Show that an equation of a plane passing through three noncolinear points
$p_1 = (x_1 , y_1 , z_1)$, $p_2 = (x_2 , y_2 , z_2)$, $p_3 = (x_3 , y_3 , z_3)$ is given by $(p − p_1) \times (p − p_2) \cdot (p − p_3) = 0$,
where $p = (x, y, z)$ is an arbitrary point of the plane and $p − p_1$, for
instance, means the vector $(x − x_1 , y − y_1 , z − z_1)$.
I have the following reasoning:
"Explicitly, the plane $\mathcal{P}$ through the point $P_0 = (x_0, y_0, z_0)$ is uniquely determined (up to a scalar multiple) by a normal vector $\mathbf n = \langle a,b,c \rangle$ according to the following: a point $P$ lies on $\mathcal{P}$ if and only if $\mathbf n$ and $\overrightarrow{P_0 P}$ are orthogonal if and only if $n \cdot \overrightarrow{P_0 P} = 0$ if and only if $$a(x – x_0) + b(y – y_0) + c(z – z_0) = 0.$$ By setting $d = ax_0 + by_0 + cz_0,$ we have $ax + by + cz = d.$"
But I have not been able to finish the problem because I must get the equation of the plane of the form $(p − p_1) \times (p − p_2) \cdot (p − p_3) = 0$. I need help to do this.
Best Answer
As noticed, we should have a $p_3$ instead of a $p$ for the cross product used to determine the normal, then we can use the triple product determinant
$$\left[(p_3 − p_1) \times (p_3 − p_2)\right] \cdot (p − p_3)=\det \begin{vmatrix}x_3-x_1&y_3-y_1&z_3-z_1\\x_3-x_2&y_3-y_2&z_3-z_2\\x-x_3&y-y_3&z-z_3 \end{vmatrix}=0$$
to get the plane equation.