Find the coordinates of ${\text{P}}$.

Show that the lines are perpendicular.

The point ${\text{Q}}(7, 5, 3)$ lies on ${L_1}$. The point ${\text{R}}$ is the reflection of ${\text{Q}}$ in the line ${L_2}$.

Find the coordinates of ${\text{R}}$.

appropriate approach     (M1)

eg     $\left( \begin{array}{c}11\\8\\2\end{array} \right) + s\left( \begin{array}{c}4\\3\\ - 1\end{array} \right) = \left( \begin{array}{c}1\\1\\ - 7\end{array} \right) + t\left( \begin{array}{c}2\\1\\11\end{array} \right)$, ${L_1} = {L_2}$

any two correct equations     A1A1

eg     $11 + 4s = 1 + 2t,{\text{ }}8 + 3s = 1 + t,{\text{ }}2 - s =  - 7 + 11t$

attempt to solve system of equations     (M1)

eg     $10 + 4s = 2(7 + 3s), \left\{ {\begin{array}{*{20}{c}} {4s - 2t =  - 10} \\  {3s - t =  - 7} \end{array}} \right.$

one correct parameter     A1

eg     $s =  - 2,{\text{ }}t = 1$

${\text{P}}(3, 2, 4)$   (accept position vector)     A1     N3

[6 marks]

choosing correct direction vectors for ${L_1}$ and ${L_2}$     (A1)(A1)

eg     $\left( {\begin{array}{*{20}{c}}   4 \\    3 \\    { - 1} \end{array}} \right),\left( {\begin{array}{*{20}{c}}   2 \\   1 \\   {11} \end{array}} \right)$ (or any scalar multiple)

evidence of scalar product (with any vectors)     (M1)

eg     $a \cdot b$, $\left( \begin{array}{c}4\\3\\ - 1\end{array} \right) \bullet \left( \begin{array}{c}2\\1\\11\end{array} \right)$

correct substitution     A1

eg     $4(2) + 3(1) + ( - 1)(11),{\text{ }}8 + 3 - 11$

calculating $a \cdot b = 0$     A1

Note: Do not award the final A1 without evidence of calculation.

vectors are perpendicular     AG     N0

[5 marks]

Note: Candidates may take different approaches, which do not necessarily involve vectors.

In particular, most of the working could be done on a diagram. Award marks in line with the markscheme.

METHOD 1

attempt to find $\overrightarrow {{\text{QP}}}$ or $\overrightarrow {{\text{PQ}}}$     (M1)

correct working (may be seen on diagram)     A1

eg     $\overrightarrow {{\text{QP}}}$ = $\left( \begin{array}{c} - 4\\ - 3\\1\end{array} \right)$, $\overrightarrow {{\text{PQ}}}$ = $\left( \begin{array}{c}7\\5\\3\end{array} \right) - \left( \begin{array}{c}3\\2\\4\end{array} \right)$

recognizing ${\text{R}}$ is on ${L_1}$ (seen anywhere)     (R1)

eg     on diagram

${\text{Q}}$ and ${\text{R}}$ are equidistant from ${\text{P}}$ (seen anywhere)     (R1)

eg     $\overrightarrow {{\text{QP}}}  = \overrightarrow {{\text{PR}}}$, marked on diagram

correct working     (A1)

eg     $\left( \begin{array}{c}3\\2\\4\end{array} \right) - \left( \begin{array}{c}7\\5\\3\end{array} \right) = \left( \begin{array}{c}x\\y\\z\end{array} \right) - \left( \begin{array}{c}3\\2\\4\end{array} \right),\left( \begin{array}{c} - 4\\ - 3\\1\end{array} \right) + \left( \begin{array}{c}3\\2\\4\end{array} \right)$

${\text{R}}(–1, –1, 5)$ (accept position vector)     A1     N3

METHOD 2 

recognizing ${\text{R}}$ is on ${L_1}$ (seen anywhere)     (R1)

eg     on diagram

${\text{Q}}$ and ${\text{R}}$ are equidistant from ${\text{P}}$ (seen anywhere)     (R1)

eg     ${\text{P}}$ midpoint of ${\text{QR}}$, marked on diagram

valid approach to find one coordinate of mid-point     (M1)

eg     ${x_p} = \frac{{{x_Q} + {x_R}}}{2},{\text{ }}2{y_p} = {y_Q} + {y_R},{\text{ }}\frac{1}{2}\left( {{z_Q} + {z_R}} \right)$

one correct substitution     A1

eg     ${x_R} = 3 + (3 - 7),{\text{ }}2 = \frac{{5 + {y_R}}}{2},{\text{ }}4 = \frac{1}{2}(z + 3)$

correct working for one coordinate     (A1)

eg     ${x_R} = 3 - 4,{\text{ }}4 - 5 = {y_R},{\text{ }}8 = (z + 3)$

${\text{R}} (-1, -1, 5)$ (accept position vector)     A1     N3

[6 marks]