How to find an orthogonal vector C in $C^3$ relative to two other (given) vectors?

Question

$A = [2,1,-i]$

$B = [i, -1, 2i]$

I need to find a C that is orthogonal to A and B.

I've tried taking AxB, but this does not work. I get the vector C = (i, 1-4i, -2-i). The problem is that $\overline{C} \cdot \overline{A} \neq 0$.

Next, I've tried to set some linear equations, namely, $\overline{C} \cdot \overline{A} = 0$, and likewise for $\overline{C} \cdot \overline{B} = 0$.

Specifically,

(a-bi)(2) + (c-di) + (e- fi)(-i) = 0

(a-bi)(-i) + (c-di)(-1) - (e-fi)(2i) = 0

This doesn't work either because there are too many unknowns and not enough equations.

How do you find an orthogonal matrix in $C^3$?

I know one answer to this problem: $(1+i,-5-3i, 1-3i)$. I just don't see how to get that answer.

Answer 1

Start with any vector $C$ that's not in the span of $\{A,B\}$ (just choose at random if you like), and then apply Gram-Schmidt to $\{A,B,C\}$. The last vector in the resulting triple will be orthogonal to both $A$ and $B$.

Alternately, the equations $A\cdot C=0$ and $B\cdot C=0$ can be written as a system of two equations in the three unknown coordinates of $C$. Then just solve that system using the usual techniques (Gaussian elimination).

Answer 2

@larry: You were close, you still have to take the conjugate. So an answer is $\overline{A\times B}= (-i, 1+4i, -2+i)$.

Note that $(-1+i) \cdot(-i, 1+4i, -2+i)= (1+i, -5-3i,1-3i)$

Any vector in $\mathbb{C}^3$ perpendicular to both $A$ and $B$ is proportional to $(-i, 1+4i, -2+i)$