Vectors Questions | Worksheets and Revision | MME

Vectors Questions, Worksheets and Revision

Level 6-7
GCSE Maths Revision Guide

Vectors

A vector is something with both magnitude and direction. On diagrams they are denoted by an arrow, where the length tells us the magnitude and the arrow tells us direction. 

  • You will need to add and subtract vectors
  • You will also need to multiply vectors and understand scalar multiples of vectors 

Make sure you are happy with the following topics before continuing:

Adding and subtracting vectors 

When we add vectors:

\textcolor{red}{\mathbf{a}+\mathbf{b}}

this takes you from the start of \mathbf{a} to the end of \mathbf{b} (right). 

The negative of a vector has the same magnitude of the original vector, it just goes in the exact opposite direction.

When we subtract vectors:

\textcolor{limegreen}{\mathbf{a}-\mathbf{b}},

we add on the negative of the vector that is being subtracted.

Level 6-7

Multiplying vectors

Multiplying vectors by a Scalar – We can multiply a vector by a number. For example, we can multiply \mathbf{a} by 3:

3\mathbf{\textcolor{red}{a}}=\mathbf{\textcolor{red}{a}}+\mathbf{\textcolor{red}{a}}+\mathbf{\textcolor{red}{a}}

This means the the vectors are added end to end. 

We can multiply more complicated vectors 

3\mathbf{(\textcolor{red}{a}+\textcolor{blue}{b})}=\mathbf{3\textcolor{red}{a}}+\mathbf{3\textcolor{blue}{b}}

Note: all vectors here are written in bold. When you’re writing this by hand, you should underline each letter that represents a vector.

Scalar multiples – Scalar multiples are all parallel to each other:

\mathbf{3\textcolor{red}{a}}+\mathbf{3\textcolor{blue}{b}} is parallel to \mathbf{\textcolor{red}{a}}+\mathbf{\textcolor{blue}{b}}

 

Level 6-7

Vector Notation

A vector from point X to point Y will be shown as \overrightarrow{XY}.

So from the diagram shown we can see that:

\overrightarrow{XY} = \mathbf{a}

\overrightarrow{YZ} = \mathbf{b}

\overrightarrow{ZX} = \mathbf{c}

If we go against the arrow the vector becomes negative. 

\overrightarrow{YX} = \mathbf{-a}

We can also combine vectors, meaning

\overrightarrow{XZ} = \mathbf{a + b} = \mathbf{-c}

Level 6-7

Example: Involving ratios

In the diagram below, we have vectors

\overrightarrow{AB}=3\mathbf{a} and \overrightarrow{AC}=4\mathbf{b} .

Point D lies on the line BC such that BD:DC=1:3.

Write the vector \overrightarrow{AD} in terms of \mathbf{a} and \mathbf{b}.

[3 marks]

To find \overrightarrow{AD}, we are going to add the vectors

\overrightarrow{AB}+\overrightarrow{BD}

We know \overrightarrow{AB}=3\mathbf{a} but we need to find \overrightarrow{BD}

We know BD:DC=1:3, so the point D is \dfrac{1}{4} of the way along \overrightarrow{BC}

\overrightarrow{BC}=-\overrightarrow{AB}+\overrightarrow{AC}=-3\mathbf{a}+4\mathbf{b}

Now we need to multiply this by \dfrac{1}{4}

\overrightarrow{BD}=\dfrac{1}{4}(3\mathbf{a}+4\mathbf{b})=-\dfrac{3}{4}\mathbf{a}+\mathbf{b}

Now we need to add \overrightarrow{AB} to the answer:

\overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{BD}=3\mathbf{a}+\left(-\dfrac{3}{4}\mathbf{a}+\mathbf{b}\right)=\dfrac{9}{4}\mathbf{a}+\mathbf{b}

Thus, we have our answer in terms of \mathbf{a} and \mathbf{b}.

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Example Questions

First we can find the vector \overrightarrow{AC}

 

\overrightarrow{AC} =\overrightarrow{AB} +\overrightarrow{BC} =3\mathbf{a} +2\mathbf{b}

 

Since M is the midpoint of AC we know that \overrightarrow{AM}=\dfrac{1}{2}\overrightarrow{AC}

 

\overrightarrow{AM}=\dfrac{1}{2}(3\mathbf{a} +2\mathbf{b}) = \dfrac{3}{2}\mathbf{a} +\mathbf{b}

The vector \overrightarrow{EF} is the same as:

 

\overrightarrow{EF}=\overrightarrow{ED}+\overrightarrow{DC}+\overrightarrow{CF}

 

First, we can find the vector \overrightarrow{ED} as \dfrac{1}{2}\mathbf{b}

 

Then we also know \overrightarrow{DC} = \overrightarrow{AB}, hence \overrightarrow{DC}=2\mathbf{a}

 

The final section \overrightarrow{CF} = \dfrac{3}{5}\times 2\mathbf{a}=\dfrac{6}{5}\mathbf{a} 

 

Hence, 

 

\overrightarrow{EF}= \dfrac{1}{2} \mathbf{b}+2\mathbf{a}+\dfrac{6}{5}\mathbf{a}=3.2\mathbf{a}+ \dfrac{1}{2} \mathbf{b}

First we can find the vector \overrightarrow{BC}

 

\overrightarrow{BC} =-\overrightarrow{AB} +\overrightarrow{AC} =-\mathbf{a} +2\mathbf{b}

 

Then we can find:

 

\overrightarrow{BD} =\dfrac{3}{5}\overrightarrow{BC} =-\dfrac{3}{5}\mathbf{a} +\dfrac{6}{5}\mathbf{b}

 

Next:

 

\begin{aligned} \overrightarrow{AD} &=\overrightarrow{AB}+\overrightarrow{BD} \\ &=\mathbf{a} +\bigg(-\dfrac{3}{5}\mathbf{a} +\dfrac{6}{5}\mathbf{b}\bigg) \\ & =\dfrac{2}{5}\mathbf{a}+\dfrac{6}{5}\mathbf{b} \end{aligned}

 

Finally:

 

\begin{aligned} \overrightarrow{AE} &=\dfrac{1}{3}\overrightarrow{AD} \\ &=\dfrac{1}{3}\bigg(\dfrac{2}{5}\mathbf{a}+\dfrac{6}{5}\mathbf{b}\bigg) \\ &= \dfrac{2}{15}\mathbf{a}+\dfrac{2}{5}\mathbf{b} \end{aligned}

We will find \overrightarrow{AC} by doing, 

 

\overrightarrow{AC}=\overrightarrow{AB}-\overrightarrow{OB}+\overrightarrow{OC}

 

Thus

 

\overrightarrow{AC} =\mathbf{b}-\mathbf{a}+\mathbf{b}= 2\mathbf{b}-\mathbf{a}

We will find \overrightarrow{EB} by doing

 

\overrightarrow{EB}=\overrightarrow{EA}+\overrightarrow{AB}

 

The first vector is straightforward, because we know \overrightarrow{AE}, and that is just the same vector in the opposite direction. So, we get:

 

\overrightarrow{EA}=-\overrightarrow{AE}=-(3\mathbf{a}-2\mathbf{b})=-3\mathbf{a}+2\mathbf{b}

 

Now we need \overrightarrow{AB}. Since B is the midpoint of AC (given in the question), we must have that \overrightarrow{AB}=\frac{1}{2} \overrightarrow{AC}. Therefore, looking at the diagram, we get that:

 

\overrightarrow{AB}=\dfrac{1}{2}\overrightarrow{AC}=\dfrac{1}{2}\left(\overrightarrow{AD}+\overrightarrow{DC}\right)

 

We’re given the second part of this, \overrightarrow{DC}=2\mathbf{a}+4\mathbf{b}, and since E is the midpoint of AD, we can also work out the first part:

 

\overrightarrow{AD}=2\overrightarrow{AE}=2(3\mathbf{a}-2\mathbf{b})=6\mathbf{a}-4\mathbf{b}

 

Now, we have everything we need and can go back through our work, filling in the gaps we find:

 

\overrightarrow{AB}=\dfrac{1}{2}\left(6\mathbf{a}-4\mathbf{b}+2\mathbf{a} +4\mathbf{b}\right)=\dfrac{1}{2}\left(8\mathbf{a}\right)=4\mathbf{a}

 

Finally we have that:

 

\overrightarrow{EB}=\overrightarrow{EA}+\overrightarrow{AB}=-3\mathbf{a}+2\mathbf{b}+4\mathbf{a}=\mathbf{a}+2\mathbf{b}

 

If \overrightarrow{EB} and \overrightarrow{DC} are parallel, then one must be a multiple of the other. Well, if we multiply \overrightarrow{EB} by 2 then, 

 

2\times\overrightarrow{EB}=2(\mathbf{a}+2\mathbf{b})=2\mathbf{a}+4\mathbf{b}=\overrightarrow{DC}

 

We have shown that 2\overrightarrow{EB}=\overrightarrow{DC}, therefore, the two lines must be parallel.

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