(1)$\mathbf{Q}$ is a matrix with orthonormal columns, $\mathbf{Q}\in\Bbb{R}^{4\times 3}$. (2) $\mathbf{Q}^T\mathbf{q}=0$. Then the column space of $\mathbf{q}$ is the orthogonal complement of the column space of $\mathbf{Q}$. By using a property of the orthogonal complement, we have $\mathbf{Q}\mathbf{Q}^T=\mathbf{I}-\mathbf{q}\mathbf{q}^T$. If $\mathbf{q}$ is already kown, how can I get $\mathbf{Q}$ with the above conditions? Thank you!
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$\def\v#1{\mathbf{#1}}\def\R{\mathbb R}$Extend $\v q$ to an orthonormal basis of $\R^4$, that is apply the Gram-Schmidt algorithm to $\v q$, $\v e_1, \ldots, \v e_4$. This will give you an orthonormal basis $\v q', \v n_1, \ldots, \v n_3$. Then you can set $\v Q = (\v n_1 \; \v n_2\; \v n_3)$.
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