Vectors: notation, arithmetic and geometric proof

A vector describes a movement with both a magnitude and a direction. For example, moving $3$ units right and $2$ units up is represented by

$$\begin{pmatrix}3\\2\end{pmatrix}.$$

Vectors turn geometric journeys into algebra. They let you calculate displacements, divide line segments in a given ratio, recognise parallel lines and prove geometric facts without relying on an accurate diagram.

Before you begin

You should be able to:

• calculate with negative numbers and fractions;
• simplify algebraic expressions;
• use coordinates and Pythagoras’ theorem;
• work with ratios;
• distinguish a point from a movement between points.

Review algebraic manipulation, coordinate geometry, ratio and proportion or Pythagoras and trigonometry if needed.

Vectors and scalars

A scalar has magnitude only. Time, mass, temperature and distance are scalars. A vector has magnitude and direction. Displacement, velocity, acceleration and force are vectors.

The distinction between distance and displacement is important. A person who walks $5$ m east and then $5$ m west has travelled a distance of $10$ m, but their overall displacement is the zero vector because they finish where they started.

A vector can be written in several ways:

$$\mathbf a, \qquad \overrightarrow{AB}, \qquad \begin{pmatrix}x\\y\end{pmatrix}.$$

• $\mathbf a$ names a vector without specifying a starting point.
• $\overrightarrow{AB}$ means the movement from $A$ to $B$.
• $\begin{pmatrix}x\\y\end{pmatrix}$ is a column vector: $x$ is the horizontal change and $y$ is the vertical change.

Thus

$$\begin{pmatrix}-4\\3\end{pmatrix}$$

means $4$ units left and $3$ units up. The entries are changes, not coordinates of a point.

Equal vectors

Two vectors are equal when they have the same magnitude and direction. Their starting positions do not matter. If several arrows all move $2$ units right and $5$ units down, each represents

$$\begin{pmatrix}2\\-5\end{pmatrix}.$$

This is the conceptual leap behind vector geometry: a vector records the journey, not where the journey begins.

The zero vector and opposite vectors

The zero vector is

$$\mathbf 0=\begin{pmatrix}0\\0\end{pmatrix}.$$

It has zero magnitude and no particular direction. The negative of $\mathbf a$ has the same magnitude as $\mathbf a$ but the opposite direction:

$$\text{if }\mathbf a=\begin{pmatrix}4\\-1\end{pmatrix}, \quad -\mathbf a=\begin{pmatrix}-4\\1\end{pmatrix}.$$

In directed line notation,

$$\overrightarrow{BA}=-\overrightarrow{AB}.$$

Direction is carried by the order

$\overrightarrow{AB}$ goes from $A$ to $B$. It is not interchangeable with $\overrightarrow{BA}$. Reversing the letters reverses the vector.

Finding a vector from coordinates

For points $A(x_1,y_1)$ and $B(x_2,y_2)$, subtract the coordinates of the starting point from those of the finishing point:

$$\boxed{\overrightarrow{AB} =\begin{pmatrix}x_2-x_1\\y_2-y_1\end{pmatrix}} =B-A.$$

The phrase finish minus start is a reliable way to remember the order.

Worked example 1: displacement between two points

Let $A(-3,4)$ and $B(5,-2)$. Find $\overrightarrow{AB}$ and $\overrightarrow{BA}$.

From $A$ to $B$,

$$\overrightarrow{AB} =\begin{pmatrix}5-(-3)\\-2-4\end{pmatrix} =\begin{pmatrix}8\\-6\end{pmatrix}.$$

This agrees with the diagrammatic meaning: move $8$ units right and $6$ units down. Reversing the journey gives

$$\overrightarrow{BA} =\begin{pmatrix}-3-5\\4-(-2)\end{pmatrix} =\begin{pmatrix}-8\\6\end{pmatrix} =-\overrightarrow{AB}.$$

Self-check 1

Given $P(7,-4)$ and $Q(-2,8)$, find $\overrightarrow{PQ}$.

Answer

$$\overrightarrow{PQ} =\begin{pmatrix}-2-7\\8-(-4)\end{pmatrix} =\begin{pmatrix}-9\\12\end{pmatrix}.$$

Adding and subtracting vectors

Vectors add component by component:

$$\begin{pmatrix}a\\b\end{pmatrix} +\begin{pmatrix}c\\d\end{pmatrix} =\begin{pmatrix}a+c\\b+d\end{pmatrix}.$$

Geometrically, addition means putting journeys head to tail. A journey from $A$ to $B$, followed by one from $B$ to $C$, has the same overall effect as going directly from $A$ to $C$:

$$\boxed{\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}}.$$

This is often called the triangle law of vector addition.

Worked example 2: combine two movements

Let

$$\mathbf a=\begin{pmatrix}3\\-4\end{pmatrix}, \qquad \mathbf b=\begin{pmatrix}-5\\7\end{pmatrix}.$$

Then

$$\mathbf a+\mathbf b =\begin{pmatrix}3+(-5)\\-4+7\end{pmatrix} =\begin{pmatrix}-2\\3\end{pmatrix}.$$

The rightward movement of $3$ and leftward movement of $5$ combine to give $2$ left. The vertical changes combine to give $3$ up.

Subtraction means adding the opposite vector:

$$\mathbf a-\mathbf b=\mathbf a+(-\mathbf b).$$

Therefore

$$\begin{aligned} \mathbf a-\mathbf b &=\begin{pmatrix}3\\-4\end{pmatrix} -\begin{pmatrix}-5\\7\end{pmatrix}\\ &=\begin{pmatrix}3-(-5)\\-4-7\end{pmatrix} =\begin{pmatrix}8\\-11\end{pmatrix}. \end{aligned}$$

Common misconception: subtracting only one component

The subtraction sign applies to the whole vector. In

$$\begin{pmatrix}2\\3\end{pmatrix} -\begin{pmatrix}-4\\5\end{pmatrix},$$

both components are subtracted, giving

$$\begin{pmatrix}2-(-4)\\3-5\end{pmatrix} =\begin{pmatrix}6\\-2\end{pmatrix}.$$

Multiplying by a scalar

Multiplying a vector by a scalar multiplies every component:

$$k\begin{pmatrix}a\\b\end{pmatrix} =\begin{pmatrix}ka\\kb\end{pmatrix}.$$

• If $k>1$, the vector keeps its direction and becomes longer.
• If $0<k<1$, it keeps its direction and becomes shorter.
• If $k<0$, its direction reverses as well as its length changing.
• If $k=0$, the result is $\mathbf 0$.

For example, if $\mathbf p=\begin{pmatrix}-2\\5\end{pmatrix}$, then

$$3\mathbf p=\begin{pmatrix}-6\\15\end{pmatrix}, \qquad -\frac12\mathbf p=\begin{pmatrix}1\\-\frac52\end{pmatrix}.$$

Worked example 3: simplify a vector expression

Given

$$\mathbf a=\begin{pmatrix}4\\-3\end{pmatrix}, \qquad \mathbf b=\begin{pmatrix}-2\\5\end{pmatrix},$$

find $2\mathbf a-3\mathbf b$.

Calculate the scalar multiples first:

$$2\mathbf a=\begin{pmatrix}8\\-6\end{pmatrix}, \qquad 3\mathbf b=\begin{pmatrix}-6\\15\end{pmatrix}.$$

Then subtract corresponding components:

$$2\mathbf a-3\mathbf b =\begin{pmatrix}8-(-6)\\-6-15\end{pmatrix} =\boxed{\begin{pmatrix}14\\-21\end{pmatrix}}.$$

A useful check is to simplify symbolically first: subtracting $3\mathbf b$, whose horizontal component is negative, should increase the horizontal result.

Self-check 2

Let $\mathbf u=\begin{pmatrix}6\\-2\end{pmatrix}$ and $\mathbf v=\begin{pmatrix}1\\4\end{pmatrix}$. Find $\frac12\mathbf u+2\mathbf v$.

Answer

$$\frac12\mathbf u+2\mathbf v =\begin{pmatrix}3\\-1\end{pmatrix} +\begin{pmatrix}2\\8\end{pmatrix} =\begin{pmatrix}5\\7\end{pmatrix}.$$

Magnitude of a vector

The magnitude of $\mathbf a$ is its length and is written $|\mathbf a|$. If

$$\mathbf a=\begin{pmatrix}x\\y\end{pmatrix},$$

then Pythagoras’ theorem gives

$$\boxed{|\mathbf a|=\sqrt{x^2+y^2}}.$$

The magnitude is a scalar and is never negative.

Worked example 4: calculate an exact magnitude

For

$$\mathbf a=\begin{pmatrix}-6\\8\end{pmatrix},$$ $$|\mathbf a| =\sqrt{(-6)^2+8^2} =\sqrt{36+64} =10.$$

The negative horizontal component affects direction, but its square contributes positively to length.

For $\mathbf b=\begin{pmatrix}4\\7\end{pmatrix}$,

$$|\mathbf b|=\sqrt{4^2+7^2}=\sqrt{65}.$$

Keep the exact surd unless a question requests a decimal approximation.

Common misconception: magnitude is not a component sum

The magnitude of $\begin{pmatrix}3\\4\end{pmatrix}$ is not $3+4=7$. The components are perpendicular movements, so Pythagoras gives

$$\sqrt{3^2+4^2}=5.$$

Parallel vectors

Two non-zero vectors are parallel precisely when one is a scalar multiple of the other. Thus $\mathbf a$ and $\mathbf b$ are parallel if

$$\mathbf a=k\mathbf b$$

for some scalar $k$.

If $k>0$, they point in the same direction. If $k<0$, they point in opposite directions. Parallel does not mean same direction only.

Worked example 5: test for parallel vectors

Determine whether

$$\mathbf p=\begin{pmatrix}6\\-9\end{pmatrix} \quad\text{and}\quad \mathbf q=\begin{pmatrix}-4\\6\end{pmatrix}$$

are parallel.

Both components have the same multiplier:

$$-\frac32\mathbf q =-\frac32\begin{pmatrix}-4\\6\end{pmatrix} =\begin{pmatrix}6\\-9\end{pmatrix} =\mathbf p.$$

Therefore the vectors are parallel and point in opposite directions.

By contrast, $\begin{pmatrix}2\\3\end{pmatrix}$ and $\begin{pmatrix}4\\5\end{pmatrix}$ are not parallel. Doubling the first horizontal component gives $4$, but doubling its vertical component gives $6$, not $5$. A single scalar must work for every component.

Worked example 6: find an unknown using parallel vectors

The vectors

$$\begin{pmatrix}3\\12\end{pmatrix} \quad\text{and}\quad \begin{pmatrix}x\\-8\end{pmatrix}$$

are parallel. Find $x$.

The multiplier taking $12$ to $-8$ is

$$k=\frac{-8}{12}=-\frac23.$$

Apply the same multiplier to the horizontal component:

$$x=-\frac23(3)=-2.$$

Hence the second vector is $-\frac23$ times the first.

Position vectors and routes through a diagram

Fix an origin $O$. The vector $\overrightarrow{OA}$ is the position vector of $A$. If

$$\overrightarrow{OA}=\mathbf a, \qquad \overrightarrow{OB}=\mathbf b,$$

then the route $A\to O\to B$ gives

$$\overrightarrow{AB} =\overrightarrow{AO}+\overrightarrow{OB} =-\mathbf a+\mathbf b =\boxed{\mathbf b-\mathbf a}.$$

This is the vector version of finish minus start.

Worked example 7: express vectors in terms of $\mathbf a$ and $\mathbf b$

Suppose $\overrightarrow{OA}=\mathbf a$ and $\overrightarrow{OB}=\mathbf b$. Point $M$ is the midpoint of $AB$. Find $\overrightarrow{AB}$ and $\overrightarrow{OM}$.

First,

$$\overrightarrow{AB}=\mathbf b-\mathbf a.$$

Since $M$ is halfway from $A$ to $B$,

$$\overrightarrow{AM}=\frac12(\mathbf b-\mathbf a).$$

Travel from $O$ to $A$, then from $A$ to $M$:

$$\begin{aligned} \overrightarrow{OM} &=\overrightarrow{OA}+\overrightarrow{AM}\\ &=\mathbf a+\frac12(\mathbf b-\mathbf a)\\ &=\frac12\mathbf a+\frac12\mathbf b\\ &=\boxed{\frac{\mathbf a+\mathbf b}{2}}. \end{aligned}$$

This is the vector form of averaging the endpoints’ coordinates.

Dividing a line in a ratio

If $P$ lies on $AB$ and

$$AP:PB=m:n,$$

then the whole line is split into $m+n$ equal parts. Consequently,

$$\overrightarrow{AP}=\frac{m}{m+n}\overrightarrow{AB}.$$

If the position vectors of $A$ and $B$ are $\mathbf a$ and $\mathbf b$, then

$$\begin{aligned} \overrightarrow{OP} &=\mathbf a+\frac{m}{m+n}(\mathbf b-\mathbf a)\\ &=\boxed{\frac{n\mathbf a+m\mathbf b}{m+n}}. \end{aligned}$$

Notice the crossed coefficients: the point closer to $A$ has more of $\mathbf a$ in its position vector.

Worked example 8: use a ratio on a line segment

The position vectors of $A$ and $B$ are $\mathbf a$ and $\mathbf b$. Point $P$ divides $AB$ so that $AP:PB=2:3$. Find $\overrightarrow{OP}$.

Since $AB$ contains $2+3=5$ parts,

$$\overrightarrow{AP}=\frac25\overrightarrow{AB} =\frac25(\mathbf b-\mathbf a).$$

Therefore

$$\begin{aligned} \overrightarrow{OP} &=\mathbf a+\frac25(\mathbf b-\mathbf a)\\ &=\mathbf a+\frac25\mathbf b-\frac25\mathbf a\\ &=\boxed{\frac35\mathbf a+\frac25\mathbf b}. \end{aligned}$$

The answer is weighted more towards $\mathbf a$ because $P$ is closer to $A$ than to $B$.

Self-check 3

Points $A$ and $B$ have position vectors $\mathbf a$ and $\mathbf b$. Point $Q$ satisfies $AQ:QB=3:1$. Find $\overrightarrow{OQ}$.

Answer

The point is three quarters of the way from $A$ to $B$:

$$\begin{aligned} \overrightarrow{OQ} &=\mathbf a+\frac34(\mathbf b-\mathbf a)\\ &=\frac14\mathbf a+\frac34\mathbf b. \end{aligned}$$

Vector proofs in geometry

A vector proof should do three things:

1. choose a route and express the required vectors using the given information;
2. simplify the expressions clearly;
3. state the geometric conclusion justified by the algebra.

Useful conclusions include:

• equal vectors have equal magnitude and direction;
• a scalar multiple proves parallelism;
• the same midpoint proves that diagonals bisect each other;
• equal position vectors identify the same point.

Worked example 9: prove a quadrilateral is a parallelogram

In quadrilateral $ABCD$,

$$\overrightarrow{AB}=\mathbf p, \qquad \overrightarrow{AD}=\mathbf q, \qquad \overrightarrow{BC}=\mathbf q.$$

Prove that $ABCD$ is a parallelogram.

Take the route $D\to A\to B\to C$:

$$\begin{aligned} \overrightarrow{DC} &=\overrightarrow{DA}+\overrightarrow{AB}+\overrightarrow{BC}\\ &=-\mathbf q+\mathbf p+\mathbf q\\ &=\mathbf p. \end{aligned}$$

But $\overrightarrow{AB}=\mathbf p$, so

$$\overrightarrow{DC}=\overrightarrow{AB}.$$

Therefore $AB$ and $DC$ are equal in length and parallel in the same direction. Hence $ABCD$ is a parallelogram.

The final sentence matters. Algebra alone is not yet a geometric proof until its implication is stated.

Worked example 10: prove points are collinear

Relative to origin $O$, points $A$, $B$ and $C$ have position vectors

$$\overrightarrow{OA}=\mathbf a, \qquad \overrightarrow{OB}=\frac13\mathbf a+\frac23\mathbf c, \qquad \overrightarrow{OC}=\mathbf c.$$

Prove that $A$, $B$ and $C$ are collinear and find $AB:BC$.

First find the vector from $A$ to $B$:

$$\begin{aligned} \overrightarrow{AB} &=\overrightarrow{OB}-\overrightarrow{OA}\\ &=\left(\frac13\mathbf a+\frac23\mathbf c\right)-\mathbf a\\ &=\frac23(\mathbf c-\mathbf a). \end{aligned}$$

Also,

$$\overrightarrow{AC}=\mathbf c-\mathbf a.$$

Thus

$$\overrightarrow{AB}=\frac23\overrightarrow{AC}.$$

Since $\overrightarrow{AB}$ is a positive scalar multiple of $\overrightarrow{AC}$, the two vectors are parallel and point in the same direction from $A$. Therefore $A$, $B$ and $C$ are collinear, with $B$ between $A$ and $C$.

Moreover, $AB$ is $\frac23$ of $AC$, leaving $BC$ as $\frac13$ of $AC$. Hence

$$\boxed{AB:BC=2:1}.$$

Parallel vectors alone may not prove collinearity

Showing that two vectors are parallel is not enough if they lie on separate parallel lines. To prove $A$, $B$ and $C$ are collinear, compare vectors that share a point, such as $\overrightarrow{AB}$ and $\overrightarrow{AC}$.

Mixed self-check

Let $\mathbf a=\begin{pmatrix}2\\-1\end{pmatrix}$ and $\mathbf b=\begin{pmatrix}5\\3\end{pmatrix}$.

1. Find $3\mathbf a-2\mathbf b$.
2. Find $|\mathbf b-\mathbf a|$.
3. Points $A$ and $B$ have position vectors $\mathbf a$ and $\mathbf b$. Find the position vector of the midpoint $M$ of $AB$.
4. Point $N$ lies on $AB$ with $AN:NB=1:2$. Find the position vector of $N$.

Answers

1. $$3\mathbf a-2\mathbf b =\begin{pmatrix}6\\-3\end{pmatrix} -\begin{pmatrix}10\\6\end{pmatrix} =\begin{pmatrix}-4\\-9\end{pmatrix}.$$
2. $$\mathbf b-\mathbf a=\begin{pmatrix}3\\4\end{pmatrix}, \qquad |\mathbf b-\mathbf a|=\sqrt{3^2+4^2}=5.$$
3. $$\overrightarrow{OM}=\frac{\mathbf a+\mathbf b}{2} =\frac12\begin{pmatrix}7\\2\end{pmatrix} =\begin{pmatrix}\frac72\\1\end{pmatrix}.$$
4. Since $N$ is one third of the way from $A$ to $B$, $$\overrightarrow{ON} =\mathbf a+\frac13(\mathbf b-\mathbf a) =\frac23\mathbf a+\frac13\mathbf b =\begin{pmatrix}3\\\frac13\end{pmatrix}.$$

What to remember

$$\begin{aligned} \overrightarrow{AB}&=B-A,\\ \overrightarrow{AB}+\overrightarrow{BC}&=\overrightarrow{AC},\\ \overrightarrow{BA}&=-\overrightarrow{AB},\\ \left|\begin{pmatrix}x\\y\end{pmatrix}\right|&=\sqrt{x^2+y^2}. \end{aligned}$$

Also remember:

• add, subtract and multiply column vectors component by component;
• a non-zero scalar multiple proves that vectors are parallel;
• use a positive scalar multiple of vectors sharing a point to prove collinearity;
• for $AP:PB=m:n$, the fraction of the journey from $A$ to $B$ is $\frac{m}{m+n}$;
• in a proof, translate the vector equality back into a geometric conclusion.

Next, strengthen the geometry behind these arguments with similarity and transformations and the logic of a complete argument with proof and reasoning. For coordinate applications, continue with straight line graphs and coordinate geometry.