# given skew lines $l$ and $m$ find the geometric locus

Suppose we have two skew lines $l$ and $m$. I want to find the geometric locus of points $P$ for which there is not line passing trough $P$ intersecting $l$ and $m$.

I know the locus of points should be a plane. But, how can we prove it?

• Maybe two planes? Here's a hint: Skew lines lie in (unique) parallel planes. – Ted Shifrin Feb 10 '16 at 7:29
• Additional hint, assuming the question takes place in $\mathbf{R}^{3}$: If $P$ is an arbitrary point and $L$ is a line not containing $P$, there exists a unique plane containing both $L$ and $P$. If that doesn't help, could you please explain what you're unable to complete in light of Ted's hint? – Andrew D. Hwang Feb 14 '16 at 0:44
• I understand now. but for $P$ to satisfy the condition, wouldnt the planes intersect through $P$? ? – user139708 Feb 15 '16 at 9:03
• Added short explanation in first para, hope it is clearer. – Narasimham Feb 15 '16 at 15:09
• i understand nothin' of what you sayin' – user139708 Feb 15 '16 at 18:31

Suppose $l=L+V$, $m=M+W$ are two skew lines in $\Bbb A^3(\mathbb{R})$ passing respectively through the points $L$, $M$ with director subspaces $V$, $W$. Since the two lines are skew, $V \neq W$ and $l\cap m= \emptyset$.

Let $U:= V \cup W$. Consider the two planes $\pi_1:=L+U$, $\pi_2:=M+U$. If $X\in \pi_1$, then $X \lor P \subset \pi_1$ for every $P\in l$. Since $\pi_1,\pi_2$ are parallel $X \lor P$ does not intersect $\pi_2$, hence $m$. The same is true for $\pi_2$ and $m$. This implies that there do not exist lines intersecting both $l$,$m$ and passing through a point of $\pi_1 \cup \pi_2$.

Can you show that a transversal through $X$ intersecting $l$, $m$ exists if $X \not \in \pi_1 \cup \pi_2$?

As pointed out in the comments, two skew lines $l$ and $m$ lie in parallel planes $H_l$ and $H_m$ respectively. Clearly $(H_l \cup H_m) \setminus (l \cup m)$ is contained in the geometric locus, since any line, which intersects $l$ and another point of $H_l$ is contained in $H_l$ and thus does not meet $m$.

To the converse let $P$ not in $(H_l \cup H_m) \setminus (l \cup m)$.

Consider the line through $P$ and $Q$ for any $Q \in l$. This line intersects $H_m$, denote the intersection point by $P_Q$. The $P_Q, Q \in l$ form a line on $H_m$, which intersects $m$, hence there is some $Q \in l$ with $P_Q \in m$, thus the line through $P$ and $Q$ intersects $m$. This shows that $P$ is not contained in the geometric locus.

Summing up, we have shown that the geometric locus is $(H_l \cup H_m) \setminus (l \cup m)$. Two planes, with the two lines removed.

EDIT 1:

You can take $any\, two$ skew lines in 3-space. There is a certain minimum distance between two lines along their common normal, wlog let us take this as lying along $y-$ axis equal to $2t$.

Let the skew lines be $z = \pm m x , y = \pm t$. Minimum separation distance is $2 t.$ Then,

$$(y \ne \pm \,t)$$

is the equation for required set of all planes parallel to X-O-Z plane, which have no line on them cutting the given skew lines.

• Why are you assuming the lines are of this form wlog? – user139708 Feb 15 '16 at 6:06
• We assume inclination between the given skew lines to be $2 tan^{−1} m$ for illustration of the particular arbitrary example taken. – Narasimham Feb 20 '16 at 19:37