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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