# Derive equation of plane through three points

In my book it is given that

Equation of a Plane through three points

Vector form: If $$A,B,C$$ are three points having P.V.'s $$\vec a,\vec b,\vec c$$ respectively, then vector equation of the plane is $$\begin{bmatrix}\vec r&\vec a&\vec b\end{bmatrix}+\begin{bmatrix}\vec r&\vec b&\vec c\end{bmatrix}+\begin{bmatrix}\vec r&\vec c&\vec a\end{bmatrix}=\begin{bmatrix}\vec a&\vec b&\vec c\end{bmatrix}$$

Cartesian form: The equation of the plane through three non-collinear points $$(x_1,y_1,z_1),(x_2,y_2,z_2)$$ and $$(x_3,y_3,z_3)$$ is $$\begin{vmatrix}x-x_1&y-y_1&z-z_1\\x_2-x_1&y_2-y_1&z_2-z_1\\x_3-x_1&y_3-y_1&z_3-z_1\end{vmatrix}=0$$

How have they written the equation of the plane? Can anyone derive it?

• What does the notation $[\vec{r} \vec{a} \vec{b}]$ mean? Dec 8 '16 at 13:02
• @snulty it is scalar triple product Dec 8 '16 at 13:28
• oh right like $[\vec{a}\vec{b}\vec{c}]=\vec{a}\cdot\vec{b}\times\vec{c}$ Dec 8 '16 at 13:35

That vector equation is the expanded form of $$[(\vec b-\vec a)\times(\vec c -\vec a)]\cdot(\vec r-\vec a)=0.$$ This is the equation of the plane through $ABC$, because $(\vec b-\vec a)\times(\vec c -\vec a)$ is a vector normal to the plane.