[Math] Vector equation of a plane from normal vector and a point (quick way)

linear algebravectors

For example: the plane passing through the point $B(1,1,1)$ with a normal vector of $(2i-j+2k)$

I know I can set up the Cartesian equation of the line, then find $2$ additional points on the plane by plugging random $x$'s and $y$'s into the Cartesian equation. Then setting up the vector equation by doing the standard method with three points.

However, is there any way to go straight to the vector equation without first having to set up the Cartesian equation and find three points?

Best Answer

For the vector equation let find two vectors orthogonal to $\vec n=(2,-1,2)$ as for example

  • $\vec v_1=(1,2,0)$
  • $\vec v_2=(0,2,1)$

then the parametric equation of the plane through the origin is given by

$$t\vec v_1+s\vec v_2$$

then find $k$ such that $$t\vec v_1+s\vec v_2+k\vec n$$

passes through $B(1,1,1)\implies k=\frac13$ then the plane equation in parametric form is

$$t\vec v_1+s\vec v_2+\frac13\vec n$$

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