Let $v_1=(-1,2,3)$ and $v_2=(5,3,-1)$. Find the equation of the plane spanned by $v_1$ and $v_2$. Also find a vector $v_3$ that can be added to the set $\{v_1,v_2\}$ to produce a basis for $\mathbb{R}^3$.
I'm stuck on both parts because everywhere I read it says I need a point in addition to these two vectors to find the equation.
For the second part I'm not sure of the method to find a vector to produce a basis, we haven't been told how to do that… I know a basis for $\mathbb{R}^3$ would have to have $3$ vectors that span $\mathbb{R}^3$ and must be L.I., so I know that $kv_1 + kv_2 + kv_3 = 0$ must be $k_1=k_2=k_3 = 0$ …but what is a concrete method to find a third vector $v_3$?
Forgive me if I missed somewhere I could have found this out, I couldn't seem to find the method.
Best Answer
Let $\begin{pmatrix}x\\y\\z\end{pmatrix}$ be a point in Span$\{v_1,v_2\}$, so there are $t,r\in\mathbb{R}$ such that \begin{align*} \begin{pmatrix}x\\y\\z\end{pmatrix}&=tv_1+rv_2\\ &=t\begin{pmatrix}-1\\2\\3\end{pmatrix}+r\begin{pmatrix}5\\3\\-1\end{pmatrix}\\ &=\begin{pmatrix}-1&5\\2&3\\3&-1\end{pmatrix}\begin{pmatrix} t \\ r\end{pmatrix}...(1) \end{align*} This system can be written as $$\begin{pmatrix}-1&5&|&x\\2&3&|&y\\3&-1&|&z\end{pmatrix}$$ By mean of elementary operations by rows we get $$\begin{pmatrix}1&0&|&4x-5y+5z\\0&1&|&x-y+z\\0&0&|&11x-14y+13z\end{pmatrix}$$ Then, (1) has solutions iff $11x-14y+13z=0$, this is the equation of the plane.