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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.

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    $\begingroup$ What, besides $\det(Q)=1$, does $Q$ being an orthogonal matrix imply? In particular about the columns of $Q$? $\endgroup$ – user137731 Nov 22 '16 at 23:33
  • $\begingroup$ $v_3$ has to be orthogonal with the other two vectors or just one of them? @Bye_World $\endgroup$ – Steve Nov 22 '16 at 23:36
  • $\begingroup$ $v_i^T v_3 = \delta_{i,3}$ ... there are only two options for $v_3$. One yields a positive determinant, the other one a negativ one. $\endgroup$ – user251257 Nov 22 '16 at 23:37
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    $\begingroup$ 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. $\endgroup$ – user137731 Nov 23 '16 at 0:02
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    $\begingroup$ There is a direct solution: take $v_3 := v_1 \times v_2$ (cross product). $\endgroup$ – Jean Marie Nov 23 '16 at 0:58
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Hint: The vector cross product of any two vectors is orthogonal to each.

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