Below is a problem I made up. I would expect there to be around $2$ vectors that meet the given criterion. However, I could not convince myself that there are $4$ vectors that meet the given criterion.
Problem:
Find all vectors that are perpendicular to the vector $(2,3)$ and have length $1$.
Answer:
If two vectors are perpendicular then there dot product is $0$. \begin{align*} (u_1,u_2) \cdot (2,3) &= 0 \\ 2u_1 + 3u_2 &= 0 \\ \sqrt{ u_1^2 + u_2^2 } &= 1 \\ u_1^2 + u_2^2 &= 1 \end{align*} Now we have two equations with two unknowns. \begin{align*} 2u_1 &= -\dfrac{3u_2}{2} \\ u_1^2 &= \sqrt{ 1 - u_2^2 } \\ \sqrt{ 1 - u_2^2 } &= -\dfrac{3u_2}{2} \\ 1 - u_2^2 &= \dfrac{9u_2^2}{4} \\ \dfrac{13u_2^2}{4} &= 1 \\ 13u_2^2 &= 4 \\ u_2 &= \pm \dfrac{ 2 } {\sqrt{13}} \\ \end{align*} We will first consider the case where $u_2$ is positive. \begin{align*} 2u_1 + 3\left( \dfrac{ 2 } {\sqrt{13}} \right) &= 0 \\ u_1 + \dfrac{ 3 } {\sqrt{13}} &= 0 \\ u_1 &= -\dfrac{ 3 } {\sqrt{13}} \end{align*} Now we will first consider the case where $u_2$ is negative. \begin{align*} 2u_1 + 3\left( \dfrac{ -2 } {\sqrt{13}} \right) &= 0 \\ u_1 - \left( \dfrac{ 3 } {\sqrt{13}} \right) &= 0 \\ u_1 &= \dfrac{ 3 } {\sqrt{13}} \end{align*} Hence our two vectors are: $\left( -\dfrac{ 3 } {\sqrt{13}}, \dfrac{ 2 } {\sqrt{13}} \right) $ and $\left( \dfrac{ 3 } {\sqrt{13}}, -\dfrac{ 2 } {\sqrt{13}} \right) $.
Is my solution correct?