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?

  • $\begingroup$ What does the notation $[\vec{r} \vec{a} \vec{b}]$ mean? $\endgroup$ – snulty Dec 8 '16 at 13:02
  • $\begingroup$ @snulty it is scalar triple product $\endgroup$ – user123733 Dec 8 '16 at 13:28
  • $\begingroup$ oh right like $[\vec{a}\vec{b}\vec{c}]=\vec{a}\cdot\vec{b}\times\vec{c}$ $\endgroup$ – snulty 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.

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