Find the position vector $\overrightarrow{\text{OS}}$ of the ship at time $t$ hours.

Find the value of $t$ when the ship will be closest to the lighthouse.

An alarm will sound if the ship travels within $20$ kilometres of the lighthouse.

State whether the alarm will sound. Give a reason for your answer.

$\overrightarrow{\text{OS}}=\left(\begin{array}{c}300\\ 100\end{array}\right)+t\left(\begin{array}{c}-12\\ 15\end{array}\right)$             A1

[1 mark]

attempt to find the vector from $\text{L}$ to $\text{S}$           (M1)

$\overrightarrow{\text{LS}}=\left(\begin{array}{c}171\\ -183\end{array}\right)+t\left(\begin{array}{c}-12\\ 15\end{array}\right)$             A1

EITHER

$\left|\overrightarrow{\text{LS}}\right|=\sqrt{{\left(171-12t\right)}^2+{\left(15t-183\right)}^2}$           (M1)(A1)

minimize to find $t$ on GDC           (M1)

OR

$\text{S}$ closest when $\overrightarrow{\text{LS}}\mathbf{·}\left(\begin{array}{c}-12\\ 15\end{array}\right)=0$           (M1)

$\left(\left(\begin{array}{c}171\\ -183\end{array}\right)+t\left(\begin{array}{c}-12\\ 15\end{array}\right)\right)\mathbf{·}\left(\begin{array}{c}-12\\ 15\end{array}\right)=0$

$-2052+144t-2745+225t=0$           (M1)(A1)

OR

$\text{S}$ closest when $\overrightarrow{\text{LS}}\mathbf{·}\left(\begin{array}{c}-12\\ 15\end{array}\right)=0$           (M1)

$\overrightarrow{\text{LS}}=\left(\begin{array}{c}5k\\ 4k\end{array}\right)$

$\overrightarrow{\text{OS}}=\left(\begin{array}{c}129+5k\\ 283+4k\end{array}\right)$           (A1)

$\left(\begin{array}{c}129+5k\\ 283+4k\end{array}\right)=\left(\begin{array}{c}300-12t\\ 100+15t\end{array}\right)$

Solving simultaneously            (M1)

THEN

$t=13$             A1

[6 marks]

the alarm will sound            A1

$\left|\overrightarrow{\text{LS}}\right|=19.2\dots <20$            R1

Note: Do not award A1R0.

[2 marks]

Part (a) was generally well answered. Following that there were several possible approaches to this question. It was clear that many candidates understood that the closest approach could be found using the scalar product to find orthogonal vectors. However, typical efforts involved attempting to find the value of $t$ by assuming that $\overrightarrow{\mathrm{OS}}$ and $\overrightarrow{\mathrm{OL}}$ are perpendicular. Other methods incorrectly applied were equating $\overrightarrow{\mathrm{OS}}$ and $\overrightarrow{\mathrm{OL}}$. Of course, this led to separate values of $t$ for each component. The method of minimizing $\left|\overrightarrow{\mathrm{LS}}\right|$ was not commonly employed.

Part (a) was generally well answered. Following that there were several possible approaches to this question. It was clear that many candidates understood that the closest approach could be found using the scalar product to find orthogonal vectors. However, typical efforts involved attempting to find the value of $t$ by assuming that $\overrightarrow{\mathrm{OS}}$ and $\overrightarrow{\mathrm{OL}}$ are perpendicular. Other methods incorrectly applied were equating $\overrightarrow{\mathrm{OS}}$ and $\overrightarrow{\mathrm{OL}}$. Of course, this led to separate values of $t$ for each component. The method of minimizing $\left|\overrightarrow{\mathrm{LS}}\right|$ was not commonly employed.

Part (a) was generally well answered. Following that there were several possible approaches to this question. It was clear that many candidates understood that the closest approach could be found using the scalar product to find orthogonal vectors. However, typical efforts involved attempting to find the value of $t$ by assuming that $\overrightarrow{\mathrm{OS}}$ and $\overrightarrow{\mathrm{OL}}$ are perpendicular. Other methods incorrectly applied were equating $\overrightarrow{\mathrm{OS}}$ and $\overrightarrow{\mathrm{OL}}$. Of course, this led to separate values of $t$ for each component. The method of minimizing $\left|\overrightarrow{\mathrm{LS}}\right|$ was not commonly employed.