How Two Lines in Space Can Neither Meet Nor Run Parallel

On a flat page, two lines either meet or are parallel. In space there is a third possibility: lines that never meet and are not parallel, like a flyover crossing above a road. Vector and Cartesian equations let us test which case applies and measure the gap between the lines.

This lesson covers direction cosines and ratios, equations of a line, coplanar and skew lines, and shortest distances.

**The line joining $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$ has direction ratios $x_2-x_1$, $y_2-y_1$ and $z_2-z_1$, and dividing each by the length PQ gives its direction cosines, which satisfy $l^2+m^2+n^2=1$.

Worked example.** For $P(1,-2,3)$ and $Q(3,1,9)$:

- Direction ratios: 2, 3 and 6
- $PQ=\sqrt{4+9+36}=7$
- Direction cosines: $\frac{2}{7}$, $\frac{3}{7}$ and $\frac{6}{7}$

Check. $\frac{4+9+36}{49}=1$.

Worked example 2. A line makes equal angles with all three axes. Then $l=m=n$ and $3l^2=1$, so each direction cosine is $\pm \frac{1}{\sqrt{3}}$, and the angle with each axis is about $54.7°$ for the positive choice.

An everyday example. A zip-line cable from a hilltop platform down to a landing point has direction ratios given by the differences between the coordinates of its two ends.

The substance. A line has two sets of direction cosines — $l,m,n$ and $-l,-m,-n$ — one for each way along it.

**A line through the point with position vector $\vec{a}$, parallel to $\vec{b}$, has vector equation $\vec{r}=\vec{a}+\lambda \vec{b}$ and Cartesian equation $\frac{x-x_1}{b_1}=\frac{y-y_1}{b_2}=\frac{z-z_1}{b_3}$; a line through two points uses the difference of their position vectors as $\vec{b}$.

Worked example (one point).** The line through $(2,-1,3)$ parallel to $\hat{i}+2\hat{j}-2\hat{k}$ is

$$\vec{r}=(2\hat{i}-\hat{j}+3\hat{k})+\lambda (\hat{i}+2\hat{j}-2\hat{k}),\frac{x-2}{1}=\frac{y+1}{2}=\frac{z-3}{-2}$$

Worked example (two points). Through $(1,0,2)$ and $(3,4,6)$, the direction is $2\hat{i}+4\hat{j}+4\hat{k}$, so the line is $\frac{x-1}{2}=\frac{y}{4}=\frac{z-2}{4}$. Setting $\lambda =1$ in $\vec{r}=\hat{i}+2\hat{k}+\lambda (2\hat{i}+4\hat{j}+4\hat{k})$ returns the second point, as a check.

Angle between lines. For direction vectors $\overrightarrow{b_1}$ and $\overrightarrow{b_2}$, $\cos \theta =\frac{|\overrightarrow{b_1}\cdot \overrightarrow{b_2}|}{|\overrightarrow{b_1}||\overrightarrow{b_2}|}$.

An everyday example. A drone flying in a straight line from a known starting point in a fixed direction has position $\vec{a}+\lambda \vec{b}$, where $\lambda$ grows with time.

The substance. The same line has many equations — any point on it, and any multiple of its direction vector, give a valid form.

**Two non-parallel lines $\vec{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}$ and $\vec{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}$ are coplanar, and therefore intersect, exactly when $(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot (\overrightarrow{b_1}\times \overrightarrow{b_2})=0$; otherwise they are skew — neither parallel nor meeting.

Cartesian test.** For lines through $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ with direction ratios $a_1,b_1,c_1$ and $a_2,b_2,c_2$, they are coplanar when

$$\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|=0$$

Worked example (intersecting). Consider $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-4}{5}=\frac{y-1}{2}=z$. The determinant with rows $(3,-1,-3)$, $(2,3,4)$ and $(5,2,1)$ is $3(3-8)+1(2-20)-3(4-15)=-15-18+33=0$, so the lines are coplanar. Solving the two sets of equations shows that they meet at $(-1,-1,-1)$.

An everyday example. A flyover and the road passing beneath it run along skew lines — they are not parallel, yet they never meet.

The substance. Not parallel does not mean intersecting — in three dimensions, two lines chosen at random are almost always skew.

**The shortest distance between skew lines $\vec{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}$ and $\vec{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}$ is $\frac{|(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot (\overrightarrow{b_1}\times \overrightarrow{b_2})|}{|\overrightarrow{b_1}\times \overrightarrow{b_2}|}$, and the distance of a point P from a line through A with direction $\vec{b}$ is $\frac{|\overrightarrow{AP}\times \vec{b}|}{|\vec{b}|}$.

Worked example (skew lines).** For $\vec{r}=(\hat{i}+\hat{j})+\lambda (2\hat{i}-\hat{j}+\hat{k})$ and $\vec{r}=(2\hat{i}+\hat{j}-\hat{k})+\mu (3\hat{i}-5\hat{j}+2\hat{k})$:

- $\overrightarrow{a_2}-\overrightarrow{a_1}=\hat{i}-\hat{k}$
- $\overrightarrow{b_1}\times \overrightarrow{b_2}=3\hat{i}-\hat{j}-7\hat{k}$, with magnitude $\sqrt{59}$
- $(\hat{i}-\hat{k})\cdot (3\hat{i}-\hat{j}-7\hat{k})=3+7=10$
- Shortest distance $=\frac{10}{\sqrt{59}}\approx 1.30$ units

Worked example (point from a line). For $P(1,2,3)$ and the line through the origin with direction $\hat{i}+\hat{j}+\hat{k}$, $\overrightarrow{AP}\times \vec{b}=-\hat{i}+2\hat{j}-\hat{k}$, so the distance is $\frac{\sqrt{6}}{\sqrt{3}}=\sqrt{2}\approx 1.41$ units.

An everyday example. Engineers checking the clearance between a power cable and a pipeline that cross at different heights calculate exactly this shortest distance.

The substance. The shortest distance runs along the common perpendicular — which is why $\overrightarrow{b_1}\times \overrightarrow{b_2}$, perpendicular to both lines, appears in the formula.