Vectors

A vector has both magnitude and direction. Vectors describe translations, displacements, velocities and forces. A scalar has magnitude only, such as mass, time, temperature or speed.

For example,

$$\mathbf a=\begin{pmatrix}3\\-2\end{pmatrix}$$

means move $3$ units in the positive $x$ direction and $2$ units in the negative $y$ direction. The vector is not the point $(3,-2)$, although the same two numbers may represent the position vector of that point.

At A level, you need to use vectors in two and three dimensions, calculate magnitudes and directions, form position and displacement vectors, divide line segments, and solve geometric and modelling problems.

Prerequisites

You should be able to:

• work confidently with negative numbers, fractions and surds;
• solve linear equations and simultaneous equations;
• use coordinates in two and three dimensions;
• apply Pythagoras’ theorem and basic trigonometry;
• rearrange formulae and compare coefficients.

Review algebraic fluency, Pythagoras and trigonometry or straight lines if these skills are uncertain.

The central idea: vectors describe changes

The arrow from point $A$ to point $B$ is written $\overrightarrow{AB}$. Its direction matters:

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

If $A(2,-1)$ and $B(7,3)$, subtract the coordinates of the starting point from those of the finishing point:

$$\overrightarrow{AB} =\begin{pmatrix}7-2\\3-(-1)\end{pmatrix} =\begin{pmatrix}5\\4\end{pmatrix}.$$

This “finish minus start” rule is one of the most useful habits in the topic.

Vectors can be translated without changing them. Any arrow with the same magnitude and direction as $\overrightarrow{AB}$ represents the same free vector. Its location on the page is irrelevant.

Worked example 1: distinguish points from vectors

Let $P(-3,4)$ and $Q(2,-5)$. Find $\overrightarrow{PQ}$ and $\overrightarrow{QP}$.

For $P$ to $Q$,

$$\overrightarrow{PQ} =\begin{pmatrix}2-(-3)\\-5-4\end{pmatrix} =\begin{pmatrix}5\\-9\end{pmatrix}.$$

Reversing the direction negates every component:

$$\overrightarrow{QP} =\begin{pmatrix}-5\\9\end{pmatrix} =-\overrightarrow{PQ}.$$

The coordinates of $Q$ were not used as the answer. They describe a point relative to the origin, whereas $\overrightarrow{PQ}$ describes a change from $P$ to $Q$.

Components, magnitude and direction

In two dimensions, a vector may be written as

$$\mathbf a=\begin{pmatrix}a_1\\a_2\end{pmatrix} =a_1\mathbf i+a_2\mathbf j,$$

where $\mathbf i$ and $\mathbf j$ are unit vectors in the positive $x$ and $y$ directions. In three dimensions,

$$\mathbf a=\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix} =a_1\mathbf i+a_2\mathbf j+a_3\mathbf k.$$

The magnitude, written $|\mathbf a|$, is found using Pythagoras:

$$\left|\begin{pmatrix}x\\y\end{pmatrix}\right| =\sqrt{x^2+y^2}, \qquad \left|\begin{pmatrix}x\\y\\z\end{pmatrix}\right| =\sqrt{x^2+y^2+z^2}.$$

A unit vector has magnitude $1$. The unit vector in the direction of a non-zero vector $\mathbf a$ is

$$\boxed{\widehat{\mathbf a}=\frac{\mathbf a}{|\mathbf a|}}.$$

Division here means multiplying every component by the scalar $1/|\mathbf a|$.

Worked example 2: magnitude and a unit vector

For

$$\mathbf a=\begin{pmatrix}-2\\3\\6\end{pmatrix},$$

find $|\mathbf a|$ and a unit vector in the direction of $\mathbf a$.

$$|\mathbf a| =\sqrt{(-2)^2+3^2+6^2} =\sqrt{4+9+36} =7.$$

Therefore

$$\widehat{\mathbf a} =\frac17\begin{pmatrix}-2\\3\\6\end{pmatrix} =\boxed{\begin{pmatrix}-2/7\\3/7\\6/7\end{pmatrix}}.$$

Check its magnitude:

$$\sqrt{\frac4{49}+\frac9{49}+\frac{36}{49}}=1.$$

The signs must remain unchanged because the unit vector must point in the same direction.

Continue with vector magnitude and direction.

Vector arithmetic has geometric meaning

Vectors are added and subtracted 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 places vectors head to tail. If a journey goes from $A$ to $B$ and then from $B$ to $C$, the overall displacement is

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

Multiplication by a scalar changes length and possibly direction. If $\lambda>0$, then $\lambda\mathbf a$ points in the same direction as $\mathbf a$ and has magnitude $\lambda|\mathbf a|$. If $\lambda<0$, its direction is reversed and its magnitude is $|\lambda||\mathbf a|$.

Worked example 3: solve a vector equation

Given

$$2\mathbf p-3\mathbf q=\begin{pmatrix}7\\-1\end{pmatrix}, \qquad \mathbf q=\begin{pmatrix}-1\\2\end{pmatrix},$$

find $\mathbf p$.

Substitute $\mathbf q$:

$$2\mathbf p-3\begin{pmatrix}-1\\2\end{pmatrix} =\begin{pmatrix}7\\-1\end{pmatrix}.$$

Hence

$$2\mathbf p =\begin{pmatrix}7\\-1\end{pmatrix} +\begin{pmatrix}-3\\6\end{pmatrix} =\begin{pmatrix}4\\5\end{pmatrix},$$

so

$$\boxed{\mathbf p=\begin{pmatrix}2\\5/2\end{pmatrix}}.$$

Vector equations are simultaneous scalar equations in compact form. You may always compare corresponding components.

Study vector arithmetic after vectors in two and three dimensions.

Position vectors turn geometry into algebra

Choose an origin $O$. The position vector of a point $A$ is

$$\mathbf a=\overrightarrow{OA}.$$

If $A$ and $B$ have position vectors $\mathbf a$ and $\mathbf b$, then

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

This is the vector form of finish minus start. It follows from the head to tail route

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

The distance between $A$ and $B$ is therefore

$$AB=|\mathbf b-\mathbf a|.$$

Worked example 4: divide a line segment in a ratio

Points $A$ and $B$ have position vectors

$$\mathbf a=\begin{pmatrix}1\\-2\end{pmatrix}, \qquad \mathbf b=\begin{pmatrix}7\\4\end{pmatrix}.$$

Point $P$ lies on $AB$ and $AP:PB=2:1$. Find the position vector of $P$.

Since $P$ is two thirds of the way from $A$ to $B$,

$$\overrightarrow{OP} =\mathbf a+\frac23(\mathbf b-\mathbf a).$$

Now

$$\mathbf b-\mathbf a =\begin{pmatrix}6\\6\end{pmatrix},$$

so

$$\overrightarrow{OP} =\begin{pmatrix}1\\-2\end{pmatrix} +\frac23\begin{pmatrix}6\\6\end{pmatrix} =\begin{pmatrix}1\\-2\end{pmatrix} +\begin{pmatrix}4\\4\end{pmatrix} =\boxed{\begin{pmatrix}5\\2\end{pmatrix}}.$$

Check the ratio:

$$\overrightarrow{AP}=\begin{pmatrix}4\\4\end{pmatrix}, \qquad \overrightarrow{PB}=\begin{pmatrix}2\\2\end{pmatrix},$$

so $\overrightarrow{AP}=2\overrightarrow{PB}$ as required.

Learn the full method in position vectors and distance.

Parallel vectors and points on a line

Two non-zero vectors are parallel if one is a scalar multiple of the other:

$$\mathbf a\parallel\mathbf b \quad\Longleftrightarrow\quad \mathbf a=\lambda\mathbf b$$

for some non-zero scalar $\lambda$. A positive $\lambda$ gives the same direction and a negative $\lambda$ gives the opposite direction.

A point $P$ lies on the line through $A$ in direction $\mathbf d$ precisely when its position vector can be written

$$\overrightarrow{OP}=\mathbf a+\lambda\mathbf d.$$

Even when a specification does not emphasise vector equations of lines, this form clarifies collinearity and geometric arguments.

Worked example 5: test collinearity

Let $A(1,2)$, $B(4,8)$ and $C(6,12)$. Are the three points collinear?

$$\overrightarrow{AB}=\begin{pmatrix}3\\6\end{pmatrix}, \qquad \overrightarrow{AC}=\begin{pmatrix}5\\10\end{pmatrix}.$$

Since

$$\overrightarrow{AC}=\frac53\overrightarrow{AB},$$

the displacements from the same point $A$ are parallel. Therefore $A$, $B$ and $C$ are collinear.

It is not enough to say that the components “look proportional”. State the scalar multiple and the geometric conclusion.

Vectors in modelling

In mechanics and other applications, vector quantities must be interpreted with units and a stated coordinate system. If east is the positive $x$ direction and north is the positive $y$ direction, then

$$\mathbf v=\begin{pmatrix}12\\-5\end{pmatrix}\text{ m s}^{-1}$$

describes a velocity of $12\text{ m s}^{-1}$ east and $5\text{ m s}^{-1}$ south. Its speed is the scalar magnitude

$$|\mathbf v|=\sqrt{12^2+(-5)^2}=13\text{ m s}^{-1}.$$

Worked example 6: displacement from constant velocity

A particle has initial position vector

$$\mathbf r_0=\begin{pmatrix}-4\\7\end{pmatrix}\text{ m}$$

and constant velocity

$$\mathbf v=\begin{pmatrix}3\\-2\end{pmatrix}\text{ m s}^{-1}.$$

Find its position after $5$ seconds and the distance it travels.

With constant velocity,

$$\mathbf r=\mathbf r_0+t\mathbf v.$$

Therefore

$$\mathbf r =\begin{pmatrix}-4\\7\end{pmatrix} +5\begin{pmatrix}3\\-2\end{pmatrix} =\begin{pmatrix}-4\\7\end{pmatrix} +\begin{pmatrix}15\\-10\end{pmatrix} =\boxed{\begin{pmatrix}11\\-3\end{pmatrix}\text{ m}}.$$

The displacement is $5\mathbf v=(15,-10)^{\mathsf T}$ metres. Since the motion is along a straight line without reversal, the distance travelled is its magnitude:

$$\sqrt{15^2+(-10)^2} =5\sqrt{13}\text{ m}.$$

The final position, displacement and distance are three different quantities. Explore applications in vectors in modelling and then connect them to two dimensional motion.

Common misconceptions

A vector is not its magnitude

$\mathbf a$ is a directed quantity, whereas $|\mathbf a|$ is a non-negative scalar. The equation $\mathbf a=5$ is normally meaningless because its two sides are different kinds of object.

Coordinates and position vectors use similar notation

The point $A(2,3)$ and its position vector $\overrightarrow{OA}=(2,3)^{\mathsf T}$ contain the same components, but they represent different objects. Subtract points conceptually by subtracting their position vectors.

Magnitudes do not usually distribute over addition

In general,

$$|\mathbf a+\mathbf b|\ne|\mathbf a|+|\mathbf b|.$$

Equality occurs only in special cases, such as when both non-zero vectors point in exactly the same direction.

Parallel does not always mean the same direction

$\mathbf a$ and $-2\mathbf a$ are parallel but point in opposite directions. If a question requires the same direction, the scalar multiple must be positive.

Magnitude requires squares of every component

For $(-3,4)^{\mathsf T}$, the magnitude is $\sqrt{(-3)^2+4^2}=5$, not $\sqrt{-3^2+4^2}$ and not $-3+4$.

Diagnostic self-check

Try these before opening the answers.

1. If $A(-2,5)$ and $B(4,1)$, find $\overrightarrow{AB}$.
2. Find the magnitude of $\begin{pmatrix}2\\-3\\6\end{pmatrix}$.
3. Find a unit vector in the direction of $\begin{pmatrix}3\\-4\end{pmatrix}$.
4. Given $\mathbf a=\begin{pmatrix}1\\5\end{pmatrix}$ and $\mathbf b=\begin{pmatrix}-2\\3\end{pmatrix}$, calculate $2\mathbf a-\mathbf b$.
5. Points $P$ and $Q$ have position vectors $\mathbf p$ and $\mathbf q$. Write $\overrightarrow{QP}$ in terms of $\mathbf p$ and $\mathbf q$.
6. Are $\begin{pmatrix}6\\-9\end{pmatrix}$ and $\begin{pmatrix}-2\\3\end{pmatrix}$ parallel? State their relative directions.
7. The midpoint of $A$ and $B$ has position vector $\mathbf m$. Express $\mathbf m$ in terms of $\mathbf a$ and $\mathbf b$.

Answers and guidance

1. Finish minus start gives

   $$\overrightarrow{AB}=\begin{pmatrix}4-(-2)\\1-5\end{pmatrix} =\boxed{\begin{pmatrix}6\\-4\end{pmatrix}}.$$

$$\sqrt{2^2+(-3)^2+6^2}=\sqrt{49}=\boxed{7}.$$

3. The original magnitude is $5$, so the required unit vector is

   $$\boxed{\begin{pmatrix}3/5\\-4/5\end{pmatrix}}.$$

$$2\mathbf a-\mathbf b =\begin{pmatrix}2\\10\end{pmatrix} -\begin{pmatrix}-2\\3\end{pmatrix} =\boxed{\begin{pmatrix}4\\7\end{pmatrix}}.$$

5. Start at $Q$ and finish at $P$:

   $$\boxed{\overrightarrow{QP}=\mathbf p-\mathbf q}.$$

6. Yes, because

   $$\begin{pmatrix}6\\-9\end{pmatrix} =-3\begin{pmatrix}-2\\3\end{pmatrix}.$$

   The negative multiplier means they point in opposite directions.

7. The midpoint is halfway from $A$ to $B$:

   $$\mathbf m=\mathbf a+\frac12(\mathbf b-\mathbf a) =\boxed{\frac{\mathbf a+\mathbf b}{2}}.$$

Recommended learning route

Work through the lessons in this order:

1. Vectors in two and three dimensions for notation, components, equality and vector types.
2. Vector arithmetic for addition, subtraction, scalar multiplication and vector equations.
3. Vector magnitude and direction for lengths, unit vectors and directional interpretation.
4. Position vectors and distance for displacement, division of a line segment and geometric proofs.
5. Vectors in modelling for motion, forces and interpreting assumptions and units.

At every stage, translate between three representations: an arrow diagram, component notation and a verbal description. If all three express the same movement, the algebra is much less likely to become a collection of disconnected rules.