Finding a vector linearly independent to other vectors.

Question

How would I find a vector that is linearly independent from two other vectors? For the exact problem I'm working with, I was giving the following vectors: $$\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}\begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix}$$

Answer 1

Let $\vec u, \vec v\in\Bbb R^3$ be nonzero and nonparallel. Then the vector $\vec w=\vec u\times\vec v$ is guaranteed to be linearly independent from $\{\vec u,\vec v\}$. This follows from the fact that $\vec u\times\vec v$ is orthogonal to both $\vec u$ and $\vec v$.

In our case we have $$ \vec w=\langle1,-1,0\rangle\times\langle 2,1,4\rangle=\langle-3,-3,3\rangle $$

Answer 2

Let $(x,y,z) \in \mathbb{R}^{3}$; then $(x,y,z)$ is not linearly independent from both of the two vectors iff $(2a+b, a-b, 3a) = (x,y,z)$ for some $a,b \in \mathbb{R}$. Now $a = b = 1$ iff $(2a+b,a-b,3a) = (3,0,3)$; so if $(x,y,z) =(3,y,3)$ for some $y \neq 0$, then $(x,y,z)$ is linearly independent from both of the two vectors.

Answer 3

The cross product will be the easiest way in this case, because you happen to have two linearly independent vectors from $\Bbb R^3$.

A more general method would be to find the kernel of the matrix $$\begin{bmatrix}1 & -1 & 0 \\ 2 & 1 & 3\end{bmatrix}$$