General equation of a plane

For every point of the plane $π$ , we can consider three parametric equations as a system of equations with two unknowns, $λ$ and $μ$ , that must have only one solution. Therefore the system is: $\begin{array}{rcl} x - a_{1} & = & λ \cdot u_{1} + μ \cdot v_{1} \\ y - a_{2} & = & λ \cdot u_{2} + μ \cdot v_{2} \\ z - a_{3} & = & λ \cdot u_{3} + μ \cdot v_{3} \end{array}}$

It has to be compatible and determined and, therefore, the following determinant must be 0: $| \begin{matrix} x - a_{1} & u_{1} & v_{1} \\ y - a_{2} & u_{2} & v_{2} \\ z - a_{3} & u_{3} & v_{3} \end{matrix} | = 0$ If we develop the previous determinant we obtain: $(u_{2} v_{3} - u_{3} v_{2}) \cdot x + (u_{3} v_{1} - u_{1} v_{3}) \cdot y + (u_{1} v_{2} - u_{2} v_{1}) \cdot z + + [- a_{1} (u_{2} v_{3} - u_{3} v_{2}) - a_{2} (u_{3} v_{1} - u_{1} v_{3}) - a_{3} (u_{1} v_{2} - u_{2} v_{1})] = 0$ And if we call $A, B$ and $C$ the coefficients of $x, y, z$ , and $D$ the independent term, we obtain the linear equation: $A x + B y + C z + D = 0$ which is known as the general, Cartesian or implicit equation of the plane.

Also the vector $\vec{v} = (A, B, C)$ is the vector perpendicular to the plane.

Example

Consider points $A = (1, - 3, 5), B = (- 2, 2, - 1)$ and $C = (1, - 1, 0)$ , and find the general equations of the plane that they determine.

The vector equation is: $(x, y, z) = (1, - 3, 5) + λ \cdot (- 3, 5, - 6) + μ \cdot (0, 2, - 5)$ and the parametric equations are: ${\begin{array}{rcl} x & = & 1 - 3 λ \\ y & = & - 3 + 5 λ + 2 μ \\ z & = & 5 - 6 λ - 5 μ \end{array}$

If we write the determinant of the system and equate it to zero we have: $| \begin{matrix} x - 1 & - 3 & 0 \\ y + 3 & 5 & 2 \\ z - 5 & - 6 & - 5 \end{matrix} | = 0$ And if we develop it: $| \begin{matrix} x - 1 & - 3 & 0 \\ y + 3 & 5 & 2 \\ z - 5 & - 6 & - 5 \end{matrix} | = - 25 (x - 1) - 6 (z - 5) - 15 (y + 3) + 12 (x - 1) = = - 25 x + 25 - 6 z + 30 - 15 y - 45 + 12 x - 12 = - 13 x - 15 y - 6 z - 2 = 0$ An important characteristic of the general equation of the plane is that it allows us to obtain a normal vector by just looking at the equation.

If the equation is $A x + B y + C z + D = 0$ then $\vec{n} = (A, B, C)$ is a normal vector of the plane. In our case $\vec{n} = (- 13, - 15, - 6)$ .