Three Dimensional Geometry

From arrows to geometry

Chapter 10 built vectors as free-floating arrows — magnitude and direction, with no fixed position in space. This chapter anchors that arrow into a real geometric object that lives in 3D space: a line.

A line is an arrow fixed at one point and let run forever in both directions. It is completely described by a point and a direction vector — those two ingredients are everything this chapter computes with.

Every question this chapter asks reduces to the same short list: do two lines meet, at what angle, how far apart are two lines that never touch. The geometry is new. The machinery computing it — the dot product for angles, the cross product for distances — is entirely Chapter 10’s.

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Direction cosines and direction ratios of a line

A line pointing through space makes some angle with each of the three axes. Call these angles $\alpha, \beta, \gamma$ against the $x$, $y$, $z$ axes respectively. The direction cosines $l, m, n$ are simply $\cos \alpha, \cos \beta, \cos \gamma$ — three numbers, one line, one fixed triple.

Squaring and adding those three cosines always gives $1$: $l^2+m^2+n^2=1$. This is not a separate fact to memorise — it is the same identity a unit vector’s components already satisfy, since $(l, m, n)$ is exactly the unit vector pointing along the line.

Direction ratios $a, b, c$ are a looser cousin: any three numbers proportional to $l, m, n$. A line has exactly one triple of direction cosines but infinitely many triples of direction ratios — $(2,4,4)$, $(1,2,2)$, $(3,6,6)$ all describe the same direction. Recovering the cosines from a given set of ratios means dividing each by the magnitude of the triple: $l = a/\sqrt{a^2+b^2+c^2}$, and likewise for $m$ and $n$.

Two points on a line already fix its direction, without needing any angle measured against the axes. For a line through $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$, the direction ratios are simply the coordinate differences: $x_2-x_1$, $y_2-y_1$, $z_2-z_1$ — how far the line moves along each axis in going from $P$ to $Q$.

*Dividing each of those three differences by the distance $P Q$ turns the ratios into the actual direction cosines*, since $P Q = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$ is exactly the magnitude the previous section divided by. No angle is ever measured directly — the two points do all the work.

A set of direction ratios like $(2,4,4)$ looks ready to use immediately: three numbers, a fixed direction, why not drop them straight into any formula that asks for direction cosines?

That shortcut skips a real requirement. Direction cosines must satisfy $l^2+m^2+n^2=1$, and $(2,4,4)$ fails it outright: $4+16+16=36 \neq 1$. A direction-ratio triple is not automatically the direction-cosine triple; it is only ever proportional to it.

Dividing $(2,4,4)$ by its own magnitude $\sqrt{2^2+4^2+4^2} = 6$ gives the true cosines $(1/3, 2/3, 2/3)$, and $1/9+4/9+4/9=1$ confirms the identity now holds. Direction ratios are any proportional triple; direction cosines are the one normalised triple pointing the same way — and normalising is never optional.

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Equation of a line in space

A line is fixed by one point on it and one direction it runs in. If $\vec{a}$ is the position vector of a known point on the line and $\vec{b}$ is a vector parallel to it, the line’s equation is $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\vec{r}$ is the position vector of a general point on the line.

$\lambda \in \mathbb{R}$ is a free parameter, not a fixed number — it is what makes the equation describe an entire line rather than one point. *Each value of $\lambda$ traces out exactly one point on the line, and every point on the line is hit by exactly one value of $\lambda$.* Setting $\lambda = 0$ recovers the starting point $\vec{a}$; every other value slides along the direction $\vec{b}$, forwards or backwards.

Writing the vector equation $\vec{r} = \vec{a} + \lambda \vec{b}$ out component by component turns it into three separate equations, one for each coordinate. Eliminating $\lambda$ between them gives a single chain of equal ratios — the symmetric, or Cartesian, form of the line.

With point $(x_1,y_1,z_1)$ and direction ratios $a, b, c$, the line reads $(x-x_1)/a = (y-y_1)/b = (z-z_1)/c$. Reading a point and a set of direction ratios straight off this equation converts it back to the vector form without any further work — the point sits in the numerators, the direction ratios sit in the denominators.

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Angle between two lines

Two lines meeting in space carry an angle between them — and that angle is entirely decided by their direction vectors, never by where the lines happen to sit. Finding it is a dot-product calculation, exactly as it was for two vectors in Chapter 10.

With direction ratios $a_1,b_1,c_1$ and $a_2,b_2,c_2$ for the two lines: $\cos \theta = \left|a_1 a_2+b_1 b_2+c_1 c_2\right|/(\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2})$. The absolute value keeps $\theta$ as the acute angle between the lines, regardless of which way each direction vector happens to point.

Two special cases fall straight out of the same formula: the lines are perpendicular exactly when $a_1 a_2+b_1 b_2+c_1 c_2 = 0$ — the dot product itself vanishes — and parallel exactly when $a_1/a_2 = b_1/b_2 = c_1/c_2$, meaning one direction is a scalar multiple of the other.

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Shortest distance between two lines

Two lines in space that neither meet nor run parallel are called skew — genuinely different from anything Class 11 plane geometry could produce, where two non-parallel lines always cross. The shortest segment joining two skew lines is perpendicular to both of them at once.

That perpendicular direction is exactly $\vec{b_1} \times \vec{b_2}$ — the cross product of the two lines' own direction vectors, since a cross product is perpendicular to both its inputs by construction. Projecting the vector joining any point on one line to any point on the other, $\vec{a_2}-\vec{a_1}$, onto that perpendicular direction gives the shortest distance.

For lines $\vec{r} = \vec{a_1}+\lambda \vec{b_1}$ and $\vec{r} = \vec{a_2}+\mu \vec{b_2}$: $d = \left|(\vec{b_1} \times \vec{b_2}) \cdot (\vec{a_2}-\vec{a_1})\right|/\left|\vec{b_1} \times \vec{b_2}\right|$. *The numerator is a scalar triple product; the denominator just makes the answer independent of how long $\vec{b_1}$ and $\vec{b_2}$ happen to be.*

Two lines that are neither parallel nor obviously crossing on a diagram can tempt a shortcut: solve their two equations together and something should still come out as an intersection point, even if it takes a bit of algebra to find it.

That shortcut assumes every pair of lines in space behaves like two lines in a plane, where non-parallel always means crossing. In three dimensions it does not: two lines can point in genuinely different directions and still never occupy the same point at the same parameter value. Solving their equations simultaneously then yields no solution at all — not a hidden one waiting to be found.

Two such lines, for instance the pair worked above, never meet; equating their coordinates gives a contradiction, not a fix-up. Skew lines still have a well-defined angle, found from their direction vectors, and a well-defined shortest distance, found from the skew-lines formula — “no intersection” does not mean “nothing left to compute”.

Two parallel lines share a direction vector $\vec{b}$ but pass through different points — no cross product between two different directions is available, since there is only one direction between them.

The skew-lines formula still answers the question, with one simplification: since $\vec{b_1}=\vec{b_2}=\vec{b}$, the triple-product numerator collapses to a plain cross product. *With points $\vec{a_1}$, $\vec{a_2}$ on the two lines, the distance between them is $d = \left|\vec{b} \times (\vec{a_2}-\vec{a_1})\right|/\left|\vec{b}\right|$* — the same machine, run on one shared direction instead of two.

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One toolkit, every question in this chapter

Every question this chapter asked — angle, distance, or whether two things meet at all — reduces to one of two operations already built in Chapter 10.

The dot product answers every angle and perpendicularity question between two lines, and it decides parallel and perpendicular as special cases along the way.

The cross product answers every distance question — the skew-lines formula folds it into a scalar triple product, and the parallel-lines formula drops that term entirely. The geometry in this chapter was new. The vector machinery computing it was entirely Chapter 10’s, run on a new set of targets: lines instead of free-floating arrows.

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Practice set

Exercise 11.1 practises finding direction cosines and direction ratios of a line — from given angles against the axes, from two points it passes through, and from arbitrary direction ratios normalised into cosines. Every answer gets checked against the identity *$l^2+m^2+n^2=1$*, the same test used throughout the chapter.

Exercise 11.2 practises writing a line’s equation in vector and Cartesian form, then finding the angle between two lines from their direction ratios. It closes with the skew-lines and parallel-lines distance formulas, applied to lines given directly in vector form.

The Miscellaneous Exercise draws on every line technique in this chapter in a single problem set, with no method named in advance. Angle and distance questions reappear — between two lines, and between skew or parallel lines — drawing on the same toolkit named in the chapter’s own recap.