Locus formed by point on a line intersecting 3 other lines in 3D

Question

I got this particular question from an old test paper...

Consider three lines given by $y-2=z+3=0$; $z-3=x+1=0$; $x-1=y+2=0$. Let $(\alpha,\beta,\gamma)$ be a point lying on a line intersecting the given three lines. Then the locus generated by $(\alpha,\beta,\gamma)$ is

a. $\quad$ $xy+3yz+2xz+6=0$

b. $\quad$ $3xy+yz+2xz+6=0$

c. $\quad$ $2xy+3yz+xy+6=0$

d. $\quad$ none of these

I suppose that the intersecting line is as

$$\frac{x-\alpha}{l}=\frac{y-\beta}{m}=\frac{z-\gamma}{n}$$

where $l,m,n$ are direction ratios.

Now I can't understand how to proceed from here. Please help me out...

Answer 1

Here's something to get you started.

First you need to find the lines that intersect all three of the given lines. To do that, fix a point on the first given line, say $(u,2,-3)$. Now you want to find the point on the second given line, say $(-1,v,3)$, such that the line through those two points intersects the third given line. Find a parameterization of that line. That parameterization will have its own parameter, say $t$, and that parameterization will include your original variable $u$. So now you have a parameterization in two variables of all the points on all those lines.

Then find which of those equations $a,b,c$ in your multiple choice will be satisfied by your parameterization. Then you are done!

---

Answer to your comment:

For a point $(x,y,z)$ to be on the third line we must have $x=1,y=-2$. Substitute those into your parameterization and you get

$$\frac{1-u}{-1-u}=\frac{-2-2}{v-2}=t$$

Solving for $v$ gives

$$v=2+\frac{4+4u}{1-u}$$

Substituting that into your parameterization and solving for $x,y,z$ gives

$$x=(-u-1)t+u, \quad y=\left(\frac{4+4u}{1-u}\right)t+2, \quad z=6t-3$$

There is the parameterization of your desired surface. Now find which of the three possible equations, if any, are satisfied by that parameterization.

Can you finish from here?

Answer 2

The general setting of this question is interesting, let me work out a general formula for the locus first.

First, any line $\ell$ on $\mathbb{R}^3$ can be described by a pair of vectors $(\vec{p}, \vec{t})$ where $\vec{p}$ is a point on $\ell$ and $\vec{t} \ne \vec{0}$ points along its tangent direction:

$$\ell = \{\; \vec{p} + \lambda \vec{t} : t \in \mathbb{R} \;\}$$

To abuse notation, for any line $\ell_?$ labelled by a index '?', we will use the notation $\vec{p}_?$ and $\vec{t}_?$ to denote an arbitrary chosen pair of vectors describing that line. Once $\vec{p}_?$ and $\vec{t}_?$ are chosen, we will use $u_?(\vec{p})$ as a shorthand for the expression $(\vec{p} - \vec{p}_?) \times \vec{t}_?$.

Given any two lines $\ell_1, \ell_2$ not parallel to each other. $\vec{t}_1 \times \vec{t}_2$ will be non-zero and perpendicular to both $\vec{t}_1$ and $\vec{t}_2$. If we look at the two lines from a direction perpendicular to this vector, the image of the two lines will become parallel. Their separation will be proportional to $(\vec{p}_1 - \vec{p}_2)\cdot \vec{t}_1 \times \vec{t}_2$.
A consequence of this is:

Two non-parallel lines $\ell_1$ and $\ell_2$ intersect if and only if $$(\vec{p}_1 - \vec{p}_2)\cdot \vec{t}_1 \times \vec{t}_2 = 0 \quad\iff\quad u_1(\vec{p}_2) \cdot \vec{t}_2 = 0 \quad\iff\quad u_2(\vec{p}_1) \cdot \vec{t}_1 = 0.$$

Given any three non-parallel lines $\ell_1, \ell_2, \ell_3$ and a point $\vec{p}$ outside the three lines. If $\vec{p}$ lies on the locus, then one can find a non-zero vector $\vec{t}$ such that $$u_1(\vec{p})\cdot \vec{t} = u_2(\vec{p})\cdot \vec{t} = u_3(\vec{p})\cdot \vec{t} = 0 \tag{*1} $$

Since $\vec{p}$ doesn't lies on these 3 lines, the 3 vectors $u_i(\vec{p})$ are non-zero. We can find a non-zero $\vec{t}$ to satisfy $(*1)$ when and only when these 3 vectors are linear dependent to each other which is equivalent to the vanishing of their triple product. To summarize,

The condition for a point $\vec{p}$ to lie on the locus for three non-parallel $\ell_1, \ell_2, \ell_3$ is $$u_1(\vec{p}) \cdot ( u_2(\vec{p}) \times u_3(\vec{p}) ) = 0$$

Back to our original problem. Let $(x,y,z)$ be the coordinates for a generic point $\vec{p}$. It is easy to see we can represent the three given lines as

$$\begin{cases} ( \vec{p}_1, \vec{t}_1 ) &= ( (0,2,-3), (1, 0, 0) )\\ ( \vec{p}_2, \vec{t}_2 ) &= ( (-1,0,3), (0, 1, 0) )\\ ( \vec{p}_3, \vec{t}_3 ) &= ( (1,-2,0), (0, 0, 1) ) \end{cases} \implies \begin{cases} u_1(\vec{p}) = ( 0, z+3, 2-y)\\ u_2(\vec{p}) = ( 3-z,0,x+1)\\ u_3(\vec{p}) = ( y+2,1-x,0) \end{cases} $$ The equation of the locus becomes

$$u_1(\vec{p}) \cdot ( u_2(\vec{p}) \times u_3(\vec{p}) ) = \left|\begin{matrix} 0 & z+3 & 2-y \\ 3-z & 0 & x+1\\ y+2 &1-x & 0 \end{matrix}\right| = 6xy + 2yz + 4xz + 12 = 0 $$ So the answer is $(b)$.