The magnitude of a vector \(\boldsymbol v\) is written in print as \(\vert\boldsymbol v \vert\) and in handwriting as \(\vert\underline v \vert\small.\)
Similarly, the magnitude of \(\small\overrightarrow{\textsf{AB}}\) is written as \(\vert\small\overrightarrow{\textsf{AB}}\normalsize\vert\small.\)
If a vector is given in component form, we can use either 2D or 3D Pythagoras to find its magnitude.
Notation
\( \small \overrightarrow{\textsf{AB}} \normalsize\) represents the vector from point A to point B.
Lower case letters are also used to represent vectors.
In handwriting, they are underlined, like this: \( \underline u \)
In print, they are in bold, like this: \( \boldsymbol u \)
Key idea
For any three points A, B and C, \( \small \overrightarrow{\textsf{AC}} = \overrightarrow{\textsf{AB}} + \overrightarrow{\textsf{BC}} \normalsize \).
Think of this as going from A to B and then continuing from B to C. In other words: from A to C via B.
It's the same resultant journey whether we go via B or not. We start at A and end at C, so the vectors are equal. Vectors only care about the end-points, not the pathway.
A great resource to prepare for your N5 Maths exams. Get £10 discount with code 'Maths.scot'. Learn more
Example 1 (non-calculator)
In the diagram below, \( \boldsymbol u \), \( \boldsymbol v \) and \( \boldsymbol w \) represent \( \small \overrightarrow{\textsf{AB}} \normalsize\), \( \small \overrightarrow{\textsf{BC}} \normalsize\) and \( \small \overrightarrow{\textsf{AC}} \normalsize\) respectively.
Express \( \boldsymbol v \) in terms of \( \boldsymbol u \) and \( \boldsymbol w \).
This simple vector pathways question uses the key idea above:
$$
\begin{eqnarray}
\small \overrightarrow{\textsf{AC}} \normalsize &=& \small\overrightarrow{\textsf{AB}} \normalsize + \small \overrightarrow{\textsf{BC}} \normalsize \\[6pt]
\underline w &=& \underline u + \underline v \\[6pt]
\underline v &=& \underline w - \underline u
\end{eqnarray}
$$
Yes, it's that simple! Questions like this are only worth one mark.
Example 2 (non-calculator)
PQRS is a trapezium with \( \small \overrightarrow{\textsf{RQ}} \normalsize = 2\tiny\ \small \overrightarrow{\textsf{SP}} \normalsize \).
\( \small \overrightarrow{\textsf{PQ}} \normalsize \) and \( \small \overrightarrow{\textsf{RQ}} \normalsize \) represent vectors \( \boldsymbol u \) and \( \boldsymbol v \) respectively.
(a) Express \( \small \overrightarrow{\textsf{RP}} \normalsize \) in terms of \( \boldsymbol u \) and \( \boldsymbol v \).
(b) Express \( \small \overrightarrow{\textsf{RS}} \normalsize \) in terms of \( \boldsymbol u \) and \( \boldsymbol v \). Give your answer in its simplest form.
Note: the reason for the minus sign is that \( \small \overrightarrow{\textsf{QP}} \) and \( \small \overrightarrow{\textsf{PQ}} \) are in opposite directions.
(b) We need to decide a route from R to S. We could go around the outside, via Q and then P. But let's "cut the corner" and just go via P:
Find \( \vert \boldsymbol v \vert \), the magnitude of vector \( \boldsymbol v = \left( \begin{matrix} \phantom{-}45 \\ -28 \end{matrix} \right) \).
The magnitude is just the length of the hypotenuse of the right-angled triangle made from 45 right and 28 down. So the method here is really just 2D Pythagoras.
Find \( \vert \boldsymbol a - \boldsymbol b \vert \), where \( \boldsymbol a = \left( \begin{matrix} \phantom{.}5\phantom{.} \\ \phantom{.}6\phantom{.} \\ \end{matrix} \right) \) and \( \boldsymbol b = \left( \begin{matrix} \phantom{.}9\phantom{.} \\ \phantom{.}4\phantom{.} \\ \end{matrix} \right) \). Express your answer as a surd in its simplest form.
We need to find \( \underline a - \underline b \) in component form, and then calculate its magnitude.
(a) \( \small \overrightarrow{\textsf{AC}} \normalsize\ = \small \overrightarrow{\textsf{AO}} \normalsize\ + \small \overrightarrow{\textsf{OC}} \normalsize\) so it involves moving 8 in the negative \(x\)-direction and 6 in the positive \(y\)-direction.
\(\small \overrightarrow{\textsf{AB}}\) represents vector \(\boldsymbol u\) and \(\small \overrightarrow{\textsf{BC}}\) represents vector \(\boldsymbol v\small.\)
Express \(\small \overrightarrow{\textsf{BD}}\) in terms of \(\boldsymbol u\) and \(\boldsymbol v\small.\)
This was a simple 1-mark question.
$$
\begin{eqnarray}
\small\overrightarrow{\textsf{BD}}\normalsize&=& \small\overrightarrow{\textsf{BA}}\normalsize + \small\overrightarrow{\textsf{AD}}\normalsize\\[6pt]
&=& \small\overrightarrow{\textsf{BA}}\normalsize + \small\overrightarrow{\textsf{BC}}\normalsize \\[6pt]
&=& -\!\underline u + \underline v
\end{eqnarray}
$$
Note that \(\boldsymbol v - \boldsymbol u\) is also correct.
Example 10 (calculator)
SQA National 5 Maths 2018 P2 Q10
In the diagram below, \(\small \overrightarrow{\textsf{AB}}\) and \(\small \overrightarrow{\textsf{EA}}\) represent the vectors \(\boldsymbol u\) and \(\boldsymbol w\) respectively.
• \(\small \overrightarrow{\textsf{ED}}=2\small\,\overrightarrow{\textsf{AB}}\)
• \(\small \overrightarrow{\textsf{EA}}=2\small\,\overrightarrow{\textsf{DC}}\)
Express \(\small \overrightarrow{\textsf{BC}}\) in terms of \(\boldsymbol u\) and \(\boldsymbol w\small.\)
Give your answer in its simplest form.
This was a 2-mark question. We need to go the long way from B to C.
$$
\begin{eqnarray}
\small\overrightarrow{\textsf{BC}}\normalsize&=& \small\overrightarrow{\textsf{BA}}\normalsize + \small\overrightarrow{\textsf{AE}}\normalsize + \small\overrightarrow{\textsf{ED}}\normalsize + \small\overrightarrow{\textsf{DC}}\normalsize \\[6pt]
&=& -\!\underline u -\underline w + 2\underline u + \small\frac{1}{2}\normalsize\underline w \\[6pt]
&=& \underline u - \small\frac{1}{2}\normalsize\underline w
\end{eqnarray}
$$
The vectors \( \boldsymbol u \) and \( \boldsymbol v \) are shown in the diagram below.
Find the resultant vector \( \boldsymbol u - \boldsymbol v\small.\)
Express your answer in component form.
This question was worth 2 marks.
First we need to read the components of \( \underline u \) and \( \underline v \) from the diagram.
\( \underline u \) is 5 right and 2 up, so \( \underline u = \left( \begin{matrix} \phantom{.}5\phantom{.} \\ \phantom{.}2\phantom{.} \end{matrix} \right)\small.\)
\( \underline v \) is 3 right and 4 down, so \( \underline v = \left( \begin{matrix} \phantom{-}3 \\ -4 \end{matrix} \right)\small.\)
Now we just calculate \( \underline u - \underline v \) in the usual way.
\(\small\overrightarrow{\textsf{AB}}\normalsize =\boldsymbol u\) and \(\small\overrightarrow{\textsf{AC}}\normalsize=\boldsymbol t\small.\)
G is the point such that CG = \(\large\frac{1}{3}\)CB.
Express \(\small\overrightarrow{\textsf{AG}}\) in terms of \(\boldsymbol u\) and \(\boldsymbol t\small.\)
Give your answer in simplest form.
This vector pathways question was worth 3 marks, as it involves two steps before we can begin expressing the vector \(\small\overrightarrow{\textsf{AG}}\) in terms of \(\boldsymbol u\) and \(\boldsymbol t\small.\)
$$
\begin{eqnarray}
\small \overrightarrow{\textsf{AG}} \normalsize &=& \small\overrightarrow{\textsf{AC}} \normalsize + \small\overrightarrow{\textsf{CG}} \normalsize \\[6pt]
&=& \small\overrightarrow{\textsf{AC}} \normalsize + \small\frac{1}{3}\overrightarrow{\textsf{CB}} \normalsize \\[6pt]
&=& \small\overrightarrow{\textsf{AC}} \normalsize + \small\frac{1}{3}\left(\overrightarrow{\textsf{CA}}\normalsize + \small\overrightarrow{\textsf{AB}}\right) \\[6pt]
&=& \underline t \normalsize + \small\frac{1}{3}\normalsize\left(-\underline t + \underline u\right) \\[6pt]
&=& \underline t \normalsize - \small\frac{1}{3}\normalsize\underline t + \small\frac{1}{3}\normalsize\underline u \\[6pt]
&=& \small\frac{2}{3}\normalsize\underline t + \small\frac{1}{3}\normalsize\underline u \\[6pt]
\end{eqnarray}
$$
Need a Nat 5 Maths tutor?
Just use this handy little widget and our partner Bark.com will help you find one.
Click here to study the vectors notes on National5.com.
You may also want to check out their Nat 5 Maths self-study course. We think it's excellent. If you decide to enrol, just use coupon code "Maths.scot" to get £10 discount.
of a 2D or 3D vector.
