Date | November 2016 | Marks available | 4 | Reference code | 16N.1.sl.TZ0.4 |
Level | SL only | Paper | 1 | Time zone | TZ0 |
Command term | Find | Question number | 4 | Adapted from | N/A |
Question
The position vectors of points P and Q are i \( + \) 2 j \( - \) k and 7i \( + \) 3j \( - \) 4k respectively.
Find a vector equation of the line that passes through P and Q.
The line through P and Q is perpendicular to the vector 2i \( + \) nk. Find the value of \(n\).
Markscheme
valid attempt to find direction vector (M1)
eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{PQ}}} ,{\text{ }}\overrightarrow {{\text{QP}}} \)
correct direction vector (or multiple of) (A1)
eg\(\,\,\,\,\,\)6i \( + \) j \( - \) 3k
any correct equation in the form r \( = \) a \( + \) tb (any parameter for \(t\)) A2 N3
where a is i \( + \) 2j \( - \) k or 7i \( + \) 3j \( - \) 4k , and b is a scalar multiple of 6i \( + \) j \( - \) 3k
eg\(\,\,\,\,\,\)r \( = \) 7i \( + \) 3j \( - \) 4k \( + \) t(6i \( + \) j \( - \) 3k), r \( = \left( {\begin{array}{*{20}{c}} {1 + 6s} \\ {2 + 1s} \\ { - 1 - 3s} \end{array}} \right),{\text{ }}r = \left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 1} \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} { - 6} \\ { - 1} \\ 3 \end{array}} \right)\)
Notes: Award A1 for the form a \( + \) tb, A1 for the form L \( = \) a \( + \) tb, A0 for the form r \( = \) b \( + \) ta.
[4 marks]
correct expression for scalar product (A1)
eg\(\,\,\,\,\,\)\(6 \times 2 + 1 \times 0 + ( - 3) \times n,{\text{ }} - 3n + 12\)
setting scalar product equal to zero (seen anywhere) (M1)
eg\(\,\,\,\,\,\)u \( \bullet \) v \( = 0,{\text{ }} - 3n + 12 = 0\)
\(n = 4\) A1 N2
[3 marks]