Is there a more algebraic way of solving this vector problem? I am asked to find a vector perpendicular to $\vec a=[4,-2,3]$. Since I know that if two vectors dot product equals zero, they are perpendicular, I made this equation:$$4x-2y+3z=0$$
Where any combination of $x,y,z$ which satisfies the equation will be the corresponding components of a vector perpendicular to $\vec a$.
My problem is that to solve this, I simply see right from the get-go that if $x=1,y=2$ and $z=2$, then the equation is true. Meaning $[1,5,2]$ is perpendicular to $\vec a$.
However, I don't think solving it this way is acceptable. So is there some way to algebraically get $[1,5,2]$ as an answer, rather than just seeing it without doing any calculations?
 A: The answer here is: not really.  Why? Because you have one equation, and three unknowns; this means that, unless you do something really goofy, you can pick ANY value for two of the three variables $x,y,z$, and you'll be able to pick a value for the third to make it work.
For instance: if $x=1$ and $y=7$, then $z=\frac{10}{3}$.
Another way you could do this is to note that for any vector $\vec{b}$, the vector $\vec{a}\times\vec{b}$ is orthogonal to both $\vec{a}$ and $\vec{b}$. Assuming that you want to find a non-zero vector (otherwise $\langle0,0,0\rangle$ always works here!), you could choose any $\vec{b}$ which is not parallel to $\vec{a}$.  For instance: if $\vec{b}=\langle1,1,1\rangle$, then
$$
\vec{a}\times\vec{b}=\langle-5,-1,6\rangle
$$
is orthogonal to both $\vec{a}$ and $\vec{b}$, and therefore in particular it is orthogonal to $\vec{a}$.
A: There is nothing wrong with doing it the way you did, guessing and then checking is part of mathematical problem solving. Alternatively, to get a perpendicular vector, simply take the cross-product with another vector and try to get a nonzero vector.
A: Part of the problem is that there is no unique solution to this equation. So there won't be a method that gives you "the" answer.
However, what you can do is the following: Consider rewriting
$$
4x - 2y + 3z = 0
$$
as
$$
2y = 4x + 3z
$$
or
$$
y =  2x + \frac{3}{2}z
$$
What this tells you is that, no matter your choice of $(x, z)$, that if you pick $y$ as above then you will obtain a solution. In your stated case, you chose $(x, z) = (1, 2)$ which yields that $y = 5$ as you found.
