Equation of a plane by given 3 points

Question

I've read of solutions, with let's say points $P(1,2,3)$, $Q(3,1,2)$, $R(2,3,1)$, where one subtracts point $P-Q$ and $P-R$ to get two vectors:

$\begin{pmatrix}1\\2\\3\end{pmatrix}-\begin{pmatrix}3\\1\\2\end{pmatrix}=\begin{pmatrix}-2\\1\\1\end{pmatrix}$

$\begin{pmatrix}1\\2\\3\end{pmatrix}-\begin{pmatrix}2\\3\\1\end{pmatrix}=\begin{pmatrix}-1\\-1\\2\end{pmatrix}$

to now get the cross-product:

$\begin{pmatrix}-2\\1\\1\end{pmatrix}\times\begin{pmatrix}-1\\-1\\2\end{pmatrix}=\begin{pmatrix}2\\-1\\2\end{pmatrix}$

hence the equation is $2x-y+2z=d$

by inserting, for example, point P we get

$2\times1-2+2\times3=d=6$

and thus

$2x-y+2z-6=0$

which can be transformed to

$z=-x+\frac12y+3$

so when I put in any point it should work right? When I put in R though, I get:

$1=\frac52$

So, what did I do wrong?

Answer 1

That's because “cross-product” doesn't mean to multiply each entry be the corresponding entry of the other vector. In this case$$\begin{pmatrix}-2\\1\\1\end{pmatrix}\times\begin{pmatrix}-1\\-1\\2\end{pmatrix}=\begin{pmatrix}3\\3\\3\end{pmatrix}.$$