16N.1.AHL.TZ0.H_8

pestleMathematicsAAHLPaper 116N· ahl-3-14-vector-equation-of-linesource ↗

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 $a$ ;

[4]

determine the coordinates of the point of intersection P.

[2]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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]

Examiners’ report

[N/A]