Cartesian equation of plane through $3$ points

Question

Let there are three points $(2,5,-3),(5,3,-3),(-2,-3,5)$ through which a plane passes. What is the equation of the plane in Cartesian form?

I know how to find it in using vector form by computing the cross product to get the normal vector and passing through any one of the given points. But I want to do it a bit differently.

We know, the equation of any plane passing through the first point is $$a(x-2)+b(y-5)+c(z+3)=0$$

This equation must satisfy the other two points. However, this given me two equations with three unknowns $a,b,c$. So can I not solve by this method?

Answer 1

After using Cramer's rule, $a=-2,b=-3,c=-4$.

Answer 2

You in fact can solve it by this method, even if it does seem like there will be infinite solutions. This is because the normal vector for a plane is not unique: if $\vec{n}$ is a normal vector to the plane, then $c\vec{n}$ is as well, provided that $c\neq 0$.

For your example, your two equations for the remaining two points yield

$$3a-2b=0$$

$$-4a-8b+8c=0$$

One way you could solve this is by multiplying the first equation by $-4$ and then adding them, giving you

$$-16a+8c=0\implies c=2a$$

Rearranging the first equation gives

$$b=\frac{3}{2}a$$

As you pointed out, there are not enough equations for a unique solution, but once we specify a value for $a$, we gain a value for $b$ and $c$. Moreover, should we scale $a$, this scales $b$ and $c$ by the same amount, which is exactly what we would expect. The normal that I would choose here is when $a=2$, so $b=3$ and $c=4$.

Answer 3

Arrange the two equations into an augmented matrix to get:

M= \begin{bmatrix} 3 & -2 & 0 & 0 \\ -4 & -8 & 8 & 0 \end{bmatrix}

The general solution of this linear system is any scalar multiple of

x = \begin{bmatrix} 2\\ 3\\ 4\\ \end{bmatrix}

This follows from that the normal vector of a plane is not uniquely determined.

You can check that if we take the coordinates of the normal as our a, b, c, then you will get the equation of the plane regardless of the choice of $c\vec{x}$.

Answer 4

Remember that the implicit Cartesian equation for a plane is not unique. If you mutiply both sides of such an equation by a nonzero constant, you get another equation for the same plane. So, having an undetermined system for the coefficients of this equation is to be expected. Choose any convenient solution of the system.