Date | May 2013 | Marks available | 2 | Reference code | 13M.1.hl.TZ1.2 |
Level | HL only | Paper | 1 | Time zone | TZ1 |
Command term | Find and Hence | Question number | 2 | Adapted from | N/A |
Question
Consider the points A(1, 2, 3), B(1, 0, 5) and C(2, −1, 4).
Find \(\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} \).
Hence find the area of the triangle ABC.
Markscheme
\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}}
1 \\
0 \\
5
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
1 \\
2 \\
3
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
0 \\
{ - 2} \\
2
\end{array}} \right)\), \(\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}}
2 \\
{ - 1} \\
4
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
1 \\
2 \\
3
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
1 \\
{ - 3} \\
1
\end{array}} \right)\) A1A1
Note: Award the above marks if the components are seen in the line below.
\(\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} = \left| {\begin{array}{*{20}{c}}
i&j&k \\
0&{ - 2}&2 \\
1&{ - 3}&1
\end{array}} \right| = \left( {\begin{array}{*{20}{c}}
4 \\
2 \\
2
\end{array}} \right)\) (M1)A1
[4 marks]
area \( = \frac{1}{2}\left| {\left( {\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} } \right)} \right|\) (M1)
\( = \frac{1}{2}\sqrt {{4^2} + {2^2} + {2^2}} = \frac{1}{2}\sqrt {24} {\text{ }}\left( { = \sqrt 6 } \right)\) A1
Note: Award M0A0 for attempts that do not involve the answer to (a).
[2 marks]
Examiners report
Candidates showed a good understanding of the vector techniques required in this question.
Candidates showed a good understanding of the vector techniques required in this question.