How to Describe the Path of a Drone Flying in a Straight Line

On a flat page, a slope and a point fix a line. In space a line can tilt any way, so we need three numbers for its direction and one point it passes through. A drone flying straight from one rooftop to another traces exactly such a line.

This part covers direction cosines and ratios, the line through two points, and the vector and Cartesian equations of a line.

**If a directed line makes angles $\alpha$, $\beta$ and $\gamma$ with the $x$, $y$ and $z$ axes, its direction cosines are $l=\cos \alpha$, $m=\cos \beta$, $n=\cos \gamma$; any three numbers proportional to them are direction ratios; and $l^2+m^2+n^2=1$.

Proof of the relation.** Take a point $P(x,y,z)$ on a line through the origin, with $OP=r$. Projecting onto the axes gives $x=lr$, $y=mr$ and $z=nr$. Since $x^2+y^2+z^2=r^2$,

$$r^2(l^2+m^2+n^2)=r^2\Rightarrow l^2+m^2+n^2=1$$

From ratios to cosines. If the direction ratios are $a$, $b$, $c$, divide each by $\sqrt{a^2+b^2+c^2}$, using the same sign throughout.

Worked example. Direction ratios $2,-1,-2$ give $\sqrt{4+1+4}=3$, so the direction cosines are

$$\pm \left(\frac{2}{3},-\frac{1}{3},-\frac{2}{3}\right)$$

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

An everyday example. A kite string stretching up from a child's hand makes definite angles with the east, north and vertical directions, captured by three direction cosines.

The substance. A line has two sets of direction cosines, one for each way along it — reversing the direction changes all three signs.

**For the line through $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$, the direction ratios are $x_2-x_1$, $y_2-y_1$, $z_2-z_1$, and dividing them by the distance $PQ$ gives the direction cosines.

The formulae:**

$$PQ=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2},l=\frac{x_2-x_1}{PQ},m=\frac{y_2-y_1}{PQ},n=\frac{z_2-z_1}{PQ}$$

Worked example. For $P(-2,4,-5)$ and $Q(1,2,3)$, the direction ratios are $3,-2,8$ and $PQ=\sqrt{9+4+64}=\sqrt{77}$, so

$$l,m,n=\frac{3}{\sqrt{77}},-\frac{2}{\sqrt{77}},\frac{8}{\sqrt{77}}$$

Collinearity test. Points $A$, $B$ and $C$ are collinear when the direction ratios of $AB$ and $BC$ are proportional. For $A(2,3,4)$, $B(-1,-2,1)$ and $C(5,8,7)$, $AB$ gives $-3,-5,-3$ and $BC$ gives $6,10,6$ — proportional, so the points are collinear.

An everyday example. A zipline from a hilltop platform to a lower landing point has its direction fixed entirely by the coordinates of its two ends.

The substance. **Direction ratios from $Q$ to $P$ are the negatives of those from $P$ to $Q$** — both describe the same line.

**A line through the point with position vector $\vec{a}$ parallel to $\vec{b}$ has vector equation $\vec{r}=\vec{a}+\lambda \vec{b}$; if the point is $(x_1,y_1,z_1)$ and $\vec{b}$ has direction ratios $a,b,c$, its Cartesian equation is $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$.

Why.** Any point $P$ on the line satisfies $\overrightarrow{AP}=\lambda \vec{b}$ for some real $\lambda$, so $\vec{r}-\vec{a}=\lambda \vec{b}$.

Worked example. The line through $(5,2,-4)$ parallel to $3\hat{i}+2\hat{j}-8\hat{k}$:

$$\vec{r}=5\hat{i}+2\hat{j}-4\hat{k}+\lambda (3\hat{i}+2\hat{j}-8\hat{k})$$

$$\frac{x-5}{3}=\frac{y-2}{2}=\frac{z+4}{-8}$$

A point on the line. Putting $\lambda =1$ gives $(8,4,-12)$, which makes all three fractions equal to $1$.

An everyday example. A laser beam leaving a pointer in a fixed direction traces $\vec{r}=\vec{a}+\lambda \vec{b}$, with $\lambda$ measuring how far along the beam you are.

The substance. A zero in a denominator of the Cartesian form is allowed — it means that coordinate stays constant along the line.

**The line through points with position vectors $\vec{a}$ and $\vec{b}$ is $\vec{r}=\vec{a}+\lambda (\vec{b}-\vec{a})$, and through $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ its Cartesian form is $\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$.

Worked example — two points.** The line through $(-1,0,2)$ and $(3,4,6)$ has direction $4\hat{i}+4\hat{j}+4\hat{k}$:

$$\vec{r}=-\hat{i}+2\hat{k}+\lambda (4\hat{i}+4\hat{j}+4\hat{k})\frac{x+1}{4}=\frac{y}{4}=\frac{z-2}{4}$$

Cartesian to vector. From $\frac{x-3}{2}=\frac{y+1}{-5}=\frac{z-4}{6}$, read the point $(3,-1,4)$ and direction $(2,-5,6)$:

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

An everyday example. A metro line running straight between two stations is fully described once the positions of the two stations are known.

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