# Finding the equation of a line entirely defined by a three variable equation.

How can you find the equation of a line that lies completely in a set defined by a three variable equation. For instance, the equation of a line entirely in the set defined by $x^2 + y^2 - z^2 = 1$

Supposedly a potential answer is the line defined by $x=1, y=t$, and $z=t$, but I don't understand how one finds that answer.

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I think the meaning might just be that, is there a line that satisfies this equation? –  Greg Ros Jan 15 '13 at 4:43
I think he's asking: given a surface, determine if it's ruled, and give equations for the family of rulings. –  user7530 Jan 15 '13 at 4:43
Oops, typed the equation a bit wrong. Fixed it now. I'm asking how I can find a line that lies entirely in the set of values defined by the above equation. Also added a little more detail on what a correct answer looks like. –  user58318 Jan 15 '13 at 4:44
In short: if you only care about finding one line, plug the line $\mathbf{q}_0 + t\mathbf{v}_0$ into the surface equation. You will get a quadratic equation in $t$, which must hold for all $t$. Equate the coefficients to $0$ to get an underdetermined system of equation for the position and velocity of the line. Solve for a solution. –  user7530 Jan 15 '13 at 4:50
@user58318 Are you able to verify that $x=1, y=t, z=t$ indeed lies in the surface? Because you sound very unsure, and knowing how to verify a solution is the first step to finding one. –  Erick Wong Jan 15 '13 at 4:51

So you have this equation for the surface: $$x^2+y^2-z^2=1.$$

What does that mean? For any point $(a,b,c)$ in $\mathbb{R}^3$, you plug it into the equation, and if you get that $a^2+b^2-c^2=1$, then the point is on the surface. Otherwise, it's not.

Now you want to find a line that's on the surface. Remember that you can represent a line by the expression $L(t)=(q_x,q_y,q_z) + t(v_x,v_y,v_z)$, where the $v$s are not all $0$ (otherwise you get just a single point). Here the $q$s and $v$s are fixed, and $t$ varies from $-\infty$ to $\infty$. Different values of $t$ give you different points on the line. You can think of $(q_x,q_y,q_z)$ as giving you the starting point of the line, and $(v_x,v_y,v_z)$ as the direction of the line.

You want the line to lie on the surface. So you plug it into the surface equation $$(q_x+tv_x)^2 + (q_y+tv_y)^2-(q_z+tv_z)^2=1.$$ Let's expand this out a bit and group like terms of $t$: $$t^2(v_x^2+v_y^2-v_z^2) + t(2q_xv_x+2q_yv_y-2q_zv_z) + (q_x^2+q_y^2-q_z^2) = 1.$$

In order for the line to lie on the surface, this equation must hold for every value of $t$. Since the left half of the equation depends on $t$, and the right half does not, the only way that can be true is if the coefficients in front of the $t$ and $t^2$ are 0. In other words, \begin{align*} v_x^2+v_y^2-v_z^2&=0\\ 2q_xv_x+2q_yv_y-2q_zv_z&=0\\ q_x^2+q_y^2-q_z^2&=1. \end{align*}

If we can find a solution to these equations, we have a line on the surface. These equations are nonlinear and there are way more unknowns than equations, so finding all solutions is somewhat bothersome.(**) If we only need one line, we can play a bit fast and loose. Let's try to satisfy the first equation first. We can do so by setting $v_x=0, v_y=1, v_z=1$ (remember that $v_x=v_y=v_z=0$ is not allowed). This gives us \begin{align*} 0+1-1&=0\\ 2q_y-2q_z&=0\\ q_x^2+q_y^2-q_z^2&=1. \end{align*} We can satisfy the second equation by setting $q_y=q_z=1$:

\begin{align*} 0+1-1&=0\\ 2-2&=0\\ q_x^2+1-1&=1. \end{align*}

Luckily, we're left with a last equation we can solve, by setting $q_x=1$ (if the last equation was unsolvable, for instance $q_x^2+2=1$, we'd have to go back and change our earlier guesses.) So our final line (in parametric form) is

$$L(t) = (1,1,1) + t(0,1,1).$$

It is a good exercise to check that plugging this line into the left hand side of the equation at the beginning gives $1$.

(**) If you did want to find all solutions, here's how to go about it. First, we'll restrict to unit vectors $v_x^2+v_y^2+v_z^2=1$ with $v_z\geq0$, since the magnitude and sign of the direction vector does not change the set of points swept out by the line. We'll parameterize our solutions by $q_x$ and $q_y$ (and some sign choices), in the region $q_x^2+q_y^2 \geq 1$. We then have $q_z = \pm\sqrt{q_x^2+q_y^2-1}$. Adding $2v_z^2$ to the first equation, we get $v_z = \sqrt{1/2}.$ We now know everything except $v_x$ and $v_y$; solving for $v_x$ in the second equation and plugging it into the first gives a quadratic equation in $v_y$ with positive discriminant, which we can solve (up to a second sign choice), and then we get $v_x$ by substitution.

Geometrically, the above tells us that for every point on the surface, there are two rulings passing through that point.

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Incredible explanation. Everything is clear to me except one part: "Since the left half of the equation depends on t, and the right half does not, the only way that can be true is if the coefficients in front of the t and t^2 are 0." I understand your reasoning here I think, but I'm not sure how making the coefficients of t zero gives you the following three separate equations.\begin{align*} v_x^2+v_y^2-v_z^2&=0\\ 2q_xv_x+2q_yv_y-2q_zv_z&=0\\ q_x^2+q_y^2-q_z^2&=1. \end{align*} Shouldn't distributing 0*t^2 and 0*t just give you \begin{align*}q_x^2+q_y^2-q_z^2&=1.\end{align*} –  user58318 Jan 15 '13 at 9:51
It's a subtle point. By moving the 1 to the LHS, we have a polynomial $f(t)$ that must equal zero for every $t$. Here are three ways to look at it, in increasing generality: 1) A polynomial that's zero for every $t$ must be the zero polynomial, hence all of its coefficients are 0. 2) If $f(t)$ is analytic and zero for all $t$, all of the terms in its Taylor series must be 0. 3) $1$, $t$, and $t^2$ are linearly independent "vectors" in the vector space of functions over $\mathbb{R}$, hence if a linear combination of them is 0, their coefficients must be zero. –  user7530 Jan 15 '13 at 16:47