wedge v$ here is equivalent to the cross product.

Any help or clarification is appreciated.

Answer 1

There are two main ways to express the equation of a plane - vector format or Cartesian equation. The vector format, in simplest form, is usually written $\vec{n}.(\vec{r}-r_0)=0$ where $\vec{n}$ is a normal vector to the plane, $\vec{r}$ is the variable vector (typically $(x,y,z)$ in 3-space) and $r_0$ is some given initial point. In the formatting given in your book, $p$ is what I'm calling $\vec{r}$, $p_3$ is what I'm calling $r_0$ and $(p- p_1)\wedge(p-p_2)$ is what I'm calling $\vec{n}$.

The other main way of writing the equations for planes is in the form $ax+by+cz=d$ where $\vec{n} = (a,b,c)$ and we have that $d=ax_0+by_0+cz_0$ for some point $(x_0, y_0, z_0)$ on the plane. This form can also be written as $a(x-x_0)+b(y-y_0)+c(z-z_0) =0$ to make the connection between the two clearer.

It is very easy to convert from the vector form to the Cartesian form by actually just evaluating the dot product and moving the constant term to the other side. What typically takes the most work is finding $\vec{n}$ in the first place. The form that you have from the book is giving you an explicit way to setup up the problem that always works. Let's try an example:

Suppose we have $p_1 = (3,4,5)$, $p_2 = (1,2,3)$ and $p_3 = (4,7,6)$. The formula says to find the cross product of the vector versions of $p_1$ and $p_2$ and then dot that with $p-p_3$. So

$(p-p_1)\wedge(p-p_2) = (x-3,y-4,z-5) \wedge (x-1,y-2,z-3) = (2y-2z+2,-2x+2z-4,2x-2y+2)$

One of the techniques often employed is that we need a normal vector. It does not have to be the exact one from the cross product but it can be any multiple. So we usually try to pick the simplest one possible. In this case, everything has a factor of 2 so we will use $(y-z+1,-x+z-2,x-y+1)$ instead. (If you skip simplifying the result here, you will be able to simplify the equation later.) Next we do

$ ((p-p_1)\wedge(p-p_2))\bullet(p-p_3) = 0 \\ \Longleftrightarrow (y-z+1,-x+z-2,x-y+1) \bullet (x-4,y-7,z-6) =0 \\ \Longleftrightarrow (y-z+1)(x-4)+(-x+z-2)(y-7)+(x-y+1)(z-6) = 0 \\ \Longleftrightarrow xy-xz+x-4y+4z-4-xy+yz-2y+7x-7z+14+xz-yz+z-6x+6y-6 = 0\\ \Longleftrightarrow 2x-2z+4 =0\\ \Longleftrightarrow x-z = -2 \\ $ or in vector form $(1,0,-1)\bullet(x-4,y-7,z-6) = 0$

In my opinion, this particular formulation is pretty hard to deal with compared to the methods I'm used to applying to these problems, which only involve vectors with numbers in the cross product to make the calculation a bit easier. Also note that I haven't actually proved it works in general but ran through the result with a specific set of numbers for you to see how it works out. I suspect you are meant to put it into $ax+by+cz =d$ form as this seems like the most natural result of the work done above.

Answer 2

The advantage of your simplified form

$$(p \wedge p_2 + p_1 \wedge p) \bullet p_3 + (p_1 \wedge p_2) \bullet (p-p_3) = 0$$

after writing it in the equivalent form

$$(p\wedge p_2)\bullet p_3 + (p_1 \wedge p)\bullet p_3 + (p_1 \wedge p_2) \bullet p = (p_1 \wedge p_2) \bullet p_3$$

is that we have a linear function of $p$ on the left and a constant on the right. Linearity can be verified by replacing $p$ with a linear combination, say $ap+bq$ and performing basic operations involving cross and dot products. For example,

$$(p_1 \wedge p_2) \bullet (ap+bq) = a(p_1 \wedge p_2)\bullet p + b(p_1 \wedge p_2) \bullet q.$$

Now and equation of the form $L(p)=c$, where $L$ is a linear function is certainly the equation of some plane, for if $p$ and $q$ are points in $\mathbb R^3$ with $L(p)=c$ and $L(q)=c$ and $r\in\mathbb R$, then $$L(rp+(1-r)q) = rL(p)+(1-r)L(q) = rc+(1-r)c=c.$$ Thus, the entire line $rp+(1-r)q$ parametrized by $r$ lies in the surface.

Finally, you can see that this is the equation of the plane you want by setting $p=p_i$ for $i=1,2,3$ in your original equation; in all cases, you get a true statement.

Note: An alternative, and more elementary, way to see that your equation is the equation of some plane is to simply set $p=\langle x,y,z \rangle$ and $p_i = \langle x_i,y_i,z_i \rangle$ and expand your equation out to get it to the form $$ax+by+cz=d.$$ If this is a class in differential geometry, I'd guess that the higher level (and computationally much shorter) approach is expected.

Incidentally, the expression $\vec{u}\cdot(\vec{v}\times\vec{w})$ is often called the vector triple product. It is well known to return the volume of the parallelepiped spanned by $\vec{u}$, $\vec{v}$, and $\vec{w}$ and is therefore zero only if the three vectors are coplanar. Setting $\vec{u}=p-p_1$, $\vec{v}=p-p_2$, and $\vec{w}=p-p_3$ provides a geometric explanation of your formula.