# Finding the basis vectors for a given plane

I was given the following question:

Find a basis for the set of vectors in $$\mathbb{R}^3$$ in the plane $$x-2y+8z=0$$. Think of this equation as a "system" of homogeneous equations.

This is what I've done:

$$(\begin{matrix}1&-2&8&0\\\end{matrix})\\x_3=t; \ x_2=s\\x_1=2s-8t\\\vec{x}=s\left(\begin{matrix}2\\1\\0\\\end{matrix}\right)+t\left(\begin{matrix}-8\\0\\1\\\end{matrix}\right)$$ The two basis vectors are $$\left(\begin{matrix}2\\1\\0\\\end{matrix}\right)$$ and $$\left(\begin{matrix}-8\\0\\1\\\end{matrix}\right)$$.

Is this correct? What is throwing me off is that it was a one-row matrix, and the terminology of the question "Think of this equation as a "system" of homogeneous equations".

• That hint doesn't make any sense to me either but your answer is correct. Maybe the hint is trying to point us to the idea that the answer is the same as finding a basis for the null space of the linear transformation $f(x,y,z)=x-2y+8z$. Commented Aug 14, 2020 at 17:06
The planar equation can be written as $$(1,-2,8)\begin{pmatrix}x\\y\\z\end{pmatrix}=0$$
So any vector that is perpendicular to $$(1,-2,8)$$ will be in the plane, such as $$(2,1,0)$$, $$(-8,0,1)$$.
More generally, given any number of linear equations representing some linear subspace, $$\begin{pmatrix}a_{11}&\cdots &a_{1n}\\ \vdots& &\vdots\\ a_{m1}&\cdots&a_{mn}\end{pmatrix}\begin{pmatrix}x_1\\\vdots\\x_n\end{pmatrix}=0$$ its vectors can be rewritten as $$\begin{pmatrix}x_1\\\vdots\\x_n\end{pmatrix}=\begin{pmatrix}b_{11}&\cdots&b_{1m}\\&\vdots\\b_{n1}&\cdots&b_{nm}\end{pmatrix}\begin{pmatrix}s_1\\\vdots\\s_m\end{pmatrix}$$ where $$AB=0$$ and $$B$$ has columns that span the null space of $$A$$.