Show that the lines ${L_1}$ and ${L_2}$ are skew.

Find the acute angle between the lines ${L_1}$ and ${L_2}$.

(i)     Find a vector perpendicular to both lines.

(ii)     Hence determine an equation of the line ${L_3}$ that is perpendicular to both ${L_1}$ and ${L_2}$ and intersects both lines.

${L_1}$ and ${L_2}$ are not parallel, since $\left( \begin{array}{l}  - 1\\ 1\\ 2 \end{array} \right) \ne k\left( \begin{array}{l} 2\\ 1\\ 6 \end{array} \right)$     R1

if they meet, then $1 - \lambda  = 1 + 2\mu$ and $2 + \lambda  = 2 + \mu$     M1

solving simultaneously $\Rightarrow \lambda  = \mu  = 0$     A1

$2 + 2\lambda  = 4 + 6\mu  \Rightarrow 2 \ne 4$ contradiction,     R1

so lines are skew     AG

 

Note:     Do not award the second R1 if their values of parameters are incorrect.

[4 marks]

$\left( \begin{array}{l}  - 1\\ 1\\ 2 \end{array} \right) \bullet \left( \begin{array}{l} 2\\ 1\\ 6 \end{array} \right)( = 11) = \sqrt 6 \sqrt {41} \cos \theta$     M1A1

$\cos \theta  = \frac{{11}}{{\sqrt {246} }}$     (A1)

$\theta  = 45.5^\circ \;\;\;(0.794{\text{ radians}})$     A1

[4 marks]

(i)     $\left( {\begin{array}{*{20}{c}}   { - 1} \\   1 \\   2 \end{array}} \right) \times \left( {\begin{array}{*{20}{c}}   2 \\   1 \\   6 \end{array}} \right) = \left( {\begin{array}{*{20}{c}}   {6 - 2} \\   {4 + 6} \\   { - 1 - 2} \end{array}} \right)$     (M1)

$= \left( {\begin{array}{*{20}{c}}   4 \\   {10} \\   { - 3} \end{array}} \right) = 4{\mathbf{i}} + 10{\mathbf{j}} - 3{\mathbf{k}}$     A1

(ii)     METHOD 1

let P be the intersection of ${L_1}$ and ${L_3}$

let Q be the intersection of ${L_2}$ and ${L_3}$

$\overrightarrow {{\text{OP}}}  = \left( {\begin{array}{*{20}{c}}   {1 - \lambda } \\   {2 + \lambda } \\   {2 + 2\lambda } \end{array}} \right)\overrightarrow {{\text{OQ}}}  = \left( {\begin{array}{*{20}{c}}   {1 + 2\mu } \\   {2 + \mu } \\   {4 + 6\mu } \end{array}} \right)$     M1

therefore $\overrightarrow {{\text{PQ}}}  = \overrightarrow {{\text{OQ}}}  - \overrightarrow {{\text{OP}}}  = \left( {\begin{array}{*{20}{c}}   {2\mu  + \lambda } \\   {\mu  - \lambda } \\   {2 + 6\mu  - 2\lambda } \end{array}} \right)$     M1A1

$\left( {\begin{array}{*{20}{c}}   {2\mu  + \lambda } \\   {\mu  - \lambda } \\   {2 + 6\mu  - 2\lambda } \end{array}} \right) = t\left( {\begin{array}{*{20}{c}}   4 \\   {10} \\   { - 3} \end{array}} \right)$     M1

$2\mu  + \lambda  - 4t = 0$
$\mu  - \lambda  - 10t = 0$
$6\mu  - 2\lambda  + 3t =  - 2$

solving simultaneously     (M1)

$\lambda  = \frac{{32}}{{125}}\;\;\;(0.256),{\text{ }}\mu  =  - \frac{{28}}{{125}}\;\;\;( - 0.224)$     A1

 

Note:     Award A1 for either correct $\lambda$ or $\mu$.

 

EITHER

therefore $\overrightarrow {{\text{OP}}}  = \left( {\begin{array}{*{20}{c}}   {1 - \lambda } \\   {2 + \lambda } \\   {2 + 2\lambda } \end{array}} \right) = \left( {\begin{array}{*{20}{c}}   {\frac{{93}}{{125}}} \\   {\frac{{282}}{{125}}} \\   {\frac{{314}}{{125}}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}}   {0.744} \\   {2.256} \\   {2.512} \end{array}} \right)$A1

therefore ${L_3}:{r_3} = \left( {\begin{array}{*{20}{c}}   {0.744} \\   {2.256} \\   {2.512} \end{array}} \right) + \alpha \left( {\begin{array}{*{20}{c}}   4 \\   {10} \\   { - 3} \end{array}} \right)$     A1

OR

therefore $\overrightarrow {{\text{OQ}}}  = \left( {\begin{array}{*{20}{c}}   {1 + 2\mu } \\   {2 + \mu } \\   {4 + 6\mu } \end{array}} \right) = \left( {\begin{array}{*{20}{c}}   {\frac{{69}}{{125}}} \\   {\frac{{222}}{{125}}} \\   {\frac{{332}}{{125}}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}}   {0.552} \\   {1.776} \\   {2.656} \end{array}} \right)$     A1

therefore ${L_3}:{r_3} = \left( {\begin{array}{*{20}{c}}   {0.552} \\   {1.776} \\   {2.656} \end{array}} \right) + \alpha \left( {\begin{array}{*{20}{c}}   4 \\   {10} \\   { - 3} \end{array}} \right)$     A1

 

Note:     Allow position vector(s) to be expressed in decimal or fractional form.

METHOD 2

${L_3}:{r_3} = \left( {\begin{array}{*{20}{c}}   a \\   b \\   c \end{array}} \right) + t\left( {\begin{array}{*{20}{c}}   4 \\   {10} \\   { - 3} \end{array}} \right)$

forming two equations as intersections with ${L_1}$ and ${L_2}$

$\left( {\begin{array}{*{20}{c}}   a \\   b \\   c \end{array}} \right) + {t_1}\left( {\begin{array}{*{20}{c}}   4 \\   {10} \\   { - 3} \end{array}} \right) = \left( {\begin{array}{*{20}{c}}   1 \\   2 \\   2 \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}   { - 1} \\   1 \\   2 \end{array}} \right)$
$\left( {\begin{array}{*{20}{c}}   a \\   b \\   c \end{array}} \right) + {t_2}\left( {\begin{array}{*{20}{c}}   4 \\   {10} \\   { - 3} \end{array}} \right) = \left( {\begin{array}{*{20}{c}}   1 \\   2 \\   4 \end{array}} \right) + \mu \left( {\begin{array}{*{20}{c}}   2 \\   1 \\   6 \end{array}} \right)$     M1A1A1

 

Note:     Only award M1A1A1 if two different parameters ${t_1},{\text{ }}{t_2}$ used.

 

attempting to solve simultaneously     M1

$\lambda  = \frac{{32}}{{125}}\;\;\;(0.256),{\text{ }}\mu  =  - \frac{{28}}{{125}}\;\;\;( - 0.224)$     A1

 

Note:     Award A1 for either correct $\lambda$ or $\mu$.
 

EITHER

$\left( {\begin{array}{*{20}{c}}   a \\   b \\   c \end{array}} \right) = \left( {\begin{array}{*{20}{c}}   {0.552} \\   {1.776} \\   {2.656} \end{array}} \right)$     A1

therefore ${L_3}:{r_3} = \left( {\begin{array}{*{20}{c}}   {0.552} \\   {1.776} \\   {2.656} \end{array}} \right) + t\left( {\begin{array}{*{20}{c}}   4 \\   {10} \\   { - 3} \end{array}} \right)$     A1A1

OR

$\left( {\begin{array}{*{20}{c}}   a \\   b \\   c \end{array}} \right) = \left( {\begin{array}{*{20}{c}}   {0.744} \\   {2.256} \\   {2.512} \end{array}} \right)$     A1

therefore ${L_3}:{r_3} = \left( {\begin{array}{*{20}{c}}   {0.744} \\   {2.256} \\   {2.512} \end{array}} \right) + t\left( {\begin{array}{*{20}{c}}   4 \\   {10} \\   { - 3} \end{array}} \right)$     A1A1
 

Note:     Allow position vector(s) to be expressed in decimal or fractional form.

10 marks

Total [18 marks]

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