Why a Flyover and the Road Beneath It Never Meet

On paper, two lines either meet or run parallel. In space there is a third possibility: a flyover and the road beneath it point in different directions yet never touch. Such skew lines need new tools — angles from direction vectors and shortest distances from vector products.

This part covers angles between lines, skew lines, and shortest distances between skew and parallel lines.

**The angle between two lines is the angle between their direction vectors, $\cos \theta =\left|\frac{\overrightarrow{b_1}\cdot \overrightarrow{b_2}}{|\overrightarrow{b_1}||\overrightarrow{b_2}|}\right|$; the lines are perpendicular when $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$ and parallel when their direction ratios are proportional.

In Cartesian form**, with direction ratios $a_1,b_1,c_1$ and $a_2,b_2,c_2$:

$$\cos \theta =\left|\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}\right|$$

Conditions:

- Perpendicular — $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$
- Parallel — $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

Worked example 1 — angle. Lines with direction ratios $3,5,4$ and $1,1,2$:

$$\cos \theta =\frac{3+5+8}{\sqrt{50}\sqrt{6}}=\frac{16}{10\sqrt{3}}=\frac{8\sqrt{3}}{15}$$

Worked example 2 — perpendicular. Find $k$ so that lines with direction ratios $-3,2k,2$ and $3k,1,-5$ are perpendicular:

$$-9k+2k-10=0\Rightarrow k=-\frac{10}{7}$$

An everyday example. A sloping staircase handrail and a corridor running beneath it make a definite angle even though they never touch — the angle comes from their directions alone.

The substance. **The absolute value keeps the angle between $0$ and $\frac{\pi }{2}$**, because two lines form an acute or right angle.

Skew lines are lines in space that are neither parallel nor intersecting, so they cannot lie in one plane; intersecting lines share a point, and parallel lines have proportional directions.

Three possibilities for two lines in space:

- Intersecting — meet at exactly one point; lie in one plane
- Parallel — proportional directions; lie in one plane; never meet unless they coincide
- Skew — directions not proportional and no common point; not coplanar

Worked example — test a pair. $L_1$: $\vec{r}=\hat{i}+\lambda (\hat{i}+\hat{j})$ and $L_2$: $\vec{r}=\hat{k}+\mu \hat{j}$.

- Directions $(1,1,0)$ and $(0,1,0)$ are not proportional, so the lines are not parallel
- Points of $L_1$ have $z=0$ and points of $L_2$ have $z=1$, so there is no common point

Hence $L_1$ and $L_2$ are skew.

In a room. An edge of the floor and a non-parallel edge of the ceiling are skew.

An everyday example. A flyover and the road passing under it at an angle are skew: different directions, different levels, no meeting point.

The substance. Non-parallel lines in a plane must meet, but in space they need not — that is exactly what makes skew lines possible.

**For skew lines $\vec{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}$ and $\vec{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}$, the shortest distance is $d=\left|\frac{(\overrightarrow{b_1}\times \overrightarrow{b_2})\cdot (\overrightarrow{a_2}-\overrightarrow{a_1})}{|\overrightarrow{b_1}\times \overrightarrow{b_2}|}\right|$, the projection of the joining vector on the common perpendicular.

Why it works.** The shortest segment between skew lines is perpendicular to both, so it points along $\overrightarrow{b_1}\times \overrightarrow{b_2}$; projecting $\overrightarrow{a_2}-\overrightarrow{a_1}$ onto that direction gives its length.

Worked example. $L_1$: $\vec{r}=\hat{i}+2\hat{j}+\hat{k}+\lambda (\hat{i}-\hat{j}+\hat{k})$ and $L_2$: $\vec{r}=2\hat{i}-\hat{j}-\hat{k}+\mu (2\hat{i}+\hat{j}+2\hat{k})$.

$$\overrightarrow{b_1}\times \overrightarrow{b_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -1& 1\\ 2& 1& 2\end{array}\right|=-3\hat{i}+3\hat{k},\overrightarrow{a_2}-\overrightarrow{a_1}=\hat{i}-3\hat{j}-2\hat{k}$$

$$d=\left|\frac{(-3)(1)+0+(3)(-2)}{\sqrt{9+9}}\right|=\frac{9}{3\sqrt{2}}=\frac{3}{\sqrt{2}}$$

An everyday example. Engineers checking the clearance between a flyover and a road crossing beneath it need exactly this shortest distance.

The substance. **If the formula gives $d=0$, the lines are not skew — they intersect.**

**For parallel lines $\vec{r}=\overrightarrow{a_1}+\lambda \vec{b}$ and $\vec{r}=\overrightarrow{a_2}+\mu \vec{b}$, the distance is $d=\left|\frac{\vec{b}\times (\overrightarrow{a_2}-\overrightarrow{a_1})}{|\vec{b}|}\right|$, and the shortest distance between two lines is zero exactly when they intersect or coincide.

Why it works.** $|\vec{b}\times (\overrightarrow{a_2}-\overrightarrow{a_1})|$ is the area of a parallelogram with base $|\vec{b}|$; dividing by the base gives its height, which is the distance between the lines.

Worked example. $\vec{r}=\hat{i}+2\hat{j}-4\hat{k}+\lambda (2\hat{i}+3\hat{j}+6\hat{k})$ and $\vec{r}=3\hat{i}+3\hat{j}-5\hat{k}+\mu (2\hat{i}+3\hat{j}+6\hat{k})$.

$$\overrightarrow{a_2}-\overrightarrow{a_1}=2\hat{i}+\hat{j}-\hat{k},\vec{b}\times (\overrightarrow{a_2}-\overrightarrow{a_1})=-9\hat{i}+14\hat{j}-4\hat{k}$$

$$d=\frac{\sqrt{81+196+16}}{\sqrt{4+9+36}}=\frac{\sqrt{293}}{7}$$

When the distance is zero. For non-parallel lines, $(\overrightarrow{b_1}\times \overrightarrow{b_2})\cdot (\overrightarrow{a_2}-\overrightarrow{a_1})=0$ means they intersect; for parallel lines, $d=0$ means they coincide.

An everyday example. Two parallel railway tracks stay the same distance apart along their whole length — the distance this formula computes.

The substance. Do not use the skew-line formula for parallel lines — $\overrightarrow{b_1}\times \overrightarrow{b_2}=\vec{0}$ makes its denominator zero.