Date | November 2008 | Marks available | 18 | Reference code | 08N.2.hl.TZ0.10 |
Level | HL only | Paper | 2 | Time zone | TZ0 |
Command term | Find, Show that, Write, and Hence | Question number | 10 | Adapted from | N/A |
Question
(a) Write the vector equations of the following lines in parametric form.
\[{r_1} = \left( {\begin{array}{*{20}{c}}
3 \\
2 \\
7
\end{array}} \right) + m\left( {\begin{array}{*{20}{c}}
2 \\
{ - 1} \\
2
\end{array}} \right)\]
\[{r_2} = \left( {\begin{array}{*{20}{c}}
1 \\
4 \\
2
\end{array}} \right) + n\left( {\begin{array}{*{20}{c}}
4 \\
{ - 1} \\
1
\end{array}} \right)\]
(b) Hence show that these two lines intersect and find the point of intersection, A.
(c) Find the Cartesian equation of the plane \(\prod \) that contains these two lines.
(d) Let B be the point of intersection of the plane \(\prod \) and the line\({r} = \left( {\begin{array}{*{20}{c}}
{ - 8} \\
{ - 3} \\
0
\end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}
3 \\
8 \\
2
\end{array}} \right)\).
Find the coordinates of B.
(e) If C is the mid-point of AB, find the vector equation of the line perpendicular to the plane \(\prod \) and passing through C.
Markscheme
(a) \(x = 3 + 2m\)
\(y = 2 - m\)
\(z = 7 + 2m\) A1
\(x = 1 + 4n\)
\(y = 4 - n\)
\(z = 2 + n\) A1
[2 marks]
(b) \(3 + 2m = 1 + 4n \Rightarrow 2m - 4n = - 2{\text{ }}({\text{i}})\)
\(2 - m = 4 - n \Rightarrow m - n = - 2{\text{ }}({\text{ii}})\) M1
\(7 + 2m = 2 + n \Rightarrow 2m - n = - 5{\text{ }}({\text{iii}})\)
\(({\text{iii}}) - ({\text{ii}}) \Rightarrow m = - 3\) A1
\( \Rightarrow n = - 1\) A1
Substitute in (i), –6 + 4 = –2 . Hence lines intersect. R1
Point of intersection A is (–3, 5,1) A1
[5 marks]
(c) \(\left| {\begin{array}{*{20}{c}}
i&j&k \\
2&{ - 1}&2 \\
4&{ - 1}&1
\end{array}} \right| = \left( {\begin{array}{*{20}{c}}
1 \\
6 \\
2
\end{array}} \right)\) M1A1
\(r \cdot \left( {\begin{array}{*{20}{c}}
1 \\
6 \\
2
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
3 \\
2 \\
7
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
1 \\
6 \\
2
\end{array}} \right)\) (M1)
\(r \cdot \left( {\begin{array}{*{20}{c}}
1 \\
6 \\
2
\end{array}} \right) = 29\)
x + 6y + 2z = 29 A1
Note: Award M1A0 if answer is not in Cartesian form.
[4 marks]
(d) \(x = - 8 + 3\lambda \)
\(y = - 3 + 8\lambda \) (M1)
\(z = 2\lambda \)
Substitute in equation of plane.
\( - 8 + 3\lambda - 18 + 48\lambda + 4\lambda = 29\) M1
\(55\lambda = 55\)
\(\lambda = 1\) A1
Coordinates of B are (–5, 5, 2) A1
[4 marks]
(e) Coordinates of C are \(\left( { - 4,{\text{ 5, }}\frac{3}{2}} \right)\) (A1)
\(r = \left( {\begin{array}{*{20}{c}}
{ - 4} \\
5 \\
{\frac{3}{2}}
\end{array}} \right) + \mu \left( {\begin{array}{*{20}{c}}
1 \\
6 \\
2
\end{array}} \right)\) M1A1
Note: Award M1A0 unless candidate writes r = or \(\left( {\begin{array}{*{20}{c}}
x \\
y \\
z
\end{array}} \right)\)
[3 marks]
Total [18 marks]
Examiners report
Most candidates found this question to their liking and many correct solutions were seen. In (b), some candidates solved two equations for m and n but then failed to show that these values satisfied the third equation. In (e), some candidates used an incorrect formula to determine the coordinates of the mid-point of AB .