GCSE Maths: How To Calculate Vectors
Introduction
You need to know how to deal with vectors at both foundation and higher level. Foundation GCSE maths questions are not very complicated and students ought to be able to achieve full if not nearly all the marks for a question.
Higher gcse maths questions can be more challenging as they can often involve the use of midpoints as well as using ratio.
A vector is a quantity that has both size and direction and it is generally shown by a straight line.
For instance:
If you consider the straight lines above then to get from the point A to the point B is represented by the vector a. You can generally represent a vector by writing it in bold font with an underscore.
Vectors Example
Take a look at the following question:
Find the vector \overrightarrow{A B} in terms of b and a.
In order to answer a question such as this you need to draw a path from A to B. Because you are given a vector from A to O and from O to B, this is the path that you must take. It is not possible to go directly from A to B.
The highlighted path shows the route that needs to be taken.
With vectors it is important to take note of the arrow. To go from A to 0 you are moving in the opposite direction of the arrow so this has the vector –a. To go from 0 to B the movement is in the direction of the arrow so the vector is b.
The vector \overrightarrow{A B}=b-a
Example
Consider the following question:
In order to go from A to B you need to take the path from A to 0 and then from 0 to B.
This will have the vector –a + b or b – a.
It is important to note that the vector from A to 0 is in the opposite direction of the arrow which is why it is negative.
\overrightarrow{O P} has been drawn and in order to obtain this vector you need to go \overrightarrow{O A} and then \overrightarrow{A P}
You are given that \overrightarrow{A P}: \overrightarrow{P B} is in the ratio 1:3 so this means that there are 4 ratio parts altogether.
\begin{gathered} \overrightarrow{O P}=\overrightarrow{O A}+\overrightarrow{A P} \\ \overrightarrow{O P}=a+\frac{3}{4} \overrightarrow{A B} \\ \overrightarrow{O P}=a+\frac{3}{4}(b-a) \\ \overrightarrow{O P}=a+\frac{3}{4} b-\frac{3}{4} a=\frac{1}{4} a+\frac{3}{4} b \end{gathered}Question Practice
Try the following questions on your own before looking at the solution:
Question Practice Solutions
(i) Here you need to read the question very carefully and to make sure that you have taken in all the necessary information. PR is highlighted below.
You are given that QR = 2PS so this means that QR = 2b
\overrightarrow{P R}=\overrightarrow{P Q}+\overrightarrow{Q R}=a+2 b(ii)
The path from S to X is highlighted below:
You are given a ratio amount which means that \overrightarrow{Q X}=\frac{3}{4}(\overrightarrow{Q R})
\begin{gathered} \overrightarrow{S X}= \\ =-b+a+\frac{3}{4}(\overrightarrow{Q R}) \\ =\frac{6}{4} b-b+a \\ =\frac{1}{2} b+a \end{gathered}Your vector revision for Higher GCSE maths needs to be consistent and if you are doing the higher paper then you need to regularly practise questions such as the ones that are seen here.
Just like any other topic you can get your easy questions and harder questions. The questions shown here would be more suited towards students looking to sit the higher paper.
In the foundation paper you are probably asked to calculate vector addition or vector subtraction which can be easily done for say 2 marks even though they may appear towards the back of the question paper.