11.1 Direction Cosines and Direction Ratios of a Line

A telecom mast is held upright by steel cables running from its top down to anchors in the ground. To order the right length of cable and to mount the clamps at the right angle, an engineer needs to know exactly which way each cable runs. In Section 10.1 you described the direction of a vector by its direction cosines — the cosines of the angles it makes with the three axes. A cable, a laser beam or the edge of a building is not a vector, though: it is a line, with no arrowhead and no fixed starting point. It runs in two opposite directions at once, and it need not pass through the origin.

This section adapts direction cosines to lines. You will see that a line has two sets of direction cosines, differing only in sign; that any three numbers proportional to them — direction ratios — describe the direction just as well and are usually much easier to work with; and that the direction ratios of the line through two points are simply the differences of their coordinates. Direction ratios then give a quick test for whether three points lie on one line. Everything in Sections 11.2 and 11.3 is built on these few ideas.

Direction Cosines of a Line

A line extends without end in two opposite directions. To measure angles with the axes we first choose one of them: a line with a chosen direction is called a directed line. The direction of a line does not depend on where the line is, so we may always slide it, without turning it, until it passes through the origin.

If we reverse the direction of $L$, each angle $\alpha$ is replaced by $\pi-\alpha$, and $\cos(\pi-\alpha)=-\cos\alpha$. So the reversed line has direction cosines $-l,-m,-n$. An ordinary (undirected) line therefore has two sets of direction cosines, $(l,m,n)$ and $(-l,-m,-n)$; we often write them together as $\pm(l,m,n)$. Unless a question fixes the direction (for example “from $P$ to $Q$” or “making an acute angle with the $z$-axis”), either set is a correct answer.

Proof. Let $P(x,y,z)$ be the point of $L_0$ described, so $OP=r$ and $\overrightarrow{OP}$ points along $L_0$. Let $A$ be the foot of the perpendicular from $P$ to the $x$-axis; then $A=(x,0,0)$ and triangle $OAP$ has a right angle at $A$, with angle $\alpha$ at $O$. Hence the signed length $OA$ along the positive $x$-axis is $r\cos\alpha$, that is $x=r\cos\alpha=lr$. (If $\alpha$ is obtuse, $x$ and $\cos\alpha$ are both negative; if $\alpha=\tfrac\pi2$, both are $0$.) In the same way $y=mr$ and $z=nr$. By the distance formula $x^2+y^2+z^2=r^2$, so $r^2(l^2+m^2+n^2)=r^2$; dividing by $r^2$ gives $l^2+m^2+n^2=1$. $\square$

So the three angles of a line with the axes are not independent: once two of them are known, the third is fixed up to the choice between $\gamma$ and $\pi-\gamma$.

Direction Ratios of a Line

Direction cosines are often awkward numbers such as $\tfrac1{\sqrt3}$. Since only the proportions $l:m:n$ matter for the direction, we usually work with any convenient multiple of them.

Proof. By Definition 11.2, $a=kl$, $b=km$, $c=kn$ with $k\ne0$. Squaring and adding, and using Theorem 11.1, $$a^2+b^2+c^2=k^2(l^2+m^2+n^2)=k^2,$$ so $k=\pm\sqrt{a^2+b^2+c^2}$ (this is not zero, because $a,b,c$ are not all zero). Then $l=\dfrac ak$, $m=\dfrac bk$, $n=\dfrac ck$, with the same $k$ in all three. $\square$

The Line through Two Points

In practice a line is usually given by two of its points — the top of the mast and the anchor of a cable, two positions of a drone, two vertices of a solid. Its direction ratios can then be written down at once.

Proof. Slide the directed line $PQ$, without turning it, so that $P$ moves to the origin: every point moves by $(-x_1,-y_1,-z_1)$, so $Q$ moves to $Q'(x_2-x_1,\ y_2-y_1,\ z_2-z_1)$. The line $OQ'$ is parallel to $PQ$ and points the same way, and $OQ'=PQ$. By Theorem 11.1, the coordinates of $Q'$ are $(l\cdot PQ,\ m\cdot PQ,\ n\cdot PQ)$, where $l,m,n$ are the direction cosines of $PQ$. Dividing by $PQ$ gives the stated cosines, and the coordinates of $Q'$, being $PQ$ times them, are direction ratios. $\square$

In the language of Chapter 10 (Theorem 10.7 in Section 10.2): $x_2-x_1,\ y_2-y_1,\ z_2-z_1$ are the components of the vector $\overrightarrow{PQ}$. Reversing the order of the points changes every sign, which is just the other direction of the same line.

Any two pairs of points taken on one line give proportional direction ratios. This turns “do three points lie on a line?” into simple arithmetic.

Why it is true. If $A,B,C$ lie on one line, then $AB$ and $BC$ are that same line, so their direction ratios are proportional. Conversely, proportional direction ratios mean that the lines $AB$ and $BC$ have the same direction cosines up to sign, so they are parallel; and two parallel lines with a common point — here $B$ — are the same line. So $A$, $B$, $C$ all lie on it. $\square$

Exercises