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Given linearly independent vectors $\boldsymbol{u_i} = (a_i, b_i, c_i)$, where $ i = \{1,2\} $, how can I rigorously find all $\boldsymbol{u_3}$ such that $\{\boldsymbol{u_1}, \boldsymbol{u_2}, \boldsymbol{u_3}\}$ is a basis of $\mathbb{R}^3$?

I know that I can use a cross product or RREF to find one such linearly independent vector, but how can I find all such vectors?

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    It's simple. Form a 3x3 matrix $M$ with each row being one of the given vectors and the last row being variables and then solve $\det(M)\neq 0$. – 2017-01-31

2 Answers 2

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The vectors $u_i$ with $i\in \{1,2,3\}$ are linearly independent iff the determinant of the matrix with $u_i$ as the i-th row is non-zero.

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Once you have a single vector perpendicular to the plane formed by ${\bf u}_1$ and ${\bf u}_2$ (call such a vector ${\bf u}_3 = {\bf u}_1 \times {\bf u}_2$), then you can span the three-dimensional space $\mathbb{R}^3$ by $\{ {\bf u}_1, {\bf u}_2, a_1 {\bf u}_1 + a_2 {\bf u}_2 + a_3 {\bf u}_3 \}$, where $a_1$ and $a_2$ are scalars and $a_3 \neq 0$.