Show that the points ${\text{O}}(0,{\text{ }}0,{\text{ }}0)$, ${\text{ A}}(6,{\text{ }}0,{\text{ }}0)$, ${\text{B}}({6,{\text{ }}- \sqrt {24} ,{\text{ }}\sqrt {12} })$, ${\text{C}}({0,{\text{ }}- \sqrt {24} ,{\text{ }}\sqrt {12}})$ form a square.

Find the coordinates of M, the mid-point of [OB].

Show that an equation of the plane ${\mathit{\Pi }}$, containing the square OABC, is $y + \sqrt 2 z = 0$.

Find a vector equation of the line $L$, through M, perpendicular to the plane ${\mathit{\Pi }}$.

Find the coordinates of D, the point of intersection of the line $L$ with the plane whose equation is $y = 0$.

Find the coordinates of E, the reflection of the point D in the plane ${\mathit{\Pi }}$.

(i)     Find the angle ${\rm{O\hat DA}}$.

(ii)     State what this tells you about the solid OABCDE.

$\left| {\overrightarrow {{\text{OA}}} } \right| = \left| {\overrightarrow {{\text{CB}}} } \right| = \left| {\overrightarrow {{\text{OC}}} } \right| = \left| {\overrightarrow {{\text{AB}}} } \right| = 6$   (therefore a rhombus)     A1A1

Note:     Award A1 for two correct lengths, A2 for all four.

Note: Award A1A0 for $\overrightarrow {{\rm{OA}}}  = \overrightarrow {{\rm{CB}}}  = \left( \begin{array}{l}6\\0\\0\end{array} \right){\rm{ or \,\,} } \overrightarrow {{\rm{OC}}}  = \overrightarrow {A{\rm{B}}}  = \left( \begin{array}{c}0\\ - \sqrt {24} \\\sqrt {12} \end{array} \right)$ if no magnitudes are shown.

$\overrightarrow {{\rm{OA}}}\,\, {\rm{ g}}\overrightarrow {{\rm{OC}}}  = \left( \begin{array}{l}6\\0\\0\end{array} \right){\rm{g}}\left( \begin{array}{c}0\\ - \sqrt {24} \\\sqrt {12} \end{array} \right) = 0$   (therefore a square)     A1

Note:     Other arguments are possible with a minimum of three conditions.

[3 marks]

${\text{M}}\left( {3,{\text{ }} - \frac{{\sqrt {24} }}{2},{\text{ }}\frac{{\sqrt {12} }}{2}} \right)\left( { = \left( {3,{\text{ }} - \sqrt 6 ,{\text{ }}\sqrt 3 } \right)} \right)$     A1

[1 mark]

METHOD 1

$\overrightarrow {{\text{OA}}}  \times \overrightarrow {{\text{OC}}}  =$$\left( \begin{array}{l}6\\0\\0\end{array} \right) \times \left( \begin{array}{c}0\\ - \sqrt {24} \\\sqrt {12} \end{array} \right) = \left( \begin{array}{c}0\\ - 6\sqrt {12} \\ - 6\sqrt {24} \end{array} \right)\left( { = \left( \begin{array}{c}0\\ - 12\sqrt 3 \\ - 12\sqrt 6 \end{array} \right)} \right)$     M1A1

Note:     Candidates may use other pairs of vectors.

equation of plane is $- 6\sqrt {12} y - 6\sqrt {24} z = d$

any valid method showing that $d = 0$     M1

$\mathit{\Pi} :y+\sqrt{2z}=0$     AG

METHOD 2

equation of plane is $ax + by + cz = d$

substituting O to find $d = 0$     (M1)

substituting two points (A, B, C or M)     M1

eg

$6a = 0,{\text{ }} - \sqrt {24} b + \sqrt {12} c = 0$     A1

$\mathit{\Pi} :y+\sqrt{2z}=0$     AG

[3 marks]

$\boldsymbol{r} = \left( \begin{array}{c}3\\ - \sqrt 6 \\\sqrt 3 \end{array} \right) + \lambda \left( \begin{array}{l}0\\1\\\sqrt 2 \end{array} \right)$     A1A1A1

Note:     Award A1 for r = , A1A1 for two correct vectors.

[3 marks]

Using $y = 0$ to find $\lambda$     M1

Substitute their $\lambda$ into their equation from part (d)     M1

D has coordinates $\left( {{\text{3, 0, 3}}\sqrt 3 } \right)$     A1

[3 marks]

$\lambda$ for point E is the negative of the $\lambda$ for point D     (M1)

Note:     Other possible methods may be seen.

E has coordinates $\left( {{\text{3, }} - 2\sqrt 6 ,{\text{ }} - \sqrt 3 } \right)$     A1A1

Note:     Award A1 for each of the y and z coordinates.

[3 marks]

(i)     $\overrightarrow {{\text{DA}}} {\text{ g}}\overrightarrow {{\text{DO}}}  =$$\left( \begin{array}{c}3\\0\\ - 3\sqrt 3 \end{array} \right){\rm{g}}\left( \begin{array}{c} - 3\\0\\ - 3\sqrt 3 \end{array} \right) = 18$     M1A1

$\cos {\rm{O\hat DA}} = \frac{{18}}{{\sqrt {36} \sqrt {36} }} = \frac{1}{2}$     M1

hence ${\rm{O\hat DA}} = 60^\circ$     A1

Note:     Accept method showing OAD is equilateral.

(ii)     OABCDE is a regular octahedron (accept equivalent description)     A2

Note:     A2 for saying it is made up of 8 equilateral triangles

     Award A1 for two pyramids, A1 for equilateral triangles.

     (can be either stated or shown in a sketch – but there must be clear indication the triangles are equilateral)

[6 marks]