Question 1 Report
The Kalu family walks between home \(H\), shop \(S\) and bank \(B\) to save fares. In km from \(O\): \(\vec{OH} = \begin{pmatrix} 0 \\ 2 \end{pmatrix}\), \(\vec{OS} = \begin{pmatrix} 5 \\ 6 \end{pmatrix}\), \(\vec{OB} = \begin{pmatrix} 9 \\ 0 \end{pmatrix}\).
This question tests using column vectors between named points to find a displacement and its magnitude, then comparing that direct distance with a given two-stage route.
(a) The vector from \(H\) to \(S\) is found by subtracting the position vector of \(H\) from that of \(S\):
\[\vec{HS}=\vec{OS}-\vec{OH}=\begin{pmatrix}5\\6\end{pmatrix}-\begin{pmatrix}0\\2\end{pmatrix}=\begin{pmatrix}5-0\\6-2\end{pmatrix}=\begin{pmatrix}5\\4\end{pmatrix} \text{ km <b>[1]</b>}\](b) The same method gives the vector from \(H\) to \(B\):
\[\vec{HB}=\vec{OB}-\vec{OH}=\begin{pmatrix}9\\0\end{pmatrix}-\begin{pmatrix}0\\2\end{pmatrix}=\begin{pmatrix}9\\-2\end{pmatrix} \text{ km}\]Its magnitude, the straight-line distance from home to the bank, is found by Pythagoras' theorem:
\[|\vec{HB}|=\sqrt{9^2+(-2)^2}=\sqrt{81+4}=\sqrt{85}=9.22 \text{ km (3 s.f.) <b>[2] (1 for } \sqrt{85}\text{, 1 for the rounded value)</b>}\](c) The saving is the difference between the two-stage route and the direct route, using the more precise unrounded value for \(|\vec{HB}|\) to avoid compounding rounding error:
\[13.6-\sqrt{85}=13.6-9.2195\ldots=4.38 \text{ km (3 s.f.) <b>[2]</b>}\]Subtracting the two vectors, as in parts (a) and (b), gives the actual displacement between two points regardless of the path taken to get there; comparing its magnitude with a longer, indirect route is exactly how "distance saved by a shortcut" problems are solved.
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