National 5 Maths
Vectors

This simple vector pathways question uses the key idea above:

Yes, it's that simple! Questions like this are only worth one mark.

(a)  Using the key idea at the top:

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:

These vectors are in "component form".

The top value is the \(x\)-component: how far left (–) or right (+). The bottom value is the \(y\)-component: how far up (+) or down (–).

The method here is just add or subtract the \(x\)- and \(y\)-components separately.

This example includes \(z\)-components: the third dimension.

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.

The method here is really just 3D Pythagoras.

We need to find \( \underline a - \underline b \) in component form, and then calculate its magnitude.

This example links with the 3D Coordinates topic.

(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.

So \( \small \overrightarrow{\textsf{AC}} \normalsize\ = \left( \begin{matrix} -8 \\ \phantom{-}6 \\ \phantom{-}0 \end{matrix} \right)\).

(b) The centre of the base is (4,3,0) so D is (4,3,7).

So the position vector \( \small \overrightarrow{\textsf{OD}} \normalsize\ = \left( \begin{matrix} 4 \\ 3 \\ 7 \end{matrix} \right)\).

\begin{eqnarray} \small \overrightarrow{\textsf{AD}} \normalsize &=& \small \overrightarrow{\textsf{AO}} \normalsize + \small \overrightarrow{\textsf{OD}} \normalsize \\[6pt] &=& \left( \begin{matrix} -8 \\ \phantom{-}0 \\ \phantom{-}0 \end{matrix} \right) + \left( \begin{matrix} 4 \\ 3 \\ 7 \end{matrix} \right)\\[6pt] &=& \left( \begin{matrix} -4 \\ \phantom{-}3 \\ \phantom{-}7 \end{matrix} \right) \end{eqnarray}

(c) Now we calculate the magnitude of our answer to the previous part.

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.

This was a 2-mark question. We need to go the long way from B to C.

$$ \begin{eqnarray} \small\overrightarrow{\textsf{BC}}\normalsize&=& \small\overrightarrow{\textsf{BA}}\normalsize + \small\overrightarrow{\textsf{AE}}\normalsize + \small\overrightarrow{\textsf{ED}}\normalsize + \small\overrightarrow{\textsf{DC}}\normalsize \\[6pt] &=& -\!\underline u -\underline w + 2\underline u + \small\frac{1}{2}\normalsize\underline w \\[6pt] &=& \underline u - \small\frac{1}{2}\normalsize\underline w \end{eqnarray} $$

(a)  This part was only worth 1 mark. We will go from P to Q via R.

$$ \begin{eqnarray} \small\overrightarrow{\textsf{PQ}}\normalsize&=& \small\overrightarrow{\textsf{PR}}\normalsize + \small\overrightarrow{\textsf{RQ}}\normalsize\\[6pt] &=& \left( \begin{matrix} \phantom{-}6 \\ -4 \end{matrix} \right) + \left( \begin{matrix} -1 \\ \phantom{-}8 \end{matrix} \right) \\[6pt] &=& \left( \begin{matrix} \phantom{.}5\phantom{.} \\ \phantom{.}4\phantom{.} \end{matrix} \right) \end{eqnarray} $$

(b)  In this 2-mark part, we could go via either P or R.

The answer below goes via R. You might like to try it the other way yourself.

$$ \begin{eqnarray} \small\overrightarrow{\textsf{MQ}}\normalsize&=& \small\overrightarrow{\textsf{MR}}\normalsize + \small\overrightarrow{\textsf{RQ}}\normalsize\\[6pt] &=& \small\frac{1}{2}\,\normalsize \small\overrightarrow{\textsf{PR}}\normalsize + \small\overrightarrow{\textsf{RQ}}\normalsize\\[6pt] &=& \small\frac{1}{2}\normalsize \left( \begin{matrix} \phantom{-}6 \\ -4 \end{matrix} \right) + \left( \begin{matrix} -1 \\ \phantom{-}8 \end{matrix} \right)\\[6pt] &=& \left( \begin{matrix} \phantom{-}3 \\ -2 \end{matrix} \right) + \left( \begin{matrix} -1 \\ \phantom{-}8 \end{matrix} \right)\\[6pt] &=& \left( \begin{matrix} \phantom{.}2\phantom{.} \\ \phantom{.}6\phantom{.} \end{matrix} \right) \end{eqnarray} $$

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.

$$ \begin{eqnarray} \underline u - \underline v &=& \left( \begin{matrix} \phantom{.}5\phantom{.} \\ \phantom{.}2\phantom{.} \end{matrix} \right) - \left( \begin{matrix} \phantom{-}3 \\ -4 \end{matrix} \right)\\[6pt] &=& \left( \begin{matrix} \phantom{.}2\phantom{.} \\ \phantom{.}6\phantom{.} \end{matrix} \right) \end{eqnarray} $$

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 \(\boldsymbol t\small.\)

$$ \begin{eqnarray} \small \overrightarrow{\textsf{AG}} \normalsize &=& \small\overrightarrow{\textsf{AC}} \normalsize + \small\overrightarrow{\textsf{CG}} \normalsize \\[6pt] &=& \small\overrightarrow{\textsf{AC}} \normalsize + \small\frac{1}{3}\overrightarrow{\textsf{CB}} \normalsize \\[6pt] &=& \small\overrightarrow{\textsf{AC}} \normalsize + \small\frac{1}{3}\left(\overrightarrow{\textsf{CA}}\normalsize + \small\overrightarrow{\textsf{AB}}\right) \\[6pt] &=& \underline t \normalsize + \small\frac{1}{3}\normalsize\left(-\underline t + \underline u\right) \\[6pt] &=& \underline t \normalsize - \small\frac{1}{3}\normalsize\underline t + \small\frac{1}{3}\normalsize\underline u \\[6pt] &=& \small\frac{2}{3}\normalsize\underline t + \small\frac{1}{3}\normalsize\underline u \\[6pt] \end{eqnarray} $$

$$ \begin{eqnarray} 3\underline a + \underline b \:&=&\: 3 \left( \begin{matrix} \phantom{-}3 \\ \phantom{-}4 \\ -1 \end{matrix} \right) + \left( \begin{matrix} 5 \\ 3 \\ 2 \end{matrix} \right) \\[6pt] &=&\: \left( \begin{matrix} \phantom{-}9 \\ \phantom{-}12 \\ -3 \end{matrix} \right) + \left( \begin{matrix} 5 \\ 3 \\ 2 \end{matrix} \right)\\[6pt] &=&\: \left( \begin{matrix} 14 \\ 15 \\ -1 \end{matrix} \right) \\[15pt] \end{eqnarray} $$