# 4 dimensional orthogonal group

Let's say I have a 4 by 4 matrix that is in the orthogonal group. The first three columns A, B, and C are known. Now, I can do a system of equations (4 equations) to solve for D, the fourth column. However, I'm interested to know if there is a much faster way to find D. Is there a formula that use A, B, and C to find D (some combination of dot products and cross products perhaps)? Thank you!

• The cross-product only exists in 3 dimensions, so it isn't going to be useful here. And 3 of those 4 equations are the fact that the dot product of the unknown column with each of the other three is $0$. The 4th column cannot be expressed as any sort of linear combination of the other 3 (otherwise the determinant of the matrix would be $0$ instead of $\pm 1$ as required). Really, I don't think you'll find anything easier than that system of 4 equations. Commented Nov 9, 2017 at 0:18
• @PaulSinclair you should add this as an official answer. I will pick it as an answer. Thanks! Commented Nov 10, 2017 at 0:20

Yes there is a formula but I would not call it much faster. Since ${\rm det\ }[A,B,C,X]$ is a linear function of column $X$, there is a vector $K$ such that $${\rm det\ }[A,B,C,X] = K\cdot X.$$ This $K$ is orthogonal to $A$ because ${\rm det\ }[A,B,C,A] = 0$, and similarly for $B$ and $C$.
In your case where $[A,B,C,D]$ is orthogonal, $D$ is plus or minus $K$ rescaled to norm 1.
This $K$ depends linearly on each of $A,B,C$ and it is known to be given by $3\times3$ determinants $$K = \begin{bmatrix} -{\rm det\ }_1[A,B,C] \\ {\rm det\ }_2[A,B,C] \\ -{\rm det\ }_3[A,B,C] \\ {\rm det\ }_4[A,B,C] \end{bmatrix}, \qquad\qquad D = \pm K/|K|,$$ where the subscript means remove that row from $[A,B,C]$. This formula is a symbolic expression for the answer to your system of equations.