In the book of Geometry that I am studying in class there is a proposition (referring to Euclidean 3-dimensional space) that is left as an exercise. I don’t know how to tackle this problem, any help is appreciated, and sorry for my broken english.
(a) Show that the geometric place of the points such that the sum of the squares of their distances to two lines that cross (which is constant) is a quadric surface.
(b) Also show that the geometric place of the points $p$ such that $d(p,r) = d(p,s)$ where $r,s$ are two lines that cross, is a quadric surface.
Edit (1)
With the help of @Robert Z I understood how to compute the problem, but I don’t know how to draw the conclusion that the final computation implies that there is a quadric.
Edit (2)
So computing $d(p,r)^2 + d(p,s)^2 = k$ I arrived to the general equation of the quadric in function of the terms $x^2, y^2, z^2, xy, yz, xz, x,y,z$: $$c_{11}x^2 + c_{22}y^2 + c_{33}z^2 + 2c_{12}xy + 2c_{13}xz + 2c_{23}yz + 2c_{01}x + 2c_{02}y + 2c_{03}z + c_{00} = 0$$ where the terms $c_{ij}$ are the constants derived from the $a_i$ from $d(p,r)^2$, and the $b_i$ are the constants from $d(p,s)^2$
Possible solution
I recalled from Lineal Algebra that the distance from a point $p = (x, y, z)$ to a hyperplane (which in this case, is a line) $r: a_1x_1 + a_2x_2 + a_3x_3 = b$ where $r = L[u]$, $u = (a_1, a_2, a_3)$ (for the purpose of the computations it is convenient to take $u$ such that $||u|| = 1)$ is $$d(p,r) = \frac{|a_1x + a_2y + a_3z - b|}{\sqrt{a_1^2 + a_2^2 +a_3^2}}$$
So, doing the same to the hyperplane $s$, we get that $$d(p,s) = \frac{|b_1x + b_2y + b_3z - d|}{\sqrt{b_1^2 + b_2^2 + b_3^2}}$$
So, now we apply the hypothesis: $$d(p,s)^2 + d(p,r)^2 = k \iff \frac{(b_1x + b_2y + b_3z - d)^2}{b_1^2 + b_2^2 + b_3^2} + \frac{(a_1x + a_2y + a_3z - b)^2}{a_1^2 + a_2^2 + a_3^2} = k \iff (b_1x + b_2y + b_3z - d)^2 + (a_1x + a_2y + a_3z - b)^2 = k$$
The factors $b_1^2 + b_2^2 + b_3^2$ and $a_1^2 + a_2^2 + a_3^2$ "vanish" because we defined $||u|| = ||(a_1, a_2, a_3)|| = a_1^2 + a_2^2 + a_3^2 = 1$ and for $||v|| = ||(b_1, b_2, b_3)|| = 1$
And from here, by expanding the sum of the squares, taking common factor on the constants that multiply the factors $x^2, y^2, z^2, xy, yz, xz, x, y, z$ and doing a change of variables, we arrive to the general form of a quadric surface. $$c_{11}x^2 + c_{22}y^2 + c_{33}z^2 + 2c_{12}xy + 2c_{13}xz + 2c_{23}yz + 2c_{01}x + 2c_{02}y + 2c_{03}z + c_{00} = 0$$
I think all the steps are correct but it is late and my brain is tired, so who knows? Please correct me if I'm wrong.
In $\mathbb{R}^3$, the distance from a point $P=(x,y,z)$ and a line $r$ is $$d(P,r)=\frac{\|(P-P_1)\times (P-P_2)\|}{\|P_2-P_1\|}$$ where $P_1=(x_1,y_1,z_1)$ and $P_2=(x_2,y_2,z_2)$ are two distinct points along $r$. Therefore, after expanding the cross product, we find that $$\begin{align}d(P,r)^2&=\frac{1}{||P_2-P_1\|^2}\left|\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ x-x_1 & y-y_1 & z-z_1 \\ x-x_2 & y-y_2 & z-z_2 \end{vmatrix}\right|^2\\ &=(a_1y+a_2z+a_3)^2+(a_4x+a_5z+a_6)^2+(a_7x+a_8y+a_9)^2 \end{align}$$ where $a_1,\dots, a_9$ are numbers which depend on $x_1,y_1,z_1,x_2,y_2,z_2$.
Now, for (a), we have that $$d(P,r)^2+d(P,s)^2=k$$ where $k$ is a given positive constant, whereas for (b), the equation to be satisfied can be written as $$d(P,r)^2 = d(P,s)^2.$$ What kind of algebraic equations (with respect to $x$, $y$, $z$) do we find in each case?