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If three lines l_1, l_2, l_3 are given as l_1={(s,0,1)|s an element of R} and l_2={(1,t,0)|t an element of R} and l_3={(0,1,u)|u an element of R} then show that there are infinitely many lines that cross l_1,l_2 and l_3 at the same time.

I tried to separate the given equations of the lines like l_1=(0,0,1)+s(1,0,0), l_2=(1,0,0)+t(0,1,0) and l_3=(0,1,0)+u(0,0,1) but it turns out that from the above equations, the direction vectors of these three lines are perpendicular to one another and the direction vectors turned out to be the standard i, j and k. I thought the three lines would be all on the same plane and not perpendicular to one another. Is there something that I am missing? Any hint or a reference to a book with similar problems will really be appreciated. Please...


2 Answers 2


Three distinct points $A, B, C$ in $\mathbb R^3$ are on the same line $\iff \exists x \in \mathbb R$ such that $A-B = x(B-C)$. Setting $A=(s,0,1), B=(1,t,0), C=(0,1,u)$ this translates to: $$s - 1 = x$$ $$-t = x(t-1)$$ $$1 = -xu$$

Solving for $s, t$ and $u$ in terms of $x$, we get that the points $(x+1,0,1), (1,\frac{x}{x+1},0)$ and $(0,1,-\frac{1}{x})$ are on the same line and also on $l_1, l_2$ and $l_3$, respectively. A moment's thought will convince you that the lines determined by all the possible choices of $x$ are all different.


Let $\ell_1$, $\ell_2$ be not sets of points but arbitrary points of that sets.
We construct the desired line as passing through $\ell_1$, $\ell_2$, i.e. $\{(s,0,1)+v((1,t,0)-(s,0,1))\}$.
Then make it intersect with $\ell_3$:
$$\begin{cases}s+v(1-s)=0,\\ 0+v(t-0)=1,\\ 1+v(0-1)=u\end{cases}$$ $$\begin{cases} s\ne 0,\\ t = \frac{s - 1}{s}, \\s - 1\ne 0, \\u = \frac{1}{1 - s}, \\v = \frac{s}{s - 1} \end{cases}$$ and show that there are infinitely many solutions of the intersection set of equations, which differs not only by the scale of direction vector (as we will have $3$ equations with $4$ variables we need to show that the system is solvable and have at least $1$ free variable).
The lines go through points $(s,0,1)\in\ell_1, (1,\frac{s - 1}{s},0)\in\ell_2, (0,1,\frac{1}{1 - s})\in\ell_3$ and none of them are the same line, so there's infinity of lines, QED.
The same as this answer with $x+1=s$.

  • $\begingroup$ I would really appreciate that. The equation that you wrote above is a parametric equation for a line passing through l_1 and l_2 right? $\endgroup$
    – John
    May 15, 2020 at 17:36
  • $\begingroup$ Right, with parameter $v$. It is already done in this answer, but the $v$ parameter is named $x$, absolutely the same idea. $\endgroup$ May 15, 2020 at 18:08
  • $\begingroup$ can you please show me the complete solution? I tried to take your hint further but I got stuck when it became system of linear equations in four variables with the additional variable being v from the above parametric equation. $\endgroup$
    – John
    May 15, 2020 at 18:11
  • $\begingroup$ OK. I want to use wolframalpha for this step. $\endgroup$ May 15, 2020 at 18:27
  • $\begingroup$ @Jmw If you got a system of linear equations, then you did something wrong. There are clearly going to be terms that are products of the variables in the resulting equation. $\endgroup$
    – amd
    May 15, 2020 at 18:29

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