[Math] Expressing a hyperplane as the span of several vectors.

linear algebra

How do I express the hyperplane $x+y=1$ as the span of two vectors or more?

P. S. We have a 3D space.

Best Answer

Using only basic analytic geometry: find three different non-collinear points on the plane, for example

$$A=(1,0,1)\;,\;\;B=(1,0,0)\;,\;\;C=(0,1,0)$$

and now construct the directed vectors

$$\vec{AB}=B-A=(0,0,-1)\;,\;\;\vec{AC}=C-A=(-1,1,-1)$$

and then the plane is

$$\pi:\;A+r\vec{AB}+s\vec{AC}=(1,0,1)+r(0,0,-1)+s(-1,1,-1)\;,\;\;r,t\in\Bbb R$$

Check: take the vectorial product of the direction vectors to get a perpendicular vector to the plane:

$$\vec{AB}\times\vec{AC}=\det\begin{vmatrix}i&j&k\\0&0&-1\\-1&1&-1\end{vmatrix}=(1,1,0)$$

and thus our plane is $\;x+y+d=0\;$ , and to find $\;d\;$ we can substitute any point on the plane here, say $\;A\;$ , to obtain

$$0+0+d=0\implies d=0$$

and the wanted plane is

$$x+y-1=0$$

which, of course, it is the same as you give. This way is just a standard form to check that what we got at the beginning is correct.

Anyway, you can look at your plane as the translation of a subspace, so;

$$\pi:\;\;\text{Span}\,\left\{\;(0,0,-1)\,,\,\,(-1,1,-1)\;\right\}+(1,0,1)$$

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