Suppose $[a, b, c]^T$ is orthogonal to both $[3, 4, -2]^T$ and $[4, 2, -3]^T$ and $a, b, c$ are not all zero. Which of the following is true?
A. $\frac{a}{b} = -5/7$
B. $\frac{b}{c} = -1$
C. $\frac{c}{a} = -1/6$
D. $\frac{a}{c} = 4/5$
E. There is not solution.
I know the solution is D, but I am having trouble coming to that answer. I tried making a system where $3a + 4b -2c = 0$ and $4a + 2b -3c = 0$ because the dot product must be zero. However, that only gives me the solution to $a$, $b$. How do I get $c$?
If the equations are $$ 3a + 4b -2c = 0\ \text{ (I)}$$ $$ 4a + 2b -3c = 0\ \text{ (II)}$$ then notice that if we multiply equation (II) by $-2$ and add it to the equation (I), we get $$-5a+4c = 0$$ which gives $\dfrac{a}{c} = \dfrac{4}{5}$.