Date | May 2009 | Marks available | 6 | Reference code | 09M.2.hl.TZ1.5 |
Level | HL only | Paper | 2 | Time zone | TZ1 |
Command term | Find | Question number | 5 | Adapted from | N/A |
Question
Find the angle between the lines \(\frac{{x - 1}}{2} = 1 - y = 2z\) and \(x = y = 3z\) .
Markscheme
consider a vector parallel to each line,
e.g. \({\boldsymbol{u}} = \left( {\begin{array}{*{20}{c}}
4 \\
{ - 2} \\
1
\end{array}} \right)\) and \({\boldsymbol{v}} = \left( {\begin{array}{*{20}{c}}
3 \\
3 \\
1
\end{array}} \right)\) A1A1
let \(\theta \) be the angle between the lines
\(\cos \theta = \frac{{\left| {{\boldsymbol{u \times v}}} \right|}}{{\left| {\boldsymbol{u}} \right|\left| {\boldsymbol{v}} \right|}} = \frac{{\left| {12 - 6 + 1} \right|}}{{\sqrt {21} \sqrt {19} }}\) M1A1
\( = \frac{7}{{\sqrt {21} \sqrt {19} }} = 0.350...\) (A1)
so \(\theta = 69.5\) \(\left( {{\text{or }}1.21{\text{ rad or }}\arccos \left( {\frac{7}{{\sqrt {21} \sqrt {19} }}} \right)} \right)\) A1 N4
Note: Allow FT from incorrect reasonable vectors.
[6 marks]
Examiners report
Most students knew how to find the angle between two vectors, although many could not find the correct two direction vectors.