| Date | November 2008 | Marks available | 4 | Reference code | 08N.1.sl.TZ0.2 |
| Level | SL only | Paper | 1 | Time zone | TZ0 |
| Command term | Find | Question number | 2 | Adapted from | N/A |
Question
A particle is moving with a constant velocity along line L . Its initial position is A(6 , −2 , 10) . After one second the particle has moved to B( 9, −6 , 15) .
(i) Find the velocity vector, \(\overrightarrow {{\rm{AB}}} \) .
(ii) Find the speed of the particle.
Write down an equation of the line L .
Markscheme
(i) evidence of approach (M1)
e.g. \(\overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OB}}} \) , \({\rm{B}} - {\rm{A}}\) , \(\left( {\begin{array}{*{20}{c}}
{9 - 6}\\
{ - 6 + 2}\\
{15 - 10}
\end{array}} \right)\)
\(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
3\\
{ - 4}\\
5
\end{array}} \right)\) (accept \(\left( {3, - 4, 5} \right)\) ) A1 N2
(ii) evidence of finding the magnitude of the velocity vector M1
e.g. \({\text{speed}} = \sqrt {{3^2} + {4^2} + {5^2}} \)
\({\text{speed}} = \sqrt {50} \) \(\left( { = 5\sqrt 2 } \right)\) A1 N1
[4 marks]
correct equation (accept Cartesian and parametric forms) A2 N2
e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
6\\
{ - 2}\\
{10}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
3\\
{ - 4}\\
5
\end{array}} \right)\) , \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
9\\
{ - 6}\\
{15}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
3\\
{ - 4}\\
5
\end{array}} \right)\)
[2 marks]
Examiners report
This question was quite well done. Marks were lost when candidates found the vector \(\overrightarrow {{\text{BA}}} \) instead of \(\overrightarrow {{\text{AB}}} \) in part (a) and for not writing their vector equation as an equation.
In part (b), a few candidates switched the position and velocity vectors or used the vectors \(\overrightarrow {{\text{OA}}} \) and \(\overrightarrow {{\text{OB}}} \) to incorrectly write the vector equation.
