Date | November 2011 | Marks available | 4 | Reference code | 11N.2.hl.TZ0.13 |
Level | HL only | Paper | 2 | Time zone | TZ0 |
Command term | Show that | Question number | 13 | Adapted from | N/A |
Question
Two planes \({\Pi _1}\) and \({\Pi _2}\) have equations \(2x + y + z = 1\) and \(3x + y - z = 2\) respectively.
Find the vector equation of L, the line of intersection of \({\Pi _1}\) and \({\Pi _2}\).
Show that the plane \({\Pi _3}\) which is perpendicular to \({\Pi _1}\) and contains L, has equation \(x - 2z = 1\).
The point P has coordinates (−2, 4, 1) , the point Q lies on \({\Pi _3}\) and PQ is perpendicular to \({\Pi _2}\). Find the coordinates of Q.
Markscheme
(a) METHOD 1
solving simultaneously (gdc) (M1)
\(x = 1 + 2z;{\text{ }}y = - 1 - 5z\) A1A1
\(L:\boldsymbol{r} = \left( {\begin{array}{*{20}{c}}
1 \\
{ - 1} \\
0
\end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}
2 \\
{ - 5} \\
1
\end{array}} \right)\) A1A1A1
Note: \({1^{{\text{st}}}}\) A1 is for r =.
[6 marks]
METHOD 2
direction of line \( = \left| {\begin{array}{*{20}{c}}
\boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k} \\
3&1&{ - 1} \\
2&1&1
\end{array}} \right|\) (last two rows swapped) M1
= 2i − 5j + k A1
putting z = 0, a point on the line satisfies \(2x + y = 1,{\text{ }}3x + y = 2\) M1
i.e. (1, −1, 0) A1
the equation of the line is
\(\left( {\begin{array}{*{20}{c}}
x \\
y \\
z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
1 \\
{ - 1} \\
0
\end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}
2 \\
{ - 5} \\
1
\end{array}} \right)\) A1A1
Note: Award A0A1 if \(\left( {\begin{array}{*{20}{c}}
x \\
y \\
z
\end{array}} \right)\) is missing.
[6 marks]
\(\left( {\begin{array}{*{20}{c}}
2 \\
1 \\
1
\end{array}} \right) \times \left( {\begin{array}{*{20}{c}}
2 \\
{ - 5} \\
1
\end{array}} \right)\) M1
= 6i − 12k A1
hence, n = i − 2k
\({\boldsymbol{n}} \cdot {\boldsymbol{a}} = \left( {\begin{array}{*{20}{c}}
1 \\
0 \\
{ - 2}
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
1 \\
{ - 1} \\
0
\end{array}} \right) = 1\) M1A1
therefore r \( \cdot \) n = a \( \cdot \) n \( \Rightarrow x - 2z = 1\) AG
[4 marks]
METHOD 1
P = (−2, 4, 1), Q = \((x,{\text{ }}y,{\text{ }}z)\)
\(\overrightarrow {{\text{PQ}}} = \left( {\begin{array}{*{20}{c}}
{x + 2} \\
{y - 4} \\
{z - 1}
\end{array}} \right)\) A1
\(\overrightarrow {{\text{PQ}}} \) is perpendicular to \(3x + y - z = 2\)
\( \Rightarrow \overrightarrow {{\text{PQ}}} \) is parallel to 3i + j − k R1
\( \Rightarrow x + 2 = 3t;{\text{ }}y - 4 = t;{\text{ }}z - 1 = - t\) A1
\(1 - z = t \Rightarrow x + 2 = 3 - 3z \Rightarrow x + 3z = 1\) A1
solving simultaneously \(x + 3z = 1;{\text{ }}x - 2z = 1\) M1
\(5z = 0 \Rightarrow z = 0;{\text{ }}x = 1,{\text{ }}y = 5\) A1
hence, Q = (1, 5, 0)
[6 marks]
METHOD 2
Line passing through PQ has equation
\({\mathbf{r}} = \begin{array}{*{20}{c}}
{ - 2} \\
4 \\
1
\end{array} + t\begin{array}{*{20}{c}}
3 \\
1 \\
{ - 1}
\end{array}\) M1A1
Meets \({\pi _3}\) when:
\( - 2 + 3t - 2(1 - t) = 1\) M1A1
t = 1 A1
Q has coordinates (1, 5, 0) A1
[6 marks]
Examiners report
Candidates generally attempted this question but with varying degrees of success. Although (a) was answered best of all the parts, quite a few did not use correct notation to designate the vector equation of a line, i.e., r =, or its equivalent. In (b) some candidates incorrectly assumed the result and worked the question from there. In (c) some candidates did not understand the necessary relationships to make a meaningful attempt.
Candidates generally attempted this question but with varying degrees of success. Although (a) was answered best of all the parts, quite a few did not use correct notation to designate the vector equation of a line, i.e., r =, or its equivalent. In (b) some candidates incorrectly assumed the result and worked the question from there. In (c) some candidates did not understand the necessary relationships to make a meaningful attempt.
Candidates generally attempted this question but with varying degrees of success. Although (a) was answered best of all the parts, quite a few did not use correct notation to designate the vector equation of a line, i.e., r =, or its equivalent. In (b) some candidates incorrectly assumed the result and worked the question from there. In (c) some candidates did not understand the necessary relationships to make a meaningful attempt.