| Date | May 2013 | Marks available | 3 | Reference code | 13M.2.sl.TZ2.8 |
| Level | SL only | Paper | 2 | Time zone | TZ2 |
| Command term | Find | Question number | 8 | Adapted from | N/A |
Question
Consider the points A(\(5\), \(2\), \(1\)) , B(\(6\), \(5\), \(3\)) , and C(\(7\), \(6\), \(a + 1\)) , \(a \in{\mathbb{R}}\) .
Let \({\rm{q}}\) be the angle between \(\overrightarrow {{\rm{AB}}} \) and \(\overrightarrow {{\rm{AC}}} \) .
Find
(i) \(\overrightarrow {{\rm{AB}}} \) ;
(ii) \(\overrightarrow {{\rm{AC}}} \) .
Find the value of \(a\) for which \({\rm{q}} = \frac{\pi }{2}\) .
i. Show that \(\cos q = \frac{{2a + 14}}{{\sqrt {14{a^2} + 280} }}\) .
ii. Hence, find the value of a for which \({\rm{q}} = 1.2\) .
Hence, find the value of a for which \({\rm{q}} = 1.2\) .
Markscheme
(i) appropriate approach (M1)
eg \(\overrightarrow {{\rm{AO}}} {\rm{ + }}\overrightarrow {{\rm{OB}}} \) , \({\rm{B}} - {\rm{A}}\)
\(\overrightarrow {{\rm{AB}}} = \left( \begin{array}{l}
1\\
3\\
2
\end{array} \right)\) A1 N2
(ii) \(\overrightarrow {{\rm{AC}}} = \left( \begin{array}{l}
2\\
4\\
a
\end{array} \right)\) A1 N1
[3 marks]
valid reasoning (seen anywhere) R1
eg scalar product is zero, \(\cos \frac{\pi }{2} = \frac{{\boldsymbol{u} \bullet \boldsymbol{v}}}{{\left| \boldsymbol{u} \right|\left| \boldsymbol{v} \right|}}\)
correct scalar product of their \(\overrightarrow {{\rm{AB}}} \) and \(\overrightarrow {{\rm{AC}}} \) (may be seen in part (c)) (A1)
eg \(1(2) + 3(4) + 2(a)\)
correct working for their \(\overrightarrow {{\rm{AB}}} \) and \(\overrightarrow {{\rm{AC}}} \) (A1)
eg \(2a + 14\) , \(2a = - 14\)
\(a = - 7\) A1 N3
[4 marks]
correct magnitudes (may be seen in (b)) (A1)(A1)
\(\sqrt {{1^2} + {3^2} + {2^2}} \left( { = \sqrt {14} } \right)\) , \(\sqrt {{2^2} + {4^2} + {a^2}} \left( { = \sqrt {20 + {a^2}} } \right)\)
substitution into formula (M1)
eg \(\cos \theta \frac{{1 \times 2 + 3 \times 4 + 2 \times a}}{{\sqrt {{1^2} + {3^2} + {2^2}} \sqrt {{2^2} + {4^2} + {a^2}} }}\) , \(\frac{{14 + 2a}}{{\sqrt {14} \sqrt {4 + 16 + {a^2}} }}\)
simplification leading to required answer A1
eg \(\cos \theta = \frac{{14 + 2a}}{{\sqrt {14} \sqrt {20 + {a^2}} }}\)
\(\cos \theta = \frac{{2a + 14}}{{\sqrt {14{a^2} + 280} }}\) AG N0
[4 marks]
correct setup (A1)
eg \(\cos 1.2 = \frac{{2a + 14}}{{\sqrt {14{a^2} + 280} }}\)
valid attempt to solve (M1)
eg sketch, \(\frac{{2a + 14}}{{\sqrt {14{a^2} + 280} }} - \cos 1.2 = 0\) , attempt to square
\(a = - 3.25\) A2 N3
[4 marks]
correct setup (A1)
eg \(\cos 1.2 = \frac{{2a + 14}}{{\sqrt {14{a^2} + 280} }}\)
valid attempt to solve (M1)
eg sketch, \(\frac{{2a + 14}}{{\sqrt {14{a^2} + 280} }} - \cos 1.2 = 0\) , attempt to square
\(a = - 3.25\) A2 N3
[4 marks]
Examiners report
The majority of candidates successfully found the vectors between the given points in part (a).
In part (b), while most candidates correctly found the value of \(a\), many unnecessarily worked with the magnitudes of the vectors, sometimes leading to algebra errors.
Some candidates showed a minimum of working in part (c)(i); in a “show that” question, candidates need to ensure that their working clearly leads to the answer given. A common error was simplifying the magnitude of vector AC to \(\sqrt {20{a^2}} \) instead of \(\sqrt {20 + {a^2}} \) .
In part (c)(ii), a disappointing number of candidates embarked on a usually fruitless quest for an algebraic solution rather than simply solving the resulting equation with their GDC. Many of these candidates showed quite weak algebra manipulation skills, with errors involving the square root occurring in a myriad of ways.
In part (c)(ii), a disappointing number of candidates embarked on a usually fruitless quest for an algebraic solution rather than simply solving the resulting equation with their GDC. Many of these candidates showed quite weak algebra manipulation skills, with errors involving the square root occurring in a myriad of ways.
