Short question about a third vector and orthonormal basis in $\mathbb R^3$

Question

I have a task that asks to complete this set of vectors $\begin{bmatrix} 2/3 \\ 2/3 \\ -1/3 \end{bmatrix},$ $\begin{bmatrix} 2/3 \\ -1/3 \\ 2/3 \end{bmatrix}$

to an orthonormal basis in $\mathbb R^3$

I solved a system that finds the third vector and obtained $2$ solutions (two vectors). But the thing is these two vectors are $\{v, -v\}$.

My question: do I have actually only one vector that allows to construct the basis(as the second vector is just the opposite) or is it two distinct vectors that allow to construct two different orthonormal bases? Thx in advance!

Answer 1

If $\{v_1,v_2,v_3\}$ is an orthonormal basis of $\mathbb R^3$, then so is $\{v_1,v_2,-v_3\}$. So, it is natural that you got two answers. It could not have been otherwise.

Answer 2

Starting with two orthogonal 3-element unnit vectors say a and b, you can construct an orthonormal 3x3 matrix whose fiirst column is a and second column is b by taking the third column as the cross-product a X b. This will give a matrix with determinant 1, or you can take the third column as -a X b to give an orthonormal matrix with determinant -1. Those are the only possibilities. Of course, you could work with rows instead.