Cross products and orthogonal complements

Question

I am having trouble with this question about cross products and orthogonality:


Let a ∈ R3 \ {0}

Show that if y ⊥ a then $\exists$ x {x ∈ R3 : a × x = y}


Could anyone explain this to me? Thanks

Answer 1

Because you can always make $v = y \times a/|a|$ and then $a \times v /|a|$ will give $y$, so $x = v/|a| = y \times a/|a|^2$

Answer 2

Write $ y = y_1e_1+y_2e_2 + y_3e_3 $ and $ a = a_1e_1 + a_2e_2 + a_3e_3 $, where $\{e_1,e_2,e_3\}$ is the canonical basis for $\mathbb{R}^3$. Consider the matrix

$$ A = \begin{pmatrix} 0 & a_3 & -a_2\\ -a_3 & 0 & a_1\\ a_2 & -a_1 & 0 \end{pmatrix} $$

and prove that the any non-trivial solution of $Ax = y$ is the $x$ you are looking for.