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.