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.
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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 \(\boldsymbol t\small.\)
$$
\begin{eqnarray}
\small \overrightarrow{\textsf{AG}} \normalsize &=& \small\overrightarrow{\textsf{AC}} \normalsize + \small\overrightarrow{\textsf{CG}} \normalsize \\[6pt]
&=& \small\overrightarrow{\textsf{AC}} \normalsize + \small\frac{1}{3}\overrightarrow{\textsf{CB}} \normalsize \\[6pt]
&=& \small\overrightarrow{\textsf{AC}} \normalsize + \small\frac{1}{3}\left(\overrightarrow{\textsf{CA}}\normalsize + \small\overrightarrow{\textsf{AB}}\right) \\[6pt]
&=& \underline t \normalsize + \small\frac{1}{3}\normalsize\left(-\underline t + \underline u\right) \\[6pt]
&=& \underline t \normalsize - \small\frac{1}{3}\normalsize\underline t + \small\frac{1}{3}\normalsize\underline u \\[6pt]
&=& \small\frac{2}{3}\normalsize\underline t + \small\frac{1}{3}\normalsize\underline u \\[6pt]
\end{eqnarray}
$$
Example 14 (non-calculator)
SQA National 5 Maths 2024 P1 Q4
Given \( \boldsymbol a = \left( \begin{matrix} \phantom{-}3 \\ \phantom{-}4 \\ -1 \end{matrix} \right) \) and \( \boldsymbol b = \left( \begin{matrix} 5 \\ 3 \\ 2 \end{matrix} \right)\small, \) find the resultant vector \(3\boldsymbol a + \boldsymbol b\small.\) Express your answer in component form.
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of a 2D or 3D vector.
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