Plane Through Three Points

Equation of the plane passing through three non-collinear points P₁(x₁,y₁,z₁), P₂, P₃. The determinant expresses the coplanarity of (x,y,z) with the three given points.

Let three non-collinear points be $P_1(x_1,y_1,z_1)$, $P_2(x_2,y_2,z_2)$, $P_3(x_3,y_3,z_3)$.

Two vectors in the plane: $\mathbf{v_1} = \overrightarrow{P_1P_2} = (x_2-x_1, y_2-y_1, z_2-z_1)$ and $\mathbf{v_2} = \overrightarrow{P_1P_3}$.

Normal to the plane: $\mathbf{n} = \mathbf{v_1}\times\mathbf{v_2}$.

Plane through $P_1$ with normal $\mathbf{n}$:

$$(\mathbf{r}-\mathbf{P_1})\cdot(\mathbf{v_1}\times\mathbf{v_2}) = 0$$

Writing this as a scalar triple product:

$$(\mathbf{r}-\mathbf{P_1})\cdot\mathbf{v_1}\cdot\mathbf{v_2} = 0$$

In determinant form (the scalar triple product of three vectors is the $3\times3$ determinant with those vectors as rows):

$$\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{vmatrix} = 0$$

When the three points are collinear: The determinant is always zero — no unique plane exists. Any plane through the line works.