| Date | May 2018 | Marks available | 4 | Reference code | 18M.2.hl.TZ1.11 |
| Level | HL only | Paper | 2 | Time zone | TZ1 |
| Command term | Write down and Show that | Question number | 11 | Adapted from | N/A |
Question
Two submarines A and B have their routes planned so that their positions at time t hours, 0 ≤ t < 20 , would be defined by the position vectors rA \( = \left( \begin{gathered}
\,2 \hfill \\
\,4 \hfill \\
- 1 \hfill \\
\end{gathered} \right) + t\left( \begin{gathered}
- 1 \hfill \\
\,1 \hfill \\
- 0.15 \hfill \\
\end{gathered} \right)\) and rB \( = \left( \begin{gathered}
\,0 \hfill \\
\,3.2 \hfill \\
- 2 \hfill \\
\end{gathered} \right) + t\left( \begin{gathered}
- 0.5 \hfill \\
\,1.2 \hfill \\
\,0.1 \hfill \\
\end{gathered} \right)\) relative to a fixed point on the surface of the ocean (all lengths are in kilometres).
To avoid the collision submarine B adjusts its velocity so that its position vector is now given by
rB \( = \left( \begin{gathered}
\,0 \hfill \\
\,3.2 \hfill \\
- 2 \hfill \\
\end{gathered} \right) + t\left( \begin{gathered}
- 0.45 \hfill \\
\,1.08 \hfill \\
\,0.09 \hfill \\
\end{gathered} \right)\).
Show that the two submarines would collide at a point P and write down the coordinates of P.
Show that submarine B travels in the same direction as originally planned.
Find the value of t when submarine B passes through P.
Find an expression for the distance between the two submarines in terms of t.
Find the value of t when the two submarines are closest together.
Find the distance between the two submarines at this time.
Markscheme
rA = rB (M1)
2 − t = − 0.5t ⇒ t = 4 A1
checking t = 4 satisfies 4 + t = 3.2 + 1.2t and − 1 − 0.15t = − 2 + 0.1t R1
P(−2, 8, −1.6) A1
Note: Do not award final A1 if answer given as column vector.
[4 marks]
\(0.9 \times \left( \begin{gathered}
- 0.5 \hfill \\
\,1.2 \hfill \\
\,0.1 \hfill \\
\end{gathered} \right) = \left( \begin{gathered}
- 0.45 \hfill \\
\,1.08 \hfill \\
\,0.09 \hfill \\
\end{gathered} \right)\) A1
Note: Accept use of cross product equalling zero.
hence in the same direction AG
[1 mark]
\(\left( \begin{gathered}
\, - 0.45t \hfill \\
3.2 + 1.08t \hfill \\
- 2 + 0.09t \hfill \\
\end{gathered} \right) = \left( \begin{gathered}
- 2 \hfill \\
\,8 \hfill \\
- 1.6 \hfill \\
\end{gathered} \right)\) M1
Note: The M1 can be awarded for any one of the resultant equations.
\( \Rightarrow t = \frac{{40}}{9} = 4.44 \ldots \) A1
[2 marks]
rA − rB = \(\left( \begin{gathered}
\,2 - t \hfill \\
\,4 + t \hfill \\
- 1 - 0.15t \hfill \\
\end{gathered} \right) - \left( \begin{gathered}
\, - 0.45t \hfill \\
3.2 + 1.08t \hfill \\
- 2 + 0.09t \hfill \\
\end{gathered} \right)\) (M1)(A1)
\( = \left( \begin{gathered}
\,2 - 0.55t \hfill \\
\,0.8 - 0.08t \hfill \\
1 - 0.24t \hfill \\
\end{gathered} \right)\) (A1)
Note: Accept rA − rB.
distance \(D = \sqrt {{{\left( {2 - 0.55t} \right)}^2} + {{\left( {0.8 - 0.08t} \right)}^2} + {{\left( {1 - 0.24t} \right)}^2}} \) M1A1
\(\left( { = \sqrt {8.64 - 2.688t + 0.317{t^2}} } \right)\)
[5 marks]
minimum when \(\frac{{{\text{d}}D}}{{{\text{d}}t}} = 0\) (M1)
t = 3.83 A1
[2 marks]
0.511 (km) A1
[1 mark]
