# General equation of a plane

For every point of the plane $$\pi$$, we can consider three parametric equations as a system of equations with two unknowns, $$\lambda$$ and $$\mu$$, that must have only one solution. Therefore the system is: $$\left. \begin{array}{rcl}x-a_1 &=&\lambda \cdot u_1 +\mu \cdot v_1 \\ y-a_2& = &\lambda \cdot u_2+\mu \cdot v_2 \\ z-a_3&=&\lambda \cdot u_3 +\mu \cdot v_3\end{array}\right\}$$$It has to be compatible and determined and, therefore, the following determinant must be 0: $$\left|\begin{matrix}x-a_1 & u_1 & v_1 \\ y-a_2 & u_2 & v_2 \\ z-a_3 & u_3 & v_3\end{matrix} \right|=0$$$ If we develop the previous determinant we obtain: $$(u_2v_3-u_3v_2)\cdot x+(u_3v_1-u_1v_3)\cdot y+(u_1v_2-u_2v_1)\cdot z+ \\ +[-a_1(u_2v_3-u_3v_2)-a_2(u_3v_1-u_1v_3)-a_3(u_1v_2-u_2v_1)]=0$$$And if we call $$A, B$$ and $$C$$ the coefficients of $$x, y, z$$, and $$D$$ the independent term, we obtain the linear equation: $$Ax + By + Cz + D = 0$$$ which is known as the general, Cartesian or implicit equation of the plane.

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

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: $$(x, y, z) = (1,-3, 5) + \lambda \cdot (-3, 5,-6) + \mu \cdot (0, 2,-5)$$$and the parametric equations are: $$\left\{\begin{array}{rcl}x&=&1-3\lambda \\ y&=&-3+5\lambda+2\mu \\ z&=&5-6\lambda-5\mu \end{array}\right.$$$

If we write the determinant of the system and equate it to zero we have: $$\left| \begin{matrix} x-1 & -3 & 0 \\ y+3 & 5 & 2 \\ z-5 & -6 & -5 \end{matrix}\right|=0$$$And if we develop it: $$\left| \begin{matrix} x-1 & -3 & 0 \\ y+3 & 5 & 2 \\ z-5 & -6 & -5 \end{matrix}\right|=-25(x-1)-6(z-5)-15(y+3)+12(x-1)=\\=-25x+25-6z+30-15y-45+12x-12=-13x-15y-6z-2=0$$$ 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 $$Ax + By + Cz + D = 0$$ then $$\overrightarrow{n}=(A,B,C)$$ is a normal vector of the plane. In our case $$\overrightarrow{n}=(-13,-15,-6)$$.