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Question
Find the angle between the lines \(\frac{{x – 1}}{2} = 1 – y = 2z\) and \(x = y = 3z\) .
Answer/Explanation
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.
Question
Find the acute angle between the planes with equations \(x + y + z = 3\) and \(2x – z = 2\).
Answer/Explanation
Markscheme
n\(_1 = \left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ 1 \end{array}} \right)\) and n\(_2 = \left( {\begin{array}{*{20}{c}} 2 \\ 0 \\ { – 1} \end{array}} \right)\) (A1)(A1)
EITHER
\(\theta = \arccos \left( {\frac{{{n_1} \bullet {n_2}}}{{\left| {{n_1}} \right|\left| {{n_2}} \right|}}} \right)\left( {\cos \theta = \frac{{{n_1} \bullet {n_2}}}{{\left| {{n_1}} \right|\left| {{n_2}} \right|}}} \right)\) (M1)
\( = \arccos \left( {\frac{{2 + 0 – 1}}{{\sqrt 3 \sqrt 5 }}} \right)\left( {\cos \theta = \frac{{2 + 0 – 1}}{{\sqrt 3 \sqrt 5 }}} \right)\) (A1)
\( = \arccos \left( {\frac{1}{{\sqrt {15} }}} \right)\left( {\cos \theta = \frac{1}{{\sqrt {15} }}} \right)\)
OR
\(\theta = \arcsin \left( {\frac{{\left| {{n_1} \times {n_2}} \right|}}{{\left| {{n_1}} \right|\left| {{n_2}} \right|}}} \right)\left( {\sin \theta = \frac{{\left| {{n_1} \times {n_2}} \right|}}{{\left| {{n_1}} \right|\left| {{n_2}} \right|}}} \right)\) (M1)
\( = \arcsin \left( {\frac{{\sqrt {14} }}{{\sqrt 3 \sqrt 5 }}} \right)\left( {\sin \theta = \frac{{\sqrt {14} }}{{\sqrt 3 \sqrt 5 }}} \right)\) (A1)
\( = \arcsin \left( {\frac{{\sqrt {14} }}{{\sqrt {15} }}} \right)\left( {\sin \theta = \frac{{\sqrt {14} }}{{\sqrt {15} }}} \right)\)
THEN
\( = 75.0^\circ {\text{ (or 1.31)}}\) A1
[5 marks]