Below is a problem I made up and did. I believe my solution is correct but I have some doubts because my solution effectively has two free variable and I expected only one free variable. Is my solution correct? I am also open to style comments.
Problem:
Find all vectors that are perpendicular to the vector $(1,2,3)$ and have length $1$.
Answer:
If two vectors are perpendicular then there dot product is $0$. \begin{align*} (u_1,u_2,u_3) \cdot ( 1,2,3) &= 0 \\ u_1 + 2u_2 + 3u_3 &= 0 \\ \sqrt{ u_1^2 + u_2^2 + u_3^2 } &= 1 \\ u_1^2 + u_2^2 + u_3^2 &= 1 \\ u_3^2 &= 1 - u_1^2 - u_2^2 \\ u_1 + 2u_2 + 3\left( \sqrt{ 1 - u_1^2 - u_2^2 }\right) &= 0 \\ \end{align*} Now we simplify the equation. \begin{align*} u_1 + 2u_2 &= -3 \sqrt{ 1 - u_1^2 - u_2^2 } \\ (u_1+2u_2)^2 &= 9( 1 - u_1^2 - u_2^2 ) \\ u_1^2 + 4u_1 u_2 + 4u_2^2 &= 9 - 9u_1^2 - 9u_2^2 \\ 10u_1^2 + 4u_1 u_2 + 4u_2^2 &= 9 - 9u_2^2 \\ 10u_1^2 + 4u_1 u_2 + 13u_2^2 &= 9 \\ \end{align*} Hence my answer is: $$ (x,y,z) \text{ where } x,y,z \in \mathbb{R} \text{ and } 10x^2 + 4xy + 13y^2 = 9 $$
Here is an updated answer: $$ (x,y, 1 - x^2 - y^2) \text{ where } x,y \in \mathbb{R} \text{ and } 10x^2 + 4xy + 13y^2 = 9 $$ Is it right now?
Here is an updated answer: $$ (x,y, -\sqrt{1 - x^2 - y^2}) \text{ where } x,y \in \mathbb{R} \text{ and } 10x^2 + 4xy + 13y^2 = 9 $$ Is it right now?
ow, I will attempt to to a partial check of the answer. Observe that: $$ (1,2,-\dfrac{5}{3}) \cdot ( 1,2,3) = 1 + 4 - 5 = 0 $$ Hence $(1,2,-\dfrac{5}{3})$ is perpendicular to $(1,2,3)$. However, it does not have length $1$. We can adjust the vector os that it does. \begin{align*} \left|(1,2,-\dfrac{5}{3})\right| &= \sqrt{ 1 + 4 + \dfrac{25}{9}} \\ \left|(1,2,-\dfrac{5}{3})\right| &= \sqrt{ \dfrac{9+36+25}{9}} = \sqrt{ \dfrac{70}{9}} \\ \left|(1,2,-\dfrac{5}{3})\right| &= \dfrac{ \sqrt{70} }{ 3 } \end{align*} Hence, we claim that $\dfrac{ (1,2,-\dfrac{5}{3}) }{\dfrac{ \sqrt{70} }{ 3 } } = \dfrac{ (3,6, -5)}{\sqrt{70}}$ is a perpendicular vector. Is it of the form of my answer? We have: \begin{align*} x &= \dfrac{3}{ \sqrt{70}} \\ y &= \dfrac{6}{ \sqrt{70}} \\ 1 - x^2 - y^2 &= 1 - \dfrac{9}{70} - \dfrac{36}{70} \\ % 1 - x^2 - y^2 &= \dfrac{ 70 - 45}{70} = \dfrac{ 25}{70} \\ -\sqrt{1-x^2-y^2} &= -\dfrac{ 5 }{ \sqrt{ 70} } \end{align*} Hence it checks in this case.
A better way is to think geometrically: let $a=(1,2,3)$. You may see $a$ as a vector with its initial point at the origin. Then the set of all vectors that are perpendicular to $a$ and whose initial point is at the origin form a plan $P$. Its equation is $x+2y+3z=0$ since $a$ has to be perpendicular to it and $(0,0,0)$ is on the plane.
Moreover, if $v=(x,y,z)$ is a unit vector, then $x^2+y^2+z^2=1$, so $(x,y,z)$ is on the unit sphere $S$ in space. Therefore, the set that describes the components $(x,y,z)$ of the unit vectors that are perpendicular to $a$ is the intersection of the plane $P$ and the sphere $S$, which is a circle. It can be described by the system
$$ \left\{ \begin{array}{rcl} x^2+y^2+z^2 & = & 1 \\ x+2y+3z & = & 0 \end{array} \right. $$
In my opinion, this is a good description since it gives you all the information you need. Loosely speaking, an equation in space describes a surface and a curve can be described as the intersection of two surfaces.
Alternatively, you may want to describe the circle using parametric equations. A figure would help to understand the construction, but it is identical to the usual parametrization of the unit circle in trigonometry: choose two perpendicular unit vectors $t$ and $n$ in $P$. Then, every unit vector perpendicular to $a$ can be described as
$$(x,y,z)=\cos(\theta)t+\sin(\theta)n$$
for some $\theta$ in $[0,2\pi)$. For example, $t=\frac{1}{\sqrt{5}}(-2,1,0)$ and $n=\frac{a\times t}{\|a\times t\|}=\frac{1}{\sqrt{70}}(-3,-6,5)$.