# Determine a vector for a $3\times 3$ matrix so it becomes positive orthogonal

Two orthonormal vectors in $\mathbb{R}^3$ are given:

$v_1=(-\frac{1}{2}\sqrt{2}, 0, \frac{1}{2}\sqrt{2})$ and $v_2=(0,1,0)$

Let $Q$ be a $3\times 3$ matrix with the first, second and third column be $v_1$, $v_2$ and $v_3$ respectively.

Determine $v_3$ so $Q$ becomes a positive orthogonal matrix.

I'm stuck. Any hints? I know that $det(Q)=1$ for it to be positive orthogonal, but I'm not sure how to get there from the given information.

• What, besides $\det(Q)=1$, does $Q$ being an orthogonal matrix imply? In particular about the columns of $Q$? – user137731 Nov 22 '16 at 23:33
• $v_3$ has to be orthogonal with the other two vectors or just one of them? @Bye_World – Steve Nov 22 '16 at 23:36
• $v_i^T v_3 = \delta_{i,3}$ ... there are only two options for $v_3$. One yields a positive determinant, the other one a negativ one. – user251257 Nov 22 '16 at 23:37
• Let $v_3 = (a,b,c)$. Then you want to solve the system of equations \begin{align}(-\frac{1}{2}\sqrt{2}, 0, \frac{1}{2}\sqrt{2})\cdot(a,b,c) &= 0 &\text{(v_1 orthogonal to v_3)} \\ (0,1,0)\cdot(a,b,c) &= 0 &\text{(v_2 orthogonal to v_3)} \\ \sqrt{a^2 + b^2 + c^2}&=1 &\text{(v_3 normalized)}\end{align} There will be two vectors which solve all of these. Choose the one that makes the determinant positive. – user137731 Nov 23 '16 at 0:02
• There is a direct solution: take $v_3 := v_1 \times v_2$ (cross product). – Jean Marie Nov 23 '16 at 0:58