The diagram shows a cube OABCDEFG.

Let O be the origin, (OA) the x-axis, (OC) the y-axis and (OD) the z-axis.

Let M, N and P be the midpoints of [FG], [DG] and [CG], respectively.

The coordinates of F are (2, 2, 2).

(a)     Find the position vectors $\overrightarrow {{\text{OM}}}$, $\overrightarrow {{\text{ON}}}$ and $\overrightarrow {{\text{OP}}}$ in component form.

(b)     Find $\overrightarrow {{\text{MP}}}  \times \overrightarrow {{\text{MN}}}$.

(c)     Hence,

  (i)     calculate the area of the triangle MNP;

  (ii)     show that the line (AG) is perpendicular to the plane MNP;

  (iii)     find the equation of the plane MNP.

(d)     Determine the coordinates of the point where the line (AG) meets the plane MNP.

(a)     $\overrightarrow {{\text{OM}}}  = \left( {\begin{array}{*{20}{c}}   1 \\   2 \\   2 \end{array}} \right)$, $\overrightarrow {{\text{ON}}}  = \left( {\begin{array}{*{20}{c}}   0 \\   1 \\   2 \end{array}} \right)$ and $\overrightarrow {{\text{OP}}}  = \left( {\begin{array}{*{20}{c}}   0 \\   2 \\   1 \end{array}} \right)$     A1A1A1

[3 marks]

(b)     $\overrightarrow {{\text{MP}}}  = \left( {\begin{array}{*{20}{c}}   { - 1} \\   0 \\   { - 1} \end{array}} \right)$ and $\overrightarrow {{\text{MN}}}  = \left( {\begin{array}{*{20}{c}}   { - 1} \\   { - 1} \\   0 \end{array}} \right)$     A1A1

$\overrightarrow {{\text{MP}}}  \times \overrightarrow {{\text{MN}}}  = \left( {\begin{array}{*{20}{c}}   i&j&k \\   { - 1}&0&{ - 1} \\   { - 1}&{ - 1}&0 \end{array}} \right) = \left( {\begin{array}{*{20}{c}}   { - 1} \\   1 \\   1 \end{array}} \right)$     (M1)A1

[4 marks]

(c)     (i)     area of MNP $= \frac{1}{2}\left| {\overrightarrow {{\text{MP}}}  \times \overrightarrow {{\text{MN}}} } \right|$     M1

$= \frac{1}{2}\left| {\left( {\begin{array}{*{20}{c}}   { - 1} \\   1 \\   1 \end{array}} \right)} \right|$

$= \frac{{\sqrt 3 }}{2}$     A1

(ii)     $\overrightarrow {{\text{OA}}}  = \left( {\begin{array}{*{20}{c}}   2 \\   0 \\   0 \end{array}} \right)$, $\overrightarrow {{\text{OG}}}  = \left( {\begin{array}{*{20}{c}}   0 \\   2 \\   2 \end{array}} \right)$

$\overrightarrow {{\text{AG}}}  = \left( {\begin{array}{*{20}{c}}   { - 2} \\   2 \\   2 \end{array}} \right)$     A1

since $\overrightarrow {{\text{AG}}} = 2(\overrightarrow {{\text{MP}}}  \times \overrightarrow {{\text{MN}}} )$ AG is perpendicular to MNP     R1

(iii)     $r \cdot \left( {\begin{array}{*{20}{c}}   { - 1} \\   1 \\   1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}}   1 \\   2 \\   2 \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}   { - 1} \\   1 \\   1 \end{array}} \right)$     M1A1

$r \cdot \left( {\begin{array}{*{20}{c}}   { - 1} \\   1 \\   1 \end{array}} \right) = 3$ (accept $- x + y + z = 3$)     A1

[7 marks]

(d)     $r = \left( {\begin{array}{*{20}{c}}   2 \\   0 \\   0 \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}   { - 2} \\   2 \\   2 \end{array}} \right)$     A1

$\left( {\begin{array}{*{20}{c}}   {2 - 2\lambda } \\   {2\lambda } \\   {2\lambda } \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}   { - 1} \\   1 \\   1 \end{array}} \right) = 3$     M1A1

$- 2 + 2\lambda + 2\lambda + 2\lambda = 3$

$\lambda = \frac{5}{6}$     A1

$r = \left( {\begin{array}{*{20}{c}}   2 \\   0 \\   0 \end{array}} \right) + \frac{5}{6}\left( {\begin{array}{*{20}{c}}   { - 2} \\   2 \\   2 \end{array}} \right)$     M1

coordinates of point $\left( {\frac{1}{3},\frac{5}{3},\frac{5}{3}} \right)$     A1

[6 marks]

Total [20 marks]

This was the most successfully answered question in part B, with many candidates achieving full marks. There were a few candidates who misread the question and treated the cube as a unit cube. The most common errors were either algebraic or arithmetic mistakes. A variety of notation forms were seen but in general were used consistently. In a few cases, candidates failed to show all the work or set it properly.