Finding the equation of a plane.

Question

How do I find the equation of a plane given by the points (0,1,1), (1,0,1) and (1,1,0)?

Graphing it, it's a triangle when you connect the points. Can I use this somehow?

Answer 1

Sounds like you want the normal vector form.

To find a normal vector to the plane, we take the cross product of 2 (linearly independent) vectors of the plane. In this case, we take the vectors $ \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} - \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} - \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}$ to obtain

$$ \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix} \times \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} -1 \\ -1 \\ -1 \end{pmatrix} $$

Then, the plane satisfies $$ \begin{pmatrix} x \\ y \\ z \end{pmatrix} \begin{pmatrix} -1 \\ -1 \\ -1 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} \begin{pmatrix} -1 \\ -1 \\ -1 \end{pmatrix} = -2,$$

or that $ - x - y - z = - 2$.

Answer 2

Take one of the points as origin and use the other two to obtain two (linearly independent? check!) vectors. Then write the vectorial equation for the plane as $$(x,y,z)=P+\lambda \vec u + \mu \vec v.$$