How to find orthogonal vector to an arbitrary 3 dimentional vector

Question

Given a vector $\begin{bmatrix}a\\b\\c\end{bmatrix}$ what is a simple solution to find any vector perpendicular to it?

Answer 1

Most of the times you can use $\begin{bmatrix}b+c\\c-a\\-a-b\end{bmatrix}$ vector. But it is zero sometimes.

Here is the verification: $\begin{bmatrix}a\\b\\c\end{bmatrix} . \begin{bmatrix}b+c\\c-a\\-a-b\end{bmatrix} = a(b+c) + b(c-a)+c(-a-b)=0$

The third orthogonal vector is found by taking a cross product of initial and derived vector.

Any linear combination of these two derived vectors are orthogonal to original one as well.

Answer 2

If you have a (non zero) $2$-dimensional vector $\begin{bmatrix}a\\b\end{bmatrix}$, the orthogonal vectors of it are the $\lambda \begin{bmatrix}-b\\a\end{bmatrix}$ where $\lambda \in \mathbf{R}$.

So if you have a (non zero) $3$ dimensional vector $\begin{bmatrix}a\\b\\c\end{bmatrix}$ you have easily a lot of orthogonal vectors : $\lambda \begin{bmatrix}-b\\a\\0\end{bmatrix} + \mu\begin{bmatrix}0\\-c\\b\end{bmatrix}+ \nu \begin{bmatrix}-c\\0\\a\end{bmatrix} $ It is possible to check you generate all the orthogonal vectors to $\begin{bmatrix}a\\b\\c\end{bmatrix}$ like this (in reality, you just need two of them, but which ones you choose depends on the eventual annulations of $a$,$b$,$c$).