Q.Prove that the straight lines $\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}$ and $\dfrac{x-2}{3}=\dfrac{y-3}{4}=\dfrac{z-4}{5}$ intersect each other. Find their point of intersection and the equation of the plane on which the lines will lie.

Skew Lines

In a plane, two straight lines have only two possibilities: they meet, or they are parallel. In three dimensions a third possibility appears — lines that neither meet nor run parallel. These are skew lines.

What Makes Lines Skew

Two lines in space are skew if they are not parallel and do not intersect. The deeper reason is that skew lines do not lie in the same plane — they are non-coplanar. Parallel lines and intersecting lines always share a plane; skew lines never do.

The Three Cases in Space

How to Test for Skew Lines

Take two lines $\vec{r} = \vec{a_1} + \lambda\vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu\vec{b_2}$.

1. Not parallel: $\vec{b_1}$ and $\vec{b_2}$ are not proportional (so $\vec{b_1}\times\vec{b_2}\neq\vec{0}$).
2. Do not intersect: no values of $\lambda,\mu$ make the points coincide.

Both conditions are captured by one scalar triple product. The lines are skew exactly when

If this value is zero, the lines are coplanar (they intersect or are parallel); if it is non-zero, they are skew.

Shortest Distance Between Skew Lines

Because skew lines miss each other, there is a well-defined shortest distance between them, measured along their common perpendicular:

$d = \frac{\left|(\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times\vec{b_2})\right|}{\left|\vec{b_1}\times\vec{b_2}\right|}.$ …