| Date | November 2016 | Marks available | 2 | Reference code | 16N.1.hl.TZ0.8 |
| Level | HL only | Paper | 1 | Time zone | TZ0 |
| Command term | Determine | Question number | 8 | Adapted from | N/A |
Question
Consider the lines \({l_1}\) and \({l_2}\) defined by
\({l_1}:\) r \( = \left( {\begin{array}{*{20}{c}} { - 3} \\ { - 2} \\ a \end{array}} \right) + \beta \left( {\begin{array}{*{20}{c}} 1 \\ 4 \\ 2 \end{array}} \right)\) and \({l_2}:\frac{{6 - x}}{3} = \frac{{y - 2}}{4} = 1 - z\) where \(a\) is a constant.
Given that the lines \({l_1}\) and \({l_2}\) intersect at a point P,
find the value of \(a\);
determine the coordinates of the point of intersection P.
Markscheme
METHOD 1
\({l_1}:\)r \( = \left( {\begin{array}{*{20}{c}} { - 3} \\ { - 2} \\ a \end{array}} \right) = \beta \left( {\begin{array}{*{20}{c}} 1 \\ 4 \\ 2 \end{array}} \right) \Rightarrow \left\{ {\begin{array}{*{20}{l}} {x = - 3 + \beta } \\ {y = - 2 + 4\beta } \\ {z = a + 2\beta } \end{array}} \right.\) M1
\(\frac{{6 - ( - 3 + \beta )}}{3} = \frac{{( - 2 + 4\beta ) - 2}}{4} \Rightarrow 4 = \frac{{4\beta }}{3} \Rightarrow \beta = 3\) M1A1
\(\frac{{6 - ( - 3 + \beta )}}{3} = 1 - (a + 2\beta ) \Rightarrow 2 = - 5 - a \Rightarrow a = - 7\) A1
METHOD 2
\(\left\{ {\begin{array}{*{20}{l}} { - 3 + \beta = 6 - 3\lambda } \\ { - 2 + 4\beta = 4\lambda + 2} \\ {a + 2\beta = 1 - \lambda } \end{array}} \right.\) M1
attempt to solve M1
\(\lambda = 2,{\text{ }}\beta = 3\) A1
\(a = 1 - \lambda - 2\beta = - 7\) A1
[4 marks]
\(\overrightarrow {{\text{OP}}} = \left( {\begin{array}{*{20}{c}} { - 3} \\ { - 2} \\ { - 7} \end{array}} \right) + 3 \bullet \left( {\begin{array}{*{20}{c}} 1 \\ 4 \\ 2 \end{array}} \right)\) (M1)
\( = \left( {\begin{array}{*{20}{c}} 0 \\ {10} \\ { - 1} \end{array}} \right)\) A1
\(\therefore {\text{P}}(0,{\text{ 10, }} - 1)\)
[2 marks]
