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Vectors

GCSELevel 6-7Level 8-9Cambridge iGCSEEdexcel iGCSE

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.

Level 6-7GCSECambridge iGCSEEdexcel iGCSE

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}

Level 6-7GCSECambridge iGCSEEdexcel iGCSE

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}

Level 6-7GCSECambridge iGCSEEdexcel iGCSE

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

Level 6-7GCSECambridge iGCSEEdexcel iGCSE

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}

Level 6-7GCSECambridge iGCSEEdexcel iGCSE

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}

 

Level 8-9GCSECambridge iGCSEEdexcel iGCSE

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}

Level 8-9GCSECambridge iGCSEEdexcel iGCSE

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}

 

Using Pythagoras,

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

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}

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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