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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?

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What do you know about equation of a plane? There are several formats, which one do you know of? 2-vector 1 point, normal vector and parametric. Sound familiar? –  Calvin Lin Jan 26 '13 at 23:29
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By inspection $x+y+z=2$ is an equation of the plane. –  André Nicolas Jan 26 '13 at 23:30
    
a(x-x0)+b(y-y0)+c(z-z0)=0. Andre got the answer right but not sure on how to get it. –  user1766888 Jan 26 '13 at 23:31

2 Answers 2

up vote 3 down vote accepted

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$.

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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 $$(x,y,z)=P+\lambda \vec u + \mu \vec v.$$

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