Vectors and coordinate geometry

MotivationH

Where does the ferry really go?

A ferry crosses the Jamuna river in Bangladesh. Its engine drives it straight across at $4$ km/h relative to the water, but the river flows downstream at $3$ km/h. Where does the ferry actually go, and how fast? The two motions combine: each hour the ferry moves $4$ km across and $3$ km downstream. Its actual velocity is an arrow of length $\sqrt{4^2 + 3^2} = 5$ km/h, pointing diagonally. To land at the jetty directly opposite, the pilot must aim upstream.

Quantities like velocity, force and displacement have a size and a direction. We call them vectors. They can be added (the ferry's velocity plus the river's) and scaled (twice as fast, or backwards). Once vectors are written in coordinates, a single operation — the dot product — gives lengths, angles, right angles and projections, and lets us prove geometric theorems with a few lines of algebra. The same ideas then work, unchanged, in three dimensions and in a thousand.

IntuitionH

Arrows that can move

Picture a vector as an arrow. What matters is its length and its direction, not where it starts: sliding an arrow without turning or stretching it gives the same vector. The displacement "3 km east, 4 km north" is the same whether you start in Dhaka or in Lima.

To add two vectors, place the tail of the second at the tip of the first; the sum goes from the first tail to the second tip ("tip to tail"). Doing it in the other order gives the other two sides of a parallelogram, with the same diagonal: $u + v = v + u$ .
To scale a vector by a number $t$ , stretch it by the factor $|t|$ , and reverse it if $t \lt 0$ .

Put the tail at the origin of a coordinate system, and the tip has coordinates: the vector "3 east, 4 north" becomes $(3, 4)$ . Now the geometry becomes arithmetic: adding arrows is adding coordinates, and stretching arrows is multiplying coordinates.

DefinitionsH

Definitions

Definition— vector

A vector in the plane $\R^2$ is an ordered pair $v = (v_1, v_2)$ of real numbers; a vector in space $\R^3$ is a triple $(v_1, v_2, v_3)$ ; in general a vector in $\R^n$ is a list of $n$ real numbers, its components. The vector from point $A$ to point $B$ is $\overrightarrow{AB} = B - A$ , computed component by component.

Definition— addition and scalar multiplication

For vectors $u, v$ in $\R^n$ and a real number $t$ (a scalar):

u + v = (u_1 + v_1, \dots, u_n + v_n), \qquad t\,v = (t v_1, \dots, t v_n).

The zero vector is $0 = (0, \dots, 0)$ , and $-v = (-1)v$ . Two vectors are parallel if one is a scalar multiple of the other.

Definition— dot product and length

The dot product of $u$ and $v$ in $\R^n$ is the number

u \cdot v = u_1 v_1 + u_2 v_2 + \cdots + u_n v_n.

The length (or norm) of $v$ is $|v| = \sqrt{v \cdot v} = \sqrt{v_1^2 + \cdots + v_n^2}$ . A unit vector has length $1$ ; for $v \ne 0$ , the vector $v / |v|$ is the unit vector in the direction of $v$ .

In the plane, $|v| = \sqrt{v_1^2 + v_2^2}$ is the Pythagorean theorem; in space, $|v|^2 = v_1^2 + v_2^2 + v_3^2$ comes from using Pythagoras twice. The distance between points $A$ and $B$ is $|B - A|$ .

Definition— angle and perpendicularity

For non-zero vectors $u, v$ , the angle $\theta$ between them, with $0 \le \theta \le \pi$ , is given by

\cos\theta = \frac{u \cdot v}{|u|\,|v|}.

The vectors are perpendicular (orthogonal), $u \perp v$ , if $u \cdot v = 0$ .

In $\R^2$ and $\R^3$ we will prove that this angle is the angle you would measure between the arrows. In higher dimensions there is no picture to measure, and the formula becomes the definition; the Cauchy–Schwarz inequality below guarantees that the right-hand side lies between $-1$ and $1$ , so the definition makes sense.

Definition— lines and planes

The line through the point $p$ with direction $d \ne 0$ is the set $\{p + t\,d : t \in \R\}$ , in $\R^2$ or $\R^3$ . Writing out components gives parametric equations, such as $x = p_1 + t d_1$ , $y = p_2 + t d_2$ .

The plane in $\R^3$ through $p$ with normal vector $n \ne 0$ is the set of points $x$ with $n \cdot (x - p) = 0$ . With $n = (a, b, c)$ this is the familiar equation $ax + by + cz = n \cdot p$ . A plane can also be written in vector form, $\{p + s\,u + t\,v : s, t \in \R\}$ , with $u$ and $v$ non-parallel vectors lying in it.

Worked examplesH

Worked examples

Example— Arithmetic, lengths and unit vectors

Let $u = (1, 2, 2)$ and $v = (2, -1, 2)$ . Then $u + v = (3, 1, 4)$ , $3u - v = (1, 7, 4)$ , $|u| = \sqrt{1 + 4 + 4} = 3$ and $|v| = 3$ . The unit vector in the direction of $u$ is $\tfrac13(1, 2, 2)$ .

Example— An angle in space

For the same $u$ and $v$ : $u \cdot v = 2 - 2 + 4 = 4$ , so $\cos\theta = \dfrac{4}{3 \cdot 3} = \dfrac49$ and $\theta = \arccos\tfrac49 \approx 63.6^\circ$ . No protractor could measure this in space — but the dot product does it in one line.

Example— Perpendicular or not?

$(2, 3) \cdot (3, -2) = 6 - 6 = 0$ , so these vectors are perpendicular. In general, $(a, b)$ is perpendicular to $(-b, a)$ : turning a vector by $90^\circ$ swaps its components and changes one sign. On the other hand $(1, 1, 1) \cdot (1, 1, -1) = 1 \ne 0$ : those diagonals of a cube are not perpendicular.

Example— Projection and work

A child pulls a sledge along level snow with a force $F = (30, 40)$ newtons (horizontal and vertical parts). The sledge moves along the displacement $d = (10, 0)$ metres. Only the part of the force along the motion does work: the work is $W = F \cdot d = 300$ joules. The projection of $F$ onto $d$ is

\operatorname{proj}_d F = \frac{F \cdot d}{d \cdot d}\, d = \frac{300}{100}(10, 0) = (30, 0),

and the rest, $F - \operatorname{proj}_d F = (0, 40)$ , is perpendicular to $d$ : it lifts the sledge a little but does no work along the ground.

Example— A line through two points, and where it meets a plane

The line through $A = (1, 0, 0)$ and $B = (2, 2, 2)$ has direction $d = B - A = (1, 2, 2)$ , so it is $\{(1 + t, 2t, 2t) : t \in \R\}$ . Where does it meet the plane $x + y + z = 11$ ? Substitute: $(1 + t) + 2t + 2t = 11$ , so $t = 2$ and the point is $(3, 4, 4)$ . In the plane $\R^2$ , the line through $(1, 2)$ with direction $(3, 1)$ has normal $(-1, 3)$ and equation $-(x - 1) + 3(y - 2) = 0$ , that is, $-x + 3y = 5$ .

Example— A non-example: lengths do not add

$u = (3, 0)$ and $v = (0, 4)$ have $|u| + |v| = 7$ , but $|u + v| = |(3, 4)| = 5$ . Walking $3$ km east and then $4$ km north leaves you only $5$ km from home. Lengths add only when the vectors point the same way.

ProofH

The dot product: theorems

Directly from the definition, by expanding components, the dot product is symmetric, $u \cdot v = v \cdot u$ , and distributes over addition and scalars: $u \cdot (v + w) = u \cdot v + u \cdot w$ and $(t u) \cdot v = t (u \cdot v)$ . Also $v \cdot v = |v|^2 \ge 0$ , with equality only for $v = 0$ . Everything below follows from these rules alone.

Theorem— Theorem (geometric meaning of the dot product)

For non-zero vectors $u, v$ in $\R^2$ or $\R^3$ that make the angle $\theta$ (measured between the arrows drawn from a common tail),

u \cdot v = |u|\,|v| \cos\theta.

Proof

Draw $u$ and $v$ from a common point $O$ , ending at $P$ and $Q$ . The third side of triangle $OPQ$ is the vector $u - v$ . By the law of cosines (see trigonometry),

|u - v|^2 = |u|^2 + |v|^2 - 2\,|u|\,|v|\cos\theta.

On the other hand, by the rules above, $|u - v|^2 = (u - v)\cdot(u - v) = u\cdot u - 2\,u\cdot v + v \cdot v = |u|^2 - 2\,u \cdot v + |v|^2$ . Comparing the two expressions gives $u \cdot v = |u|\,|v|\cos\theta$ . (If $u$ and $v$ are parallel, the "triangle" is flat, and $\theta$ is $0$ or $\pi$ ; the same computation works directly.) $\blacksquare$

∎

Theorem— Corollary

Non-zero vectors $u$ and $v$ are perpendicular exactly when $u \cdot v = 0$ . Moreover, $|u + v|^2 = |u|^2 + |v|^2$ if and only if $u \cdot v = 0$ — Pythagoras' theorem and its converse in vector form.

Proof

$\cos\theta = 0$ exactly when $\theta = 90^\circ$ . For the second statement, expand: $|u + v|^2 = |u|^2 + 2\,u\cdot v + |v|^2$ . $\blacksquare$

∎

Theorem— Cauchy–Schwarz inequality

For all vectors $u, v$ in $\R^n$ : $\;|u \cdot v| \le |u|\,|v|$ . Equality holds exactly when $u$ and $v$ are parallel.

Proof

In $\R^2$ and $\R^3$ this follows from the last theorem, since $|\cos\theta| \le 1$ . But in $\R^n$ we want to define angles using Cauchy–Schwarz, so we need a proof that does not use angles.

If $v = 0$ both sides are $0$ . Otherwise, for every real number $t$ ,

0 \le |u - t v|^2 = |u|^2 - 2t\,(u \cdot v) + t^2 |v|^2.

The right-hand side is a quadratic in $t$ with positive leading coefficient $|v|^2$ that is never negative, so it has at most one real root and its discriminant is at most zero (see quadratics):

4 (u \cdot v)^2 - 4 |u|^2 |v|^2 \le 0, \qquad\text{that is,}\qquad (u \cdot v)^2 \le |u|^2|v|^2.

Taking square roots gives the inequality. Equality means the discriminant is zero, so the quadratic has a root $t_0$ , and then $|u - t_0 v| = 0$ , i.e. $u = t_0 v$ . Conversely, if $u = t v$ both sides equal $|t|\,|v|^2$ . $\blacksquare$

∎

In the plane there is an even more direct proof, Lagrange's identity:

|u|^2|v|^2 - (u \cdot v)^2 = (u_1^2 + u_2^2)(v_1^2 + v_2^2) - (u_1 v_1 + u_2 v_2)^2 = (u_1 v_2 - u_2 v_1)^2 \ge 0,

which you can check by expanding. The gap is a perfect square, and it vanishes exactly when $u_1 v_2 = u_2 v_1$ , i.e. when $u$ and $v$ are parallel.

Theorem— Triangle inequality

For all vectors $u, v$ in $\R^n$ : $\;|u + v| \le |u| + |v|$ , with equality exactly when one vector is a non-negative multiple of the other.

Proof

Using Cauchy–Schwarz in the middle step,

|u + v|^2 = |u|^2 + 2\,u\cdot v + |v|^2 \le |u|^2 + 2\,|u|\,|v| + |v|^2 = (|u| + |v|)^2.

Both $|u + v|$ and $|u| + |v|$ are non-negative, so $|u + v| \le |u| + |v|$ . Equality holds exactly when $u \cdot v = |u|\,|v|$ : by the equality case of Cauchy–Schwarz the vectors are parallel, and $u \cdot v \ge 0$ forces them to point the same way (or one to be zero). $\blacksquare$

∎

Theorem— Projection

For $v \ne 0$ , every vector $u$ splits as $u = p + w$ with $p = \dfrac{u \cdot v}{v \cdot v}\,v$ parallel to $v$ and $w = u - p$ perpendicular to $v$ .

Proof

$w \cdot v = u \cdot v - \dfrac{u\cdot v}{v \cdot v}\,(v \cdot v) = 0$ . $\blacksquare$

∎

ProofH

Geometry proved with vectors

Vectors give short proofs of classical theorems. The trick is to name the points by their position vectors $a, b, c, \ldots$ (from a fixed origin) and to translate each geometric statement into an equation.

Theorem— The diagonals of a parallelogram bisect each other

If $ABCD$ is a parallelogram, the midpoints of the diagonals $AC$ and $BD$ coincide.

Proof

" $ABCD$ is a parallelogram" means that the sides $AB$ and $DC$ are the same vector: $b - a = c - d$ . Rearranging, $a + c = b + d$ . The midpoint of $AC$ is $\tfrac12(a + c)$ and the midpoint of $BD$ is $\tfrac12(b + d)$ , and these are equal. $\blacksquare$

∎

Theorem— The medians of a triangle meet at the centroid

The three medians of a triangle $ABC$ (each joining a vertex to the midpoint of the opposite side) pass through the point $g = \tfrac13(a + b + c)$ , which divides each median in the ratio $2 : 1$ .

Proof

The midpoint of $BC$ is $m = \tfrac12(b + c)$ . The point two thirds of the way from $A$ to $m$ is

a + \tfrac23\,(m - a) = a + \tfrac13(b + c) - \tfrac23 a = \tfrac13(a + b + c).

This expression does not change if we permute $a$ , $b$ , $c$ , so the same point is two thirds of the way along each median. $\blacksquare$

∎

Theorem— The diagonals of a rhombus are perpendicular

If $ABCD$ is a parallelogram with all sides equal, its diagonals are perpendicular.

Proof

Let $u = b - a$ and $v = d - a$ , so the diagonals are $\overrightarrow{AC} = u + v$ and $\overrightarrow{DB} = u - v$ . Then $(u + v)\cdot(u - v) = |u|^2 - |v|^2 = 0$ , because $|u| = |v|$ . $\blacksquare$

∎

Compare these with the congruent-triangle proofs in Euclidean geometry and proof: the vector proofs are shorter, but they hide the geometry inside the algebra. The rhombus argument, read backwards, is also a proof of Thales' theorem: if $|u| = |v|$ then $(u + v) \perp (u - v)$ .

Note

Lines in space can be skew. In the plane, two distinct lines either meet in one point or are parallel. In space there is a third possibility: lines such as $\{(t, 0, 0)\}$ and $\{(0, t, 1)\}$ are not parallel (their directions are $(1, 0, 0)$ and $(0, 1, 0)$ ) and never meet (the first has $z = 0$ , the second $z = 1$ ). Such lines are called skew.

Formal mathematicsF

Vectors in Lean

We start in coordinates. Here is the dot product of two vectors in the plane, Lagrange's identity, and the Cauchy–Schwarz inequality with its equality case, exactly as in the lesson:

dot2Lean 4Formally verified

/-- The dot product of two vectors in the plane, given by coordinates:
`(u₁, u₂) · (v₁, v₂) = u₁ v₁ + u₂ v₂`. -/
def dot2 (u v : ℝ × ℝ) : ℝ := u.1 * v.1 + u.2 * v.2