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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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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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