Find a vector equation for the line, \({L_1}\), which passes through the points P and Q.

The line \({L_2}\) has equation

\[r = \left( {\begin{array}{*{20}{c}}
  { - 4} \\
  0 \\
  4
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
  5 \\
  2 \\
  0
\end{array}} \right){\text{.}}\]

Show that \({L_1}\) and \({L_2}\) intersect at the point R(1, 2, 4).

Find the acute angle between \({L_1}\) and \({L_2}\).

Let S be a point on \({L_2}\) such that \(\left| {\overrightarrow {{\text{RP}}} } \right| = \left| {\overrightarrow {{\text{RS}}} } \right|\). 

Show that one of the possible positions for S is \({{\text{S}}_1}\)(−4, 0, 4) and find the coordinates of the other possible position, \({{\text{S}}_2}\).

Let S be a point on \({L_2}\) such that \(\left| {\overrightarrow {{\text{RP}}} } \right| = \left| {\overrightarrow {{\text{RS}}} } \right|\).

Find a vector equation of the line which passes through R and bisects \({\rm{P\hat R}}{{\text{S}}_1}\).

\(\overrightarrow {{\text{PQ}}}  = \left( {\begin{array}{*{20}{c}}
  8 \\
  6 \\
  4
\end{array}} \right)\)     (A1)

equation of line: \(r = \left( {\begin{array}{*{20}{c}}
  { - 3} \\
  { - 1} \\
  2
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
  8 \\
  6 \\
  4
\end{array}} \right)\) (or equivalent)     M1A1

Note: Award M1A0 if r = is omitted.

[3 marks]

METHOD 1

\(x:{\text{     }}- 4 + 5s = - 3 + 8t\)

\(y:{\text{     }}2s = - 1 + 6t\)

\(z:{\text{     }}4 = 2 + 4t\)     M1

solving any two simultaneously     M1

t = 0.5, s = 1 (or equivalent)     A1

verification that these values give R when substituted into both equations (or that the three equations are consistent and that one gives R)     R1

METHOD 2

(1, 2, 4) is given by t = 0.5 for \({L_1}\) and s = 1 for \({L_2}\)     M1A1A1

because (1, 2, 4) is on both lines it is the point of intersection of the two lines     R1

[4 marks]

\(\left( {\begin{array}{*{20}{c}}
  5 \\
  2 \\
  0
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
  4 \\
  3 \\
  2
\end{array}} \right) = 26 = \sqrt {29}  \times \sqrt {29} \cos \theta \)     M1

\(\cos \theta  = \frac{{26}}{{29}}\)     (A1)

\(\theta  = 0.459\) or 26.3°     A1

[3 marks]

\(\overrightarrow {{\text{RP}}}  = \left( {\begin{array}{*{20}{c}}
  { - 3} \\
  { - 1} \\
  2
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
  1 \\
  2 \\
  4
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  { - 4} \\
  { - 3} \\
  { - 2}
\end{array}} \right)\), \(\left| {\overrightarrow {{\text{RP}}} } \right| = \sqrt {29} \)     (M1)A1

Note: This could also be obtained from \(\left| {0.5\left( {\begin{array}{*{20}{c}}
  8 \\
  6 \\
  4
\end{array}} \right)} \right|\)

 EITHER

\(\overrightarrow {{\text{R}}{{\text{S}}_{\text{1}}}}  = \left( {\begin{array}{*{20}{c}}
  { - 4} \\
  0 \\
  4
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
  1 \\
  2 \\
  4
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  { - 5} \\
  { - 2} \\
  0
\end{array}} \right)\), \(\left| {\overrightarrow {{\text{R}}{{\text{S}}_{\text{1}}}} } \right| = \sqrt {29} \)     A1

\(\therefore \overrightarrow {{\text{O}}{{\text{S}}_2}}  = \overrightarrow {{\text{O}}{{\text{S}}_{\text{1}}}}  + 2\overrightarrow {{{\text{S}}_{\text{1}}}{\text{R}}}  = \left( {\begin{array}{*{20}{c}}
  { - 4} \\
  0 \\
  4
\end{array}} \right) + 2\left( {\begin{array}{*{20}{c}}
  5 \\
  2 \\
  0
\end{array}} \right)\)     M1A1

\(\left( {{\text{or }}\overrightarrow {{\text{O}}{{\text{S}}_2}}  = \overrightarrow {{\text{OR}}}  + \overrightarrow {{{\text{S}}_{\text{1}}}{\text{R}}}  = \left( {\begin{array}{*{20}{c}}
  1 \\
  2 \\
  4
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
  5 \\
  2 \\
  0
\end{array}} \right)} \right)\)

\( = \left( {\begin{array}{*{20}{c}}
  6 \\
  4 \\
  4
\end{array}} \right)\)

\({{\text{S}}_2}\) is (6, 4, 4)     A1 

OR

\(\left( {\begin{array}{*{20}{c}}
  { - 4 + 5s} \\
  {2s} \\
  4
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
  1 \\
  2 \\
  4
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  {5s - 5} \\
  {2s - 2} \\
  0
\end{array}} \right)\)     M1

\({(5s - 5)^2} + {(2s - 2)^2} = 29\)     M1A1

\(29{s^2} - 58s + 29 = 29\)

\(s(s - 2) = 0,{\text{ }}s = 0,{\text{ 2}}\)

\((6,{\text{ }}4,{\text{ 4}}){\text{ }}\left( {{\text{and (}} - 4,{\text{ }}0,{\text{ }}4{\text{)}}} \right)\)     A1

Note: There are several geometrical arguments possible using information obtained in previous parts, depending on what forms the previous answers had been given.

[6 marks]

EITHER

midpoint of \([{\text{P}}{{\text{S}}_1}]\) is M(–3.5, –0.5, 3)     M1A1

\(\overrightarrow {{\text{RM}}}  = \left( {\begin{array}{*{20}{c}}
  { - 4.5} \\
  { - 2.5} \\
  { - 1}
\end{array}} \right)\)     A1 

OR

\(\overrightarrow {{\text{R}}{{\text{S}}_1}}  = \left( {\begin{array}{*{20}{c}}
  { - 5} \\
  { - 2} \\
  0
\end{array}} \right)\)     M1

the direction of the line is \({\overrightarrow {{\text{RS}}} _1} + \overrightarrow {{\text{RP}}} \)

\(\left( {\begin{array}{*{20}{c}}
  { - 5} \\
  { - 2} \\
  0
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
  { - 4} \\
  { - 3} \\
  { - 2}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  { - 9} \\
  { - 5} \\
  { - 2}
\end{array}} \right)\)     M1A1 

THEN

the equation of the line is:

\(r = \left( {\begin{array}{*{20}{c}}
  1 \\
  2 \\
  4
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
  9 \\
  5 \\
  2
\end{array}} \right)\) or equivalent     A1

Note: Marks cannot be awarded for methods involving halving the angle, unless it is clear that the candidate considers also the equation of the plane of \({L_1}\) and \({L_2}\) to reduce the number of parameters involved to one (to obtain the vector equation of the required line).

[4 marks]

There were many good answers to part (a) showing a clear understanding of finding the vector equation of a line. Unfortunately this understanding was marred by many students failing to write the equation properly resulting in just 2 marks out of the 3. The most common response was of the form \({L_1} = \left( {\begin{array}{*{20}{c}}
  { - 3} \\
  { - 1} \\
  2
\end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}
  4 \\
  3 \\
  2
\end{array}} \right)\) which seemed a waste of a mark.

In part (b) many students failed to verify that the lines do indeed intersect.

Part (c) was very well done.

In part (d) most candidates were able to obtain the first three marks, but few were able to find the second point.

There were few correct answers to part (e).