# Vectors

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## 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.

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## 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 $\boldsymbol{-a}$:

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:

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

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

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## 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:

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

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## 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:

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

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## 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,

$\boldsymbol{a}-\boldsymbol{b}$

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## 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 $\overrightarrow{BC}$,

$\overrightarrow{AD}=\boldsymbol{a}$

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## 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,

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

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

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## Vectors Example Questions

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)

$\overrightarrow{AD}-\overrightarrow{AB}$

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

$=-5\boldsymbol{a}$

Gold Standard Education

Using Pythagoras,

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

Gold Standard Education

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,

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

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

Gold Standard Education

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 $\dfrac{1}{3}\overrightarrow{AC}$

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

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