# Vectors

## Vectors Revision

**Vectors**

**Vectors** are quantities that are defined by both **magnitude** and **direction**. They are often used to describe where **points are**, or how to get from one corner of a shape to another.

You will need to use and understand **vectors**, as well as be able to **add, subtract**, and **multiply** them.

**Vector Notation**

There are several ways a **vector** can be represented, so it’s important you are familiar with them all to prepare for your exams.

**Bold / Underlined**

The most common way you’ll see is a letter written in **bold**, for example, these two lines are **vectors**:

For example, the **vector** \boldsymbol{b} is in the direction of the pink arrow.

**Note:** when you are writing in your exam, as you can’t write in bold, you will be expected to underline **vectors**, like: \underline{a}

**Column Notation**

**Vectors** may also be displayed in columns:

**Example:** The column **vector** \begin{pmatrix}2 \\ 1\end{pmatrix} means 2 spaces to the right and 1 space up.

If you had \begin{pmatrix}-2 \\ -1\end{pmatrix}, this would mean 2 spaces to the left and 1 space down.

**Arrow**

Finally, you may see **vectors** written like:

\overrightarrow{AB} which is the **vector** between point A and point B

\overrightarrow{OC} which is the **vector** between the origin and point C

**Note:** The point O almost always represents the origin in vector questions.

**Adding and Subtracting Vectors**

When **adding** and **subtracting** using diagrams, it is important to pay attention to the direction the **vector** arrow is going.

A vector \boldsymbol{a} is positive in one direction, and if you flip the direction (point the arrow towards the opposite direction), the vector is now \boldsymbol{-a}:

**Adding**

If we have two **vectors**, \boldsymbol{a} and \boldsymbol{b}, these can be **added**:

\boldsymbol{a}+\boldsymbol{b}

\boldsymbol{a}+\boldsymbol{b} will follow a route from the start of the \boldsymbol{a} arrow, to the end of the \boldsymbol{b} arrow, both in the positive direction:

**Subtracting**

For **subtraction**, for example,

\boldsymbol{a}-\boldsymbol{b}

The \boldsymbol{b} is negative, so either of these diagrams correctly portray this **subtraction**:

**Adding and Subtracting Column Vectors**

Column vectors are easy to **add** and **subtract**, because you just treat the top row and the bottom row as separate sums, for example:

\begin{pmatrix}5 \\ 2\end{pmatrix}+\begin{pmatrix}3 \\ 1\end{pmatrix}=\begin{pmatrix}5+3 \\ 2+1\end{pmatrix}=\begin{pmatrix}8 \\ 3\end{pmatrix}

\begin{pmatrix}6 \\ 4\end{pmatrix}-\begin{pmatrix}2 \\ 3\end{pmatrix}=\begin{pmatrix}6-2 \\ 4-3\end{pmatrix}=\begin{pmatrix}4 \\ 1\end{pmatrix}

**Multiplying Vectors by a Scalar**

**Multiplying** **vectors** by a scalar simply means **multiplying** **vectors** by a number (not a vector).

For example,

\boldsymbol{a}\times3=3\boldsymbol{a}

This is represented in the diagram:

If we have a **vector** containing more than one letter, we just **multiply** each letter by the scalar:

3\times(\boldsymbol{a}+\boldsymbol{b})=3\boldsymbol{a}+3\boldsymbol{b}

Similarly, in column **vector** form:

4\times\begin{pmatrix}3 \\ 2\end{pmatrix}=\begin{pmatrix}4\times3 \\ 4\times2\end{pmatrix}=\begin{pmatrix}12 \\ 8\end{pmatrix}

**Vector Magnitudes**

The **magnitude** of a **vector** is the length of a **vector**.

**Magnitude** has no direction, so is always positive.

**Example:**

Find the **magnitude** of the vector

\overrightarrow{XY}=\begin{pmatrix}3 \\ 4\end{pmatrix}

Looking at the diagram, we can work out the length of the vector \overrightarrow{XY} by treating it as a right-angled triangle and using Pythagoras’ theorem:

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

**Example 1: Finding Vectors**

Express the vector \overrightarrow{XZ} in terms of \boldsymbol{a} and \boldsymbol{b}

**[2 marks]**

So, let’s start at X. We will have to go through Y to get to Z.

To get from X to Y, we have to go in the negative direction of \boldsymbol{b}, so

\boldsymbol{-b}

Then, we will have to go from Y to Z, which is \boldsymbol{a} in the positive direction,

\boldsymbol{+a}

So, in total:

\boldsymbol{-b}+\boldsymbol{a}

Or,

\boldsymbol{a}-\boldsymbol{b}

**Example 2: Vectors on Parallel Lines**

Express the vectors \overrightarrow{AB} and \overrightarrow{AD} in terms of \boldsymbol{a} and \boldsymbol{b} in this parallelogram.

**[2 marks]**

With **vectors**, parallel lines of the same length are equal, as they have the same** direction** and **magnitude**.

So, in this parallelogram \overrightarrow{AB} is equal to \overrightarrow{DC}, and \overrightarrow{AD} is equal to \overrightarrow{BC}.

So, to find \overrightarrow{AB}…

We know this is the same as \overrightarrow{DC}, which, in this direction, equals \boldsymbol{-b}. So,

\overrightarrow{AB}=\boldsymbol{-b}

And as \overrightarrow{AD} is equal to \overrightarrow{BC},

\overrightarrow{AD}=\boldsymbol{a}

**Example 3: Using Ratio**

ABC is a triangle.

M lies on the line AB such that AM:MB is 1:1.

Express the vector AM in terms of \boldsymbol{a} and \boldsymbol{b}

**[3 marks]**

The ratio 1:1 means M is the midpoint of \overrightarrow{AB}.

Firstly, finding \overrightarrow{AB},

\overrightarrow{AB}=4\boldsymbol{a}+6\boldsymbol{b}

\overrightarrow{AM} is half of \overrightarrow{AB}, so,

\overrightarrow{AM}=\dfrac{1}{2}(4\boldsymbol{a}+6\boldsymbol{b})

\overrightarrow{AM}=2\boldsymbol{a}+3\boldsymbol{b}

## Vectors Example Questions

**Question 1**:

a) Find \overrightarrow{AC}+\overrightarrow{AB}

b) Find 2\times\overrightarrow{AD}

c) Find \overrightarrow{AD}-\overrightarrow{AB}

**[6 marks]**

a)

\overrightarrow{AC}+\overrightarrow{AB}

=12\boldsymbol{b}+3\boldsymbol{a-b}

=3\boldsymbol{a}+11\boldsymbol{b}

b)

2\times\overrightarrow{AD}

=2\times (-2\boldsymbol{a-b})

=-4\boldsymbol{a}-2\boldsymbol{b}

c)

\overrightarrow{AD}-\overrightarrow{AB}

=-2\boldsymbol{a-b} - (3\boldsymbol{a-b})

=-5\boldsymbol{a}

**Question 2**: Find the magnitude of vector \begin{pmatrix}12 \\ 16\end{pmatrix}

**[3 marks]**

Using Pythagoras,

\sqrt{12^2+16^2}=20

**Question 3**: ABCDEF is a regular hexagon.

Find \overrightarrow{FB} in terms of \boldsymbol{a} and \boldsymbol{b}

**[3 marks]**

As this is a regular hexagon, \overrightarrow{FA} and \overrightarrow{DC} are parallel, \overrightarrow{AB} and \overrightarrow{ED} are parallel, and \overrightarrow{EF} and \overrightarrow{CB} are parallel.

To find \overrightarrow{FB}, we need \overrightarrow{FA} plus \overrightarrow{AB}.

\overrightarrow{FA} is the parallel to \overrightarrow{DC}, but in the opposite direction, so,

\overrightarrow{FA}=-5\boldsymbol{a}-\boldsymbol{b}

To find \overrightarrow{AB} , we can start by working out \overrightarrow{OC} which is parallel.

\overrightarrow{OC}=-3\boldsymbol{a}-\boldsymbol{b} -(5\boldsymbol{b}+\boldsymbol{b})=-8\boldsymbol{a}-2\boldsymbol{b}.

Thus,

\overrightarrow{OC}=-8\boldsymbol{a}-2\boldsymbol{b}

So,

\overrightarrow{FB}=-5\boldsymbol{a}-\boldsymbol{b}-8\boldsymbol{a}-2\boldsymbol{b}

\overrightarrow{FB}=-13\boldsymbol{a}-3\boldsymbol{b}

**Question 4**: M lies on the line \overrightarrow{AC} such that \overrightarrow{AM}:\overrightarrow{MC} is 2:1

Find \overrightarrow{MC} in terms of \boldsymbol{a} and \boldsymbol{b}

**[3 marks]**

Let’s firstly find \overrightarrow{AC}

\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}

\overrightarrow{AC}=5\boldsymbol{a}+\boldsymbol{b}+4\boldsymbol{a}-4\boldsymbol{b}

\overrightarrow{AC}=9\boldsymbol{a}-3\boldsymbol{b}

As \overrightarrow{AM}:\overrightarrow{MC} is 2:1, \overrightarrow{MC} is \dfrac{1}{3}\overrightarrow{AC}

\overrightarrow{AC}=\dfrac{1}{3}\times (9\boldsymbol{a}-3\boldsymbol{b})

\overrightarrow{AC}=3\boldsymbol{a}-\boldsymbol{b}

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