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Suppose that $a,b$ are two orthogonal unit vectors in $\mathbb R^3$, want to find a unit vector $c$ orthogonal to both $a$ and $b$. And the matrix formed by using $a$, $b$, $c$ as row vectors has determinant 1.

One strategy would be to suppose that $c = (x_1, x_2, x_3)$, and write down three equations using given conditions. Then solve it. But this would be tedious. Are there any other methods?

joseph
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Take their cross product $a \times b$. Since $a$ and $b$ both have magnitude $1$ and are orthogonal, the result is also a unit vector.

ameed
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  • I am not familiar with the cross product, how will it guarantee that the determinant is 1? – joseph Nov 27 '16 at 04:26
  • Also I am looking for a formula that involves only $a,b$, since I am trying to show that it is continuous. – joseph Nov 27 '16 at 04:27
  • How can I express the vector that is orthogonal to the plane spanned by $a,b$ explicitly? – joseph Nov 27 '16 at 04:29
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    @joseph The cross product $a\times b = (a_2b_3 - a_3b_2, a_3b_1-a_1b_3,a_1b_2-a_2b_1)$ will be orthogonal to both $a$ and $b$ and, because a and $b$ are unit vectors, $a\times b$ will be a unit vector. Putting them in a matrix you'll have a matrix with orthonormal columns - i.e. an orthogonal matrix. All orthogonal matrices have determinants of either 1 or -1. Check it and if it's negative, you can either switch two columns or negate the column with $a\times b$ in it. –  Nov 27 '16 at 04:32
  • How to show that it is the unique solution? – joseph Nov 27 '16 at 04:35
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    @joseph If $a\times b$ is one vector that's orthogonal to $a$ and $b$ then $k(a\times b)$ where $k$ is a real number is the form that every other orthogonal vector has. So plug $k(a\times b)$ into the matrix and see that $1$ (or $-1$, whichever makes the determinant positive) is the only value of $k$ that makes the determinant exactly $+1$. –  Nov 27 '16 at 04:38