Vectors

A cable car moves along a cable from A to B:

Draw the cable car when it arrives at B.

The cable car's move can be read on the grid: $3$ squares to the right and $3$ squares up. Two numbers are enough to describe the whole displacement — and the arrow from $A$ to $B$ gathers them into a single object: a vector.

Moving a figure by translation means sliding the figure without rotating it. To describe this movement, we use an arrow $\vec{u}$ called a vector: every point of the figure slides along the same direction, in the same sense, by the same length — the ones given by the arrow.

Definition — Vector

A non-zero vector $\vec{u}$ is defined geometrically by a direction, a sense (orientation along that direction) and a magnitude (length).

A vector does not depend on where it is drawn: any two arrows with the same direction, sense and magnitude represent the same vector.

Example

Describe the vector $\vec{w}$ drawn below: give its direction, its sense and its length.

The vector $\vec{w}$ has an oblique direction (the line it sits on), its sense is towards the bottom right, and its length is the length of the arrow — located by its moves on the grid: $4$ squares to the right and $2$ squares down (with Pythagoras' theorem, $\sqrt{4^2 + 2^2} \approx 4.5$ squares). Any arrow with these three features represents the same vector $\vec{w}$.

Definition — Vector Components

A vector $\vec{u}$ can also be defined algebraically, as an ordered pair of components $\begin{pmatrix} \textcolor{colordef}{x-\text{step}}\\\textcolor{colorprop}{y-\text{step}}\end{pmatrix}$: how many units to the right or left, and how many units up or down.

The $x$-step is positive if the movement is to the right and negative if to the left.
The $y$-step is positive if the movement is upwards and negative if downwards.

Example

$\quad\quad\quad$

Skills to practice

Identifying Vectors
Finding the Components of a Vector

I.2 Two-Point Notation

Definition — Two-Point Notation

The notation $\Vect{AB}$ names the vector with origin $A$ and endpoint $B$: the arrow that starts at $A$ and ends at $B$.

Proposition — Components of $\Vect{AB}$

Let $A(x_A,\,y_A)$ and $B(x_B,\,y_B)$ be two points. The components of the vector $\Vect{AB}$ are:$$\Vect{AB} = \begin{pmatrix}x_B - x_A \\ y_B - y_A\end{pmatrix}$$ The components $x_B - x_A$ and $y_B - y_A$ are exactly the $x$-step and $y$-step from $A$ to $B$.

Displacement Vector and Position Vector

Two names borrowed from physics make the two roles of a vector visible.

The vector $\Vect{AB}$ can be called the displacement vector from $A$ to $B$: it records the move from $A$ to $B$ — its coordinates $x_B - x_A$ and $y_B - y_A$ are the horizontal and the vertical steps of that move. Two moves with the same steps, from different starting points, are the same displacement: the same vector.
Once an origin $O$ is chosen, the vector $\Vect{OM}$ can be called the position vector of the point $M$ (relative to $O$). Its coordinates are $x_M - 0 = x_M$ and $y_M - 0 = y_M$: the coordinates of the position vector are the coordinates of the point. A point is thus located by one vector, and the displacement from $A$ to $B$ is read from the two position vectors: the coordinates of $\Vect{AB}$ are those of $\Vect{OB}$ minus those of $\Vect{OA}$.

On the figure, $\Vect{OA}\begin{pmatrix} 2 \\ 1 \end{pmatrix}$ and $\Vect{OB}\begin{pmatrix} 6 \\ 4 \end{pmatrix}$ are the position vectors of $A$ and $B$, and $\Vect{AB}\begin{pmatrix} 6 - 2 \\ 4 - 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \end{pmatrix}$ is the displacement from $A$ to $B$.

Skills to practice

Calculating the Components of a Vector from Two Points

I.3 Equal Vectors

Two vectors are said to be equal if they represent the same translation.

Definition — Geometric Equality of Vectors

Two non-zero vectors are equal if they have:

the same direction (parallel lines),
the same sense (same orientation),
and the same magnitude (same length).

$\textcolor{olive}{\Vect{u}} = \textcolor{olive}{\Vect{v}}$

Definition — Algebraic Equality of Vectors

Two vectors $\begin{pmatrix}x\\y\end{pmatrix}$ and $\begin{pmatrix}x'\\y'\end{pmatrix}$ are equal if and only if $x = x'$ and $y = y'$.

Proposition — Parallelogram criterion

Two vectors $\Vect{AB}$ and $\Vect{DC}$ are equal if and only if $ABCD$ is a parallelogram (possibly flattened).

Skills to practice

Finding a Point from a Vector
Constructing a Parallelogram

II Operations on Vectors

II.1 Addition of Vectors

Suppose we translate the point $A$ to $B$, then we translate the point $B$ to $C$. What single move replaces the two?

Translating the point $A$ directly to $C$ gives the same result. In vector terms, this is written as the sum $\Vect{AB}+\Vect{BC}=\Vect{AC}$, where the $+$ symbol can be read as “followed by”.

Definition — Vector Addition

The sum of two vectors $\vec{u}$ and $\vec{v}$, written $\vec{u} + \vec{v}$, is the vector corresponding to performing the translation by $\vec{u}$ followed by the translation by $\vec{v}$.

The geometric operations on vectors — sum, opposite, difference, multiplication by a real number — can all be carried out directly on the coordinates: each rule below computes, coordinate by coordinate, the vector that the geometric construction would produce.

Definition — Algebraic Vector Addition

$$\begin{pmatrix}x\\ y\end{pmatrix}+\begin{pmatrix}x'\\ y'\end{pmatrix}=\begin{pmatrix}x+x'\\ y +y' \\ \end{pmatrix}$$

Proposition — Chasles' Relation

For any points $A$, $B$, and $C$,$$\Vect{AB} + \Vect{BC} = \Vect{AC}$$ This is Chasles' relation (also called the triangle law).

Definition — Zero Vector

The vector associated with the translation that transforms any point into itself is the zero vector, denoted $\Vect{0}$.$$\Vect{v} + \Vect{0} = \Vect{0} + \Vect{v} = \Vect{v}$$

Proposition — Vector from a Point to Itself

For any point $A$:$$\Vect{AA} = \Vect{0}$$

Proposition — The Sum Does Not Depend on the Order

For all vectors $\vec{u}$ and $\vec{v}$: $$\vec{u} + \vec{v} = \vec{v} + \vec{u}$$ Starting from a point $A$, the path “$\vec{u}$ then $\vec{v}$” and the path “$\vec{v}$ then $\vec{u}$” reach the same point $D$: the two paths close a parallelogram (possibly flattened), and the sum $\Vect{AD}$ is its diagonal from $A$ — the parallelogram rule.

Skills to practice

Recognising Sums of Vectors
Calculating Sums of Vectors

II.2 Subtraction of Vectors

Suppose we translate the point $A$ to $B$, then we translate the point $B$ back to $A$. What single move replaces the two?

No move at all: the point $A$ comes back to itself. In vector terms, $\Vect{AB} + \Vect{BA} = \Vect{AA} = \Vect{0}$, the zero vector: the vector $\Vect{BA}$ undoes the vector $\Vect{AB}$.

Definition — Negative Vector

The negative vector of a non-zero vector $\vec{u}$ is the vector with the same direction, the same length, but the opposite sense (orientation). It is denoted by $-\vec{u}$. The negative of the zero vector is the zero vector.

Definition — Algebraic Negative Vector

$$-\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}-x\\ -y \\ \end{pmatrix}$$

Proposition — Negative of $\Vect{AB}$

The vector $\Vect{BA}$ (from $B$ to $A$) is the negative of the vector $\Vect{AB}$ (from $A$ to $B$):$$\Vect{BA} = -\Vect{AB}$$

Proposition — Sum of a Vector and its Negative

For any vector $\vec{u}$:$$\vec{u} + \left(-\vec{u}\right) = \Vect{0}$$ In particular, for any points $A$ and $B$: $\Vect{AB} + \Vect{BA} = \Vect{AA} = \Vect{0}$ (Chasles' relation).

To subtract one vector from another, we simply add its negative (opposite).

Definition — Subtraction of Vectors

The subtraction of vectors $\Vect{u} - \Vect{v}$ is defined as adding the negative of the second vector:$$\Vect{u}-\Vect{v} = \Vect{u} + (-\Vect{v})$$

Example

Calculate $\Vect{AB}-\Vect{CB}$.

$$\begin{aligned}\Vect{AB}-\Vect{CB} &= \Vect{AB} + (-\Vect{CB})\\ &= \Vect{AB} + \Vect{BC}\\ &= \Vect{AC}\end{aligned}$$

Definition — Algebraic Vector Subtraction

$$\begin{pmatrix}x\\ y\end{pmatrix}-\begin{pmatrix}x'\\ y'\end{pmatrix}=\begin{pmatrix}x-x'\\ y-y' \\ \end{pmatrix}$$

Skills to practice

Recognising the Negative of a Vector
Calculating Differences of Vectors

II.3 Multiplying a Vector by a Scalar

Let $\Vect{v} = \begin{pmatrix} 2\\1\end{pmatrix}$. The notation $2\Vect{v}$ means $\Vect{v}+\Vect{v}$. Find the coordinates of the vectors:

$\textcolor{olive}{2} \Vect{v}$
$\textcolor{olive}{3} \Vect{v}$
$\vdots$
$\textcolor{olive}{k} \Vect{v}$

$\begin{aligned}[t] \textcolor{olive}{2} \Vect{v} &= \Vect{v} + \Vect{v} \\ &= \begin{pmatrix} 2 \\ 1 \end{pmatrix} + \begin{pmatrix} 2 \\ 1 \end{pmatrix} \\ &= \begin{pmatrix} \textcolor{olive}{2} \times 2 \\ \textcolor{olive}{2} \times 1 \end{pmatrix} \\ &= \begin{pmatrix} 4 \\ 2 \end{pmatrix} \end{aligned}$
$\begin{aligned}[t] \textcolor{olive}{3} \Vect{v} &= \Vect{v} + \Vect{v} + \Vect{v} \\ &= \begin{pmatrix} 2 \\ 1 \end{pmatrix} + \begin{pmatrix} 2 \\ 1 \end{pmatrix} + \begin{pmatrix} 2 \\ 1 \end{pmatrix} \\ &= \begin{pmatrix} \textcolor{olive}{3} \times 2 \\ \textcolor{olive}{3} \times 1 \end{pmatrix} \\ &= \begin{pmatrix} 6 \\ 3 \end{pmatrix} \end{aligned}$
$\vdots$
$\begin{aligned}[t] \textcolor{olive}{k} \Vect{v} &= \Vect{v} + \Vect{v} + \dots + \Vect{v} \quad \text{($k$ times)} \\ &= \begin{pmatrix} 2 \\ 1 \end{pmatrix} + \begin{pmatrix} 2 \\ 1 \end{pmatrix} + \dots + \begin{pmatrix} 2 \\ 1 \end{pmatrix} \\ &= \begin{pmatrix} \textcolor{olive}{k} \times 2 \\ \textcolor{olive}{k} \times 1 \end{pmatrix} \end{aligned}$

This can be generalised to the multiplication of a vector by any real number.

Definition — Geometric Scalar Multiplication

Let $k$ be a non-zero real number and $\vec{u}$ a non-zero vector. The vector $k\vec{u}$ is defined by:

its direction: the same as $\vec{u}$;
its sense: the same as $\vec{u}$ if $k > 0$, the opposite if $k < 0$;
its length: $|k|$ times the length of $\vec{u}$.

If $k = 0$ or $\vec{u} = \vec{0}$, then $k\vec{u} = \vec{0}$.

Example

The points $A$, $B$ and $C$ are collinear, in this order, with $AB = 2$~cm and $BC = 4$~cm. Express $\Vect{AC}$ and $\Vect{BA}$ as multiples of $\Vect{AB}$.

$AC = AB + BC = 6$~cm, and $\Vect{AC}$ has the same direction and sense as $\Vect{AB}$: so $\Vect{AC} = 3\Vect{AB}$. $\Vect{BA}$ has the same length as $\Vect{AB}$ but the opposite sense: $\Vect{BA} = -\Vect{AB}$.

Definition — Scalar Multiplication

Let $k$ be a real number. The product $k\Vect{v}$ is defined by:$$k\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}kx\\ ky\end{pmatrix}$$

Remark

$2\Vect{v}$ has the same direction and sense as $\Vect{v}$, and is twice as long.
$-2\Vect{v}$ has the same direction but the opposite sense, and is twice as long as $\Vect{v}$.
If $0 < k < 1$, the vector $k\Vect{v}$ has the same direction and sense as $\Vect{v}$, but is shorter.
$0\Vect{v} = \Vect{0}$, the zero vector.

Example

Let $\Vect{u}\begin{pmatrix} -2 \\ 3 \end{pmatrix}$ and $\Vect{v}\begin{pmatrix} 4 \\ -1 \end{pmatrix}$. Calculate the coordinates of $\Vect{u} + \Vect{v}$, of $-\Vect{v}$, and of $2\Vect{u} - 3\Vect{v}$.

$$\Vect{u} + \Vect{v} = \begin{pmatrix} -2+4 \\ 3+(-1) \end{pmatrix} = \begin{pmatrix} 2 \\ 2 \end{pmatrix}, \qquad -\Vect{v} = \begin{pmatrix} -4 \\ 1 \end{pmatrix},$$ $$2\Vect{u} - 3\Vect{v} = \begin{pmatrix} 2\times(-2) - 3\times 4 \\ 2\times 3 - 3\times(-1) \end{pmatrix} = \begin{pmatrix} -16 \\ 9 \end{pmatrix}.$$ Watch the signs: $2 \times (-2) - 3 \times 4 = -4 - 12 = -16$ and $6 - (-3) = 9$.

Definition — Orthonormal Basis

Let $\Vect{i}$ and $\Vect{j}$ be two vectors of the plane that are perpendicular and of length $1$ (unit vectors). The pair $(\Vect{i},\,\Vect{j})$ is called an orthonormal basis of the plane. Together with a point $O$, the triple $(O,\,\Vect{i},\,\Vect{j})$ is an orthonormal coordinate system.

Proposition — Coordinates in a Basis

Every vector $\Vect{u}$ of the plane can be written in a unique way as $$\Vect{u} = x\,\Vect{i} + y\,\Vect{j},$$ where $x$ and $y$ are real numbers: the coordinates of $\Vect{u}$ in the basis, written $\Vect{u}\begin{pmatrix} x \\ y \end{pmatrix}$.

Example

On the grid, the basis $(\Vect{i},\,\Vect{j})$ and the vector $\Vect{v}$ are drawn. Write $\Vect{v}$ in the form $x\,\Vect{i} + y\,\Vect{j}$ and give its coordinates.

To follow $\Vect{v}$, one moves $3$ squares in the direction of $\Vect{i}$ and $2$ squares in the direction of $\Vect{j}$: $$\Vect{v} = 3\,\Vect{i} + 2\,\Vect{j}, \qquad \Vect{v}\begin{pmatrix} 3 \\ 2 \end{pmatrix}.$$

Skills to practice

Recognising Multiples of a Vector
Calculating Linear Combinations of Vectors

III Midpoints and Magnitudes

III.1 Midpoint of a Segment

Proposition — Vector Characterisation of the Midpoint

Let $A$ and $B$ be two distinct points and $I$ a point of the plane. The following statements are equivalent:

$I$ is the midpoint of the segment $\Segment{AB}$;
$\Vect{AI} = \Vect{IB}$;
$\Vect{AI} = \dfrac{1}{2}\Vect{AB}$;
$\Vect{IA} + \Vect{IB} = \Vect{0}$.

Moreover, when $I$ is the midpoint of $\Segment{AB}$, for every point $M$ of the plane: $$\textcolor{colorprop}{\Vect{MA} + \Vect{MB} = 2\,\Vect{MI}}.$$

(When $A = B$, the segment is the single point $A$ and each of the four statements says $I = A$; we assume $A \neq B$, so that the line $\Line{AB}$ exists.)

(1) $\Leftrightarrow$ (2). If $I$ is the midpoint of $\Segment{AB}$, the vectors $\Vect{AI}$ and $\Vect{IB}$ have the same direction (the line $\Line{AB}$), the same sense (from $A$ towards $B$) and the same length ($AI = IB$): they are equal. Conversely, if $\Vect{AI} = \Vect{IB}$, the two arrows have the same direction, so $I$ lies on the line $\Line{AB}$; the same sense, so $I$ is between $A$ and $B$; and the same length, so $AI = IB$: $I$ is the midpoint.
(2) $\Leftrightarrow$ (3). By Chasles' relation, $\Vect{AB} = \Vect{AI} + \Vect{IB}$. If $\Vect{AI} = \Vect{IB}$, then $\Vect{AB} = 2\,\Vect{AI}$, that is $\Vect{AI} = \dfrac{1}{2}\Vect{AB}$. Conversely, if $\Vect{AI} = \dfrac{1}{2}\Vect{AB}$, then $\Vect{IB} = \Vect{AB} - \Vect{AI} = \Vect{AB} - \dfrac{1}{2}\Vect{AB} = \dfrac{1}{2}\Vect{AB} = \Vect{AI}$.
(2) $\Leftrightarrow$ (4). Since $\Vect{IA} = -\Vect{AI}$, the relation $\Vect{IA} + \Vect{IB} = \Vect{0}$ means $-\Vect{AI} + \Vect{IB} = \Vect{0}$, that is $\Vect{AI} = \Vect{IB}$.
The relation with $M$. Splitting each vector at $I$ by Chasles' relation: $$\Vect{MA} + \Vect{MB} = (\Vect{MI} + \Vect{IA}) + (\Vect{MI} + \Vect{IB}) = 2\,\Vect{MI} + (\Vect{IA} + \Vect{IB}) = 2\,\Vect{MI},$$ since $\Vect{IA} + \Vect{IB} = \Vect{0}$ by (4).

Example

$ABCD$ is a parallelogram, so that $\Vect{AB} = \Vect{DC}$. Let $I$ be the midpoint of the diagonal $\Segment{AC}$. Show, with vectors, that $I$ is also the midpoint of the other diagonal $\Segment{BD}$.

Since $I$ is the midpoint of $\Segment{AC}$, $\Vect{IA} + \Vect{IC} = \Vect{0}$ (characterisation (4)). We compute $\Vect{IB} + \Vect{ID}$ by passing through $A$ and $C$: $$\begin{aligned} \Vect{IB} + \Vect{ID} &= (\Vect{IA} + \Vect{AB}) + (\Vect{IC} + \Vect{CD}) && \text{(Chasles' relation)} \\ &= (\Vect{IA} + \Vect{IC}) + (\Vect{AB} + \Vect{CD}) && \text{(regroup)} \\ &= \Vect{0} + (\Vect{AB} - \Vect{DC}) && \text{($I$ midpoint of $\Segment{AC}$; $\Vect{CD} = -\Vect{DC}$)} \\ &= \Vect{0}. && \text{($\Vect{AB} = \Vect{DC}$)} \end{aligned}$$ By the characterisation (4), $I$ is the midpoint of $\Segment{BD}$: the diagonals of a parallelogram have the same midpoint.

Proposition — Coordinates of the Midpoint

Let $A(x_A,\,y_A)$ and $B(x_B,\,y_B)$ be two points. The midpoint $M$ of the segment $\Segment{AB}$ has coordinates $$M\left(\dfrac{x_A+x_B}{2},\ \dfrac{y_A+y_B}{2}\right).$$

$M$ is the midpoint of $\Segment{AB}$ exactly when $\Vect{AM} = \dfrac{1}{2}\Vect{AB}$. Reading the first coordinates: $$x_M - x_A = \dfrac{1}{2}(x_B - x_A) \quad\text{so}\quad x_M = x_A + \dfrac{x_B - x_A}{2} = \dfrac{x_A+x_B}{2},$$ and the same computation on the second coordinates gives $y_M = \dfrac{y_A+y_B}{2}$.

Example

Let $A(1,\ 5)$ and $B(7,\ 1)$. Calculate the coordinates of the midpoint $M$ of $\Segment{AB}$.

$$M\left(\dfrac{1+7}{2},\ \dfrac{5+1}{2}\right) = M(4,\ 3).$$

Skills to practice

Justifying with Midpoints and Vectors

III.2 Magnitude of a Vector

By Pythagoras' theorem, the magnitude is the length of the hypotenuse of the right-angled triangle formed by the two components:

Proposition — Magnitude of a Vector

If $\Vect{v}=\begin{pmatrix}x\\y\end{pmatrix}$ in an orthonormal basis, the magnitude (length) of $\Vect{v}$, written $|\Vect{v}|$ (also written $\|\Vect{v}\|$), is $$|\Vect{v}|=\sqrt{x^2+y^2}$$

Example

Calculate the magnitude of $\Vect{v}=\begin{pmatrix}4\\3\end{pmatrix}$.

$$\begin{aligned}[t]|\Vect{v}| & =\sqrt{4^2+3^2}\\ & =\sqrt{16+9}\\ & =\sqrt{25}\\ &=5 \text{ units}\\ \end{aligned}$$

Proposition — Distance between Two Points

The distance between points $A(x_A, y_A)$ and $B(x_B, y_B)$ in an orthonormal coordinate system is the magnitude of the vector $\Vect{AB}$:$$AB = |\Vect{AB}| = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}$$

Example

Calculate the distance between the points $A(1,\ 2)$ and $B(5,\ 5)$.

$$\begin{aligned}[t]AB &= |\Vect{AB}|\\ &= \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} \\ &= \sqrt{(5 - 1)^2 + (5 - 2)^2} \\ &= \sqrt{4^2 + 3^2} \\ &= \sqrt{16 + 9} \\ &= \sqrt{25} \\ &= 5\ \text{units}\end{aligned}$$

Skills to practice

Calculating the Magnitude of a Vector

IV Collinearity

IV.1 Collinear Vectors

Definition — Collinear Vectors

Two non-zero vectors $\Vect{u}$ and $\Vect{v}$ are said to be collinear if there exists a real number $k$ such that:$$\Vect{u} = k \Vect{v}$$ By convention, the zero vector is collinear with every vector.

Proposition — Collinearity and Geometry

Two lines $\Line{AB}$ and $\Line{CD}$ are parallel if and only if the vectors $\Vect{AB}$ and $\Vect{CD}$ are collinear.
Three points $A$, $B$, and $C$ are collinear (lie on the same straight line) if and only if the vectors $\Vect{AB}$ and $\Vect{AC}$ are collinear.

Example

On the grid, the vectors $\Vect{AB}$ and $\Vect{CD}$ are drawn. Are the lines $\Line{AB}$ and $\Line{CD}$ parallel?

$\Vect{AB}$ goes $4$ right and $2$ up; $\Vect{CD}$ goes $2$ left and $1$ down: same direction, half as long, opposite sense — $\Vect{CD} = -\dfrac{1}{2}\Vect{AB}$. The vectors are collinear, so the lines $\Line{AB}$ and $\Line{CD}$ are parallel.

Definition — Determinant of Two Vectors

Let $\Vect{u} = \begin{pmatrix} x \\ y \end{pmatrix}$ and $\Vect{v} = \begin{pmatrix} x' \\ y' \end{pmatrix}$ be two vectors in the plane. The determinant of $\Vect{u}$ and $\Vect{v}$, denoted $\det(\Vect{u},\,\Vect{v})$, is defined as:$$\det(\Vect{u},\,\Vect{v}) =\begin{vmatrix}x & x' \\ y & y'\end{vmatrix}= x \cdot y' - y \cdot x'$$

Proposition — Collinearity and Determinant

Two vectors $\Vect{u}$ and $\Vect{v}$ are collinear if and only if$$\det(\Vect{u},\,\Vect{v}) = 0$$

Let $\Vect{u}\begin{pmatrix} x \\ y \end{pmatrix}$ and $\Vect{v}\begin{pmatrix} x' \\ y' \end{pmatrix}$. If $\Vect{u} = \Vect{0}$, both statements are true: the zero vector is collinear with every vector, and $\det(\Vect{u},\Vect{v}) = 0 \cdot y' - 0 \cdot x' = 0$. Suppose now $\Vect{u} \neq \Vect{0}$.

If the vectors are collinear, then $\Vect{v} = k\,\Vect{u}$ for some real $k$, so $x' = kx$ and $y' = ky$, and $$\det(\Vect{u},\,\Vect{v}) = x \cdot ky - y \cdot kx = k(xy - yx) = 0.$$
Conversely, suppose $\det(\Vect{u},\,\Vect{v}) = x\,y' - y\,x' = 0$. Since $\Vect{u} \neq \Vect{0}$, one of its coordinates is not zero.
- If $x \neq 0$: set $k = \dfrac{x'}{x}$. Then $x' = kx$, and $x\,y' = y\,x' = ykx$ gives $y' = ky$: so $\Vect{v} = k\,\Vect{u}$.
- If $x = 0$, then $y \neq 0$ and $0 = x\,y' - y\,x' = -y\,x'$ gives $x' = 0$: set $k = \dfrac{y'}{y}$, then $\Vect{v} = k\,\Vect{u}$.
In both cases the vectors are collinear.

Example

Let $A(1,\ 2)$, $B(5,\ 4)$, $C(-1,\ -1)$, and $D(5,\ 2)$.
Are the lines $\Line{AB}$ and $\Line{CD}$ parallel?

First, compute the vectors:$$\Vect{AB} = \begin{pmatrix}5-1 \\ 4-2\end{pmatrix} = \begin{pmatrix}4 \\ 2\end{pmatrix}, \quad\Vect{CD} = \begin{pmatrix}5-(-1) \\ 2-(-1)\end{pmatrix} = \begin{pmatrix}6 \\ 3\end{pmatrix}$$ Calculate the determinant:$$\det(\Vect{AB},\,\Vect{CD}) =\begin{vmatrix}4 & 6 \\ 2 & 3\end{vmatrix}= 4\times 3 - 6\times 2 = 12 - 12 = 0$$ The determinant is zero: by the criterion $\det = 0$, the vectors are collinear, so the lines are parallel. (Double-check: $\frac{6}{4} = \frac{3}{2} = 1.5$, indeed $\Vect{CD} = 1.5\,\Vect{AB}$.)

Proposition — Collinearity and Proportional Coordinates

Let $\Vect{u}\begin{pmatrix} x \\ y \end{pmatrix}$ and $\Vect{v}\begin{pmatrix} x' \\ y' \end{pmatrix}$ be two non-zero vectors. The following statements are equivalent:

$\Vect{u}$ and $\Vect{v}$ are collinear;
the coordinates of $\Vect{v}$ are proportional to those of $\Vect{u}$: there is a real number $k$ such that $x' = kx$ and $y' = ky$;
the cross products are equal: $x\,y' = y\,x'$.

In a proportionality table whose rows are the two vectors, the row of $\Vect{v}$ is $k$ times the row of $\Vect{u}$ — $k$ is the coefficient of the table — and the equality of the cross products is the rule of that table. It is exactly $\det(\Vect{u},\,\Vect{v}) = x\,y' - y\,x' = 0$.

	first coordinate	second coordinate
$\Vect{u}$	$x$	$y$
$\Vect{v}$	$x' = kx$	$y' = ky$

(1) $\Rightarrow$ (2). If $\Vect{u}$ and $\Vect{v}$ are collinear and non-zero, then by definition there is a real number $k'$ such that $\Vect{u} = k'\,\Vect{v}$; since $\Vect{u} \neq \Vect{0}$, $k' \neq 0$, and $\Vect{v} = \dfrac{1}{k'}\Vect{u} = k\,\Vect{u}$ with $k = \dfrac{1}{k'}$. The coordinates of $k\,\Vect{u}$ are $kx$ and $ky$, so $x' = kx$ and $y' = ky$.
(2) $\Rightarrow$ (1). If $x' = kx$ and $y' = ky$, then $\Vect{v}$ and $k\,\Vect{u}$ have the same coordinates: $\Vect{v} = k\,\Vect{u}$. Since $\Vect{v} \neq \Vect{0}$, $k \neq 0$ and $\Vect{u} = \dfrac{1}{k}\Vect{v}$: the vectors are collinear.
(2) $\Rightarrow$ (3). If $x' = kx$ and $y' = ky$, then $x\,y' = x \cdot ky = y \cdot kx = y\,x'$: the cross products are equal.
(3) $\Rightarrow$ (2). Suppose $x\,y' = y\,x'$. Since $\Vect{u} \neq \Vect{0}$, one of its coordinates is not zero.
- If $x \neq 0$, set $k = \dfrac{x'}{x}$: then $x' = kx$, and $y' = \dfrac{x\,y'}{x} = \dfrac{y\,x'}{x} = ky$.
- If $x = 0$, then $y \neq 0$, and the equality gives $y\,x' = 0$, so $x' = 0$; set $k = \dfrac{y'}{y}$: then $x' = 0 = kx$ and $y' = ky$.
In both cases the coordinates are proportional. (The equality $x\,y' = y\,x'$ is the criterion $\det(\Vect{u},\,\Vect{v}) = 0$ of the Proposition “Collinearity and Determinant”, read on the cross products.)

Example

In each case, decide whether the two vectors are collinear by comparing the cross products, and give the coefficient $k$ such that $\Vect{v} = k\,\Vect{u}$ when they are.

$\Vect{u}\begin{pmatrix} 6 \\ -4 \end{pmatrix}$ and $\Vect{v}\begin{pmatrix} -9 \\ 6 \end{pmatrix}$;
$\Vect{u}\begin{pmatrix} 0 \\ 3 \end{pmatrix}$ and $\Vect{v}\begin{pmatrix} 0 \\ -5 \end{pmatrix}$;
$\Vect{u}\begin{pmatrix} 2 \\ 5 \end{pmatrix}$ and $\Vect{v}\begin{pmatrix} 4 \\ 9 \end{pmatrix}$.

Cross products: $x\,y' = 6 \times 6 = 36$ and $y\,x' = (-4) \times (-9) = 36$. They are equal: the vectors are collinear. The coefficient is $k = \dfrac{x'}{x} = \dfrac{-9}{6} = -\dfrac{3}{2}$, and indeed $-\dfrac{3}{2} \times (-4) = 6 = y'$: $\Vect{v} = -\dfrac{3}{2}\Vect{u}$.
Cross products: $0 \times (-5) = 0$ and $3 \times 0 = 0$: equal, the vectors are collinear (both are vertical). Here $k$ cannot be read on the first coordinates ($\dfrac{0}{0}$ is not a number): it is read on the second ones, $k = \dfrac{-5}{3}$, and $\Vect{v} = -\dfrac{5}{3}\Vect{u}$. The cross products decide even when a coordinate is zero.
Cross products: $2 \times 9 = 18$ and $5 \times 4 = 20$. They differ: the vectors are not collinear. (Equivalently, $\det(\Vect{u},\,\Vect{v}) = 18 - 20 = -2 \neq 0$.)

Proposition — Collinear Points and Parallel Lines

Let $A$, $B$, $C$, $D$ be points, with $A \neq B$ and $C \neq D$.

The points $A$, $B$, $C$ are collinear if and only if $\det(\Vect{AB},\,\Vect{AC}) = 0$.
The lines $\Line{AB}$ and $\Line{CD}$ are parallel if and only if $\det(\Vect{AB},\,\Vect{CD}) = 0$.

Proposition — Decomposition on Two Non-Collinear Vectors

Let $\Vect{u}$ and $\Vect{v}$ be two non-collinear vectors of the plane (neither is the zero vector, and they do not have the same direction). Every vector $\Vect{w}$ of the plane can be written in a unique way as $$\textcolor{colorprop}{\Vect{w} = a\,\Vect{u} + b\,\Vect{v}}, \qquad a \text{ and } b \text{ real numbers.}$$ The pair $(\Vect{u},\,\Vect{v})$ is a basis of the plane and $(a, b)$ are the coordinates of $\Vect{w}$ in that basis. The orthonormal basis $(\Vect{i},\,\Vect{j})$ and the writing $\Vect{w} = x\,\Vect{i} + y\,\Vect{j}$ are the special case where the two vectors are perpendicular and of length $1$.

Existence. Draw representatives from a common point $O$: $\Vect{u} = \Vect{OA}$, $\Vect{v} = \Vect{OB}$ and $\Vect{w} = \Vect{OM}$. The lines $\Line{OA}$ and $\Line{OB}$ intersect at $O$, since $\Vect{u}$ and $\Vect{v}$ do not have the same direction. If $M$ is on the line $\Line{OA}$, then $\Vect{OM}$ is collinear with $\Vect{OA} \neq \Vect{0}$, so $\Vect{OM} = a\,\Vect{u}$ for some real $a$, and $\Vect{w} = a\,\Vect{u} + 0\,\Vect{v}$; likewise, if $M$ is on $\Line{OB}$, then $\Vect{w} = 0\,\Vect{u} + b\,\Vect{v}$ (and $\Vect{w} = 0\,\Vect{u} + 0\,\Vect{v}$ when $M = O$). Otherwise, the parallel to $\Line{OB}$ through $M$ meets $\Line{OA}$ at a point $P$, and the parallel to $\Line{OA}$ through $M$ meets $\Line{OB}$ at a point $Q$: $OPMQ$ is a parallelogram, so $\Vect{PM} = \Vect{OQ}$ (parallelogram criterion), and $\Vect{OM} = \Vect{OP} + \Vect{PM} = \Vect{OP} + \Vect{OQ}$ (Chasles' relation). Now $P$ is on the line $\Line{OA}$, so $\Vect{OP} = a\,\Vect{u}$ for some real $a$; likewise $\Vect{OQ} = b\,\Vect{v}$. Hence $\Vect{w} = a\,\Vect{u} + b\,\Vect{v}$.
Uniqueness. Suppose $\Vect{w} = a\,\Vect{u} + b\,\Vect{v} = a'\,\Vect{u} + b'\,\Vect{v}$. Then $(a - a')\,\Vect{u} = (b' - b)\,\Vect{v}$. If $a \neq a'$, dividing by $a - a'$ gives $\Vect{u} = \dfrac{b' - b}{a - a'}\,\Vect{v}$: the vectors $\Vect{u}$ and $\Vect{v}$ would be collinear, which is excluded. So $a = a'$, and then $(b' - b)\,\Vect{v} = \Vect{0}$ with $\Vect{v} \neq \Vect{0}$ forces $b' = b$.

Method — Expressing a Vector in Terms of Two Non-Collinear Vectors

In a figure, two non-collinear vectors — typically two sides of a triangle or of a parallelogram from the same vertex — serve as a basis: every vector of the figure is a combination of them, and comparing two such combinations proves that points are collinear or that lines are parallel.

Choose the two reference vectors, for instance $\Vect{AB}$ and $\Vect{AD}$ from the vertex $A$, and check that they are not collinear.
Walk from the origin to the endpoint of the wanted vector through known points, with Chasles' relation: $\Vect{AI} = \Vect{AB} + \Vect{BI}$.
Replace each step by its expression: a side of the parallelogram ($\Vect{BC} = \Vect{AD}$), a midpoint ($\Vect{BI} = \dfrac{1}{2}\Vect{BC}$), a given multiple ($\Vect{AE} = \dfrac{2}{3}\Vect{AB}$), an opposite ($\Vect{DA} = -\Vect{AD}$).
Gather the terms in $\Vect{AB}$ and in $\Vect{AD}$: the result is $a\,\Vect{AB} + b\,\Vect{AD}$, and it is unique.

Example

On the grid, the vectors $\Vect{u}$, $\Vect{v}$ and $\Vect{w}$ are drawn. Write $\Vect{w}$ as a combination of $\Vect{u}$ and $\Vect{v}$, then check with the coordinates.

Reading the grid: $\Vect{u}\begin{pmatrix} 2 \\ 1 \end{pmatrix}$, $\Vect{v}\begin{pmatrix} -1 \\ 2 \end{pmatrix}$ and $\Vect{w}\begin{pmatrix} 4 \\ 7 \end{pmatrix}$; $\Vect{u}$ and $\Vect{v}$ are not collinear ($2 \times 2 \neq 1 \times (-1)$). From the origin of $\Vect{w}$, walking $3$ times $\Vect{u}$ leads $6$ right and $3$ up; the remaining move to the endpoint of $\Vect{w}$ is $2$ left and $4$ up, which is $2\,\Vect{v}$: $$\Vect{w} = 3\,\Vect{u} + 2\,\Vect{v}.$$

Check with the coordinates: $3\,\Vect{u} + 2\,\Vect{v} = \begin{pmatrix} 3 \times 2 + 2 \times (-1) \\ 3 \times 1 + 2 \times 2 \end{pmatrix} = \begin{pmatrix} 6 - 2 \\ 3 + 4 \end{pmatrix} = \begin{pmatrix} 4 \\ 7 \end{pmatrix} = \Vect{w}$. The coordinates of $\Vect{w}$ in the basis $(\Vect{u},\,\Vect{v})$ are $(3, 2)$.

Example

$ABCD$ is a parallelogram and $I$ is the midpoint of $\Segment{BC}$. Express the vectors $\Vect{AI}$ and $\Vect{DI}$ in terms of $\Vect{AB}$ and $\Vect{AD}$.