18M.1.AHL.TZ1.H_10

pestleMathematicsAAHLPaper 118M· ahl-3-17-vector-equations-of-a-planesource ↗

The following figure shows a square based pyramid with vertices at O(0, 0, 0), A(1, 0, 0), B(1, 1, 0), C(0, 1, 0) and D(0, 0, 1).

The Cartesian equation of the plane ${\Pi _2}$ , passing through the points B , C and D , is $y + z = 1$ .

The plane ${\Pi _3}$ passes through O and is normal to the line BD.

${\Pi _3}$ cuts AD and BD at the points P and Q respectively.

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

[3]

Find the angle between the faces ABD and BCD.

[4]

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

[3]

Show that P is the midpoint of AD.

[4]

Find the area of the triangle OPQ.

[5]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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

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

Examiners’ report

[N/A]