| Date | May 2016 | Marks available | 2 | Reference code | 16M.2.hl.TZ1.1 |
| Level | HL only | Paper | 2 | Time zone | TZ1 |
| Command term | Hence | Question number | 1 | Adapted from | N/A |
Question
The points A and B have position vectors \(\overrightarrow {{\text{OA}}} = \left\{ {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 2} \end{array}} \right\}\) and \(\overrightarrow {{\text{OB}}} = \left\{ {\begin{array}{*{20}{c}} 1 \\ 0 \\ 2 \end{array}} \right\}\).
Find \(\overrightarrow {{\text{OA}}} \times \overrightarrow {{\text{OB}}} \).
Hence find the area of the triangle OAB.
Markscheme
\(\overrightarrow {{\text{OA}}} \times \overrightarrow {{\text{OB}}} = \left( {\begin{array}{*{20}{c}} 4 \\ { - 4} \\ { - 2} \end{array}} \right)\) (M1)A1
Note: M1A0 can be awarded for attempt at a correct method shown, or correct method implied by the digits 4, 4, 2 found in the correct order.
[2 marks]
\({\text{area}} = \frac{1}{2}\sqrt {{4^2} + {4^2} + {2^2}} = 3\) M1A1
[2 marks]
Examiners report
Generally well done. Most students were able to obtain full marks on this question. Most of the errors made were due to careless mistakes.
Generally well done. Most students were able to obtain full marks on this question. Most of the errors made were due to careless mistakes. A few students did not take notice of the “hence” in part (b) and were consequently not able to obtain the marks.
