Find the Cartesian equation of the plane ${\Pi _1}$, passing through the points A , B and D.

Find the angle between the faces ABD and BCD.

Find the Cartesian equation of ${\Pi _3}$.

Show that P is the midpoint of AD.

Find the area of the triangle OPQ.

recognising normal to plane or attempting to find cross product of two vectors lying in the plane      (M1)

for example, $\mathop {{\text{AB}}}\limits^ \to \,\, \times \mathop {{\text{AD}}}\limits^ \to = \left( \begin{gathered} 0 \hfill \\ 1 \hfill \\ 0 \hfill \\ \end{gathered} \right) \times \left( \begin{gathered} - 1 \hfill \\ \,0 \hfill \\ \,1 \hfill \\ \end{gathered} \right) = \left( \begin{gathered} 1 \hfill \\ 0 \hfill \\ 1 \hfill \\ \end{gathered} \right)$     (A1)

${\Pi _1}\,{\text{:}}\,\,x + z = 1$     A1

[3 marks]

EITHER

$\left( \begin{gathered} 1 \hfill \\ 0 \hfill \\ 1 \hfill \\ \end{gathered} \right) \bullet \left( \begin{gathered} 0 \hfill \\ 1 \hfill \\ 1 \hfill \\ \end{gathered} \right) = 1 = \sqrt 2 \sqrt 2 \,{\text{cos}}\,\theta$     M1A1

OR

$\left| {\left( \begin{gathered} 1 \hfill \\ 0 \hfill \\ 1 \hfill \\ \end{gathered} \right) \times \left( \begin{gathered} 0 \hfill \\ 1 \hfill \\ 1 \hfill \\ \end{gathered} \right)} \right| = \sqrt 3 = \sqrt 2 \sqrt 2 \,{\text{sin}}\,\theta$     M1A1

Note: M1 is for an attempt to find the scalar or vector product of the two normal vectors.

$\Rightarrow \theta  = 60^\circ \left( { = \frac{\pi }{3}} \right)$     A1

angle between faces is $20^\circ \left( { = \frac{{2\pi }}{3}} \right)$     A1

[4 marks]

$\mathop {{\text{DB}}}\limits^ \to = \left( \begin{gathered} \,1 \hfill \\ \,1 \hfill \\ - 1 \hfill \\ \end{gathered} \right)$ or $\mathop {{\text{BD}}}\limits^ \to = \left( \begin{gathered} - 1 \hfill \\ - 1 \hfill \\ \,1 \hfill \\ \end{gathered} \right)$     (A1)

${\Pi _3}\,{\text{:}}\,\,x + y - z = k$     (M1)

${\Pi _3}\,{\text{:}}\,\,x + y - z = 0$     A1

[3 marks]

METHOD 1

line AD : (r =)$\left( \begin{gathered} 0 \hfill \\ 0 \hfill \\ 1 \hfill \\ \end{gathered} \right) + \lambda \left( \begin{gathered} \,1 \hfill \\ \,0 \hfill \\ - 1 \hfill \\ \end{gathered} \right)$     M1A1

intersects ${\Pi _3}$ when $\lambda  - \left( {1 - \lambda } \right) = 0$     M1

so $\lambda  = \frac{1}{2}$     A1

hence P is the midpoint of AD      AG

METHOD 2

midpoint of AD is (0.5, 0, 0.5)      (M1)A1

substitute into $x + y - z = 0$     M1

0.5 + 0.5 − 0.5 = 0     A1

hence P is the midpoint of AD     AG

[4 marks]

METHOD 1

${\text{OP}} = \frac{1}{{\sqrt 2 }},\,\,{\text{O}}\mathop {\text{P}}\limits^ \wedge  {\text{Q}} = 90^\circ ,\,\,{\text{O}}\mathop {\text{Q}}\limits^ \wedge  {\text{P}} = 60^\circ$      A1A1A1

${\text{PQ}} = \frac{1}{{\sqrt 6 }}$     A1

area $= \frac{1}{{2\sqrt {12} }} = \frac{1}{{4\sqrt 3 }} = \frac{{\sqrt 3 }}{{12}}$     A1

METHOD 2

line BD : (r  =)$\left( \begin{gathered} 1 \hfill \\ 1 \hfill \\ 0 \hfill \\ \end{gathered} \right) + \lambda \left( \begin{gathered} - 1 \hfill \\ - 1 \hfill \\ \,1 \hfill \\ \end{gathered} \right)$

$\Rightarrow \lambda  = \frac{2}{3}$     (A1)

$\mathop {{\text{OQ}}}\limits^ \to = \left( \begin{gathered} \frac{1}{3} \hfill \\ \frac{1}{3} \hfill \\ \frac{2}{3} \hfill \\ \end{gathered} \right)$    A1

area = $\frac{1}{2}\left| {\mathop {{\text{OP}}}\limits^ \to  \, \times \mathop {{\text{OQ}}}\limits^ \to  } \right|$     M1

$\mathop {{\text{OP}}}\limits^ \to = \left( \begin{gathered} \frac{1}{2} \hfill \\ 0 \hfill \\ \frac{1}{2} \hfill \\ \end{gathered} \right)$    A1

Note: This A1 is dependent on M1.

area = $\frac{{\sqrt 3 }}{{12}}$     A1

[5 marks]