| Date | May 2018 | Marks available | 1 | Reference code | 18M.1.hl.TZ2.9 |
| Level | HL only | Paper | 1 | Time zone | TZ2 |
| Command term | Explain | Question number | 9 | Adapted from | N/A |
Question
The points A, B, C and D have position vectors a, b, c and d, relative to the origin O.
It is given that \(\mathop {{\text{AB}}}\limits^ \to = \mathop {{\text{DC}}}\limits^ \to \).
The position vectors \(\mathop {{\text{OA}}}\limits^ \to \), \(\mathop {{\text{OB}}}\limits^ \to \), \(\mathop {{\text{OC}}}\limits^ \to \) and \(\mathop {{\text{OD}}}\limits^ \to \) are given by
a = i + 2j − 3k
b = 3i − j + pk
c = qi + j + 2k
d = −i + rj − 2k
where p , q and r are constants.
The point where the diagonals of ABCD intersect is denoted by M.
The plane \(\Pi \) cuts the x, y and z axes at X , Y and Z respectively.
Explain why ABCD is a parallelogram.
Using vector algebra, show that \(\mathop {{\text{AD}}}\limits^ \to = \mathop {{\text{BC}}}\limits^ \to \).
Show that p = 1, q = 1 and r = 4.
Find the area of the parallelogram ABCD.
Find the vector equation of the straight line passing through M and normal to the plane \(\Pi \) containing ABCD.
Find the Cartesian equation of \(\Pi \).
Find the coordinates of X, Y and Z.
Find YZ.
Markscheme
a pair of opposite sides have equal length and are parallel R1
hence ABCD is a parallelogram AG
[1 mark]
attempt to rewrite the given information in vector form M1
b − a = c − d A1
rearranging d − a = c − b M1
hence \(\mathop {{\text{AD}}}\limits^ \to = \mathop {{\text{BC}}}\limits^ \to \) AG
Note: Candidates may correctly answer part i) by answering part ii) correctly and then deducing there
are two pairs of parallel sides.
[3 marks]
EITHER
use of \(\mathop {{\text{AB}}}\limits^ \to = \mathop {{\text{DC}}}\limits^ \to \) (M1)
\(\left( \begin{gathered}
2 \hfill \\
- 3 \hfill \\
p + 3 \hfill \\
\end{gathered} \right) = \left( \begin{gathered}
q + 1 \hfill \\
1 - r \hfill \\
4 \hfill \\
\end{gathered} \right)\) A1A1
OR
use of \(\mathop {{\text{AD}}}\limits^ \to = \mathop {{\text{BC}}}\limits^ \to \) (M1)
\(\left( \begin{gathered}
- 2 \hfill \\
r - 2 \hfill \\
1 \hfill \\
\end{gathered} \right) = \left( \begin{gathered}
q - 3 \hfill \\
2 \hfill \\
2 - p \hfill \\
\end{gathered} \right)\) A1A1
THEN
attempt to compare coefficients of i, j, and k in their equation or statement to that effect M1
clear demonstration that the given values satisfy their equation A1
p = 1, q = 1, r = 4 AG
[5 marks]
attempt at computing \(\mathop {{\text{AB}}}\limits^ \to \, \times \mathop {{\text{AD}}}\limits^ \to \) (or equivalent) M1
\(\left( \begin{gathered}
- 11 \hfill \\
- 10 \hfill \\
- 2 \hfill \\
\end{gathered} \right)\) A1
area \( = \left| {\mathop {{\text{AB}}}\limits^ \to \, \times \mathop {{\text{AD}}}\limits^ \to } \right|\left( { = \sqrt {225} } \right)\) (M1)
= 15 A1
[4 marks]
valid attempt to find \(\mathop {{\text{OM}}}\limits^ \to = \left( {\frac{1}{2}\left( {a + c} \right)} \right)\) (M1)
\(\left( \begin{gathered}
1 \hfill \\
\frac{3}{2} \hfill \\
- \frac{1}{2} \hfill \\
\end{gathered} \right)\) A1
the equation is
r = \(\left( \begin{gathered}
1 \hfill \\
\frac{3}{2} \hfill \\
- \frac{1}{2} \hfill \\
\end{gathered} \right) + t\left( \begin{gathered}
11 \hfill \\
10 \hfill \\
2 \hfill \\
\end{gathered} \right)\) or equivalent M1A1
Note: Award maximum M1A0 if 'r = …' (or equivalent) is not seen.
[4 marks]
attempt to obtain the equation of the plane in the form ax + by + cz = d M1
11x + 10y + 2z = 25 A1A1
Note: A1 for right hand side, A1 for left hand side.
[3 marks]
putting two coordinates equal to zero (M1)
\({\text{X}}\left( {\frac{{25}}{{11}},\,0,\,0} \right),\,\,{\text{Y}}\left( {0,\,\frac{5}{2},\,0} \right),\,\,{\text{Z}}\left( {0,\,0,\,\frac{{25}}{2}} \right)\) A1
[2 marks]
\({\text{YZ}} = \sqrt {{{\left( {\frac{5}{2}} \right)}^2} + {{\left( {\frac{{25}}{2}} \right)}^2}} \) M1
\( = \sqrt {\frac{{325}}{2}} \left( { = \frac{{5\sqrt {104} }}{4} = \frac{{5\sqrt {26} }}{2}} \right)\) A1
[4 marks]
