| Date | May 2018 | Marks available | 3 | Reference code | 18M.1.hl.TZ1.10 |
| Level | HL only | Paper | 1 | Time zone | TZ1 |
| Command term | Find | Question number | 10 | Adapted from | N/A |
Question
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.
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.
Markscheme
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]
