| Date | November 2012 | Marks available | 4 | Reference code | 12N.1.hl.TZ0.6 |
| Level | HL only | Paper | 1 | Time zone | TZ0 |
| Command term | Find | Question number | 6 | Adapted from | N/A |
Question
Consider the following equations, where a , \(b \in \mathbb{R}:\)
\(x + 3y + (a - 1)z = 1\)
\(2x + 2y + (a - 2)z = 1\)
\(3x + y + (a - 3)z = b.\)
If each of these equations defines a plane, show that, for any value of a , the planes do not intersect at a unique point.
Find the value of b for which the intersection of the planes is a straight line.
Markscheme
METHOD 1
\(\det \left( {\begin{array}{*{20}{c}}
1&3&{a - 1} \\
2&2&{a - 2} \\
3&1&{a - 3}
\end{array}} \right)\) M1
\( = 1\left( {2(a - 3) - (a - 2)} \right) - 3\left( {2(a - 3) - 3(a - 2)} \right) + (a - 1)(2 - 6)\)
(or equivalent) A1
= 0 (therefore there is no unique solution) A1
[3 marks]
METHOD 2
\(\left( {\begin{array}{*{20}{c}}
1&3&{a - 1}\\
2&2&{a - 2}\\
3&1&{a - 3}
\end{array}} \right|\left. {\begin{array}{*{20}{c}}
1\\
1\\
b
\end{array}} \right):\left( {\begin{array}{*{20}{c}}
1&3&{a - 1}\\
0&{ - 4}&{ - a}\\
0&{ - 8}&{ - 2a}
\end{array}} \right|\left. {\begin{array}{*{20}{c}}
1\\
{ - 1}\\
{b - 3}
\end{array}} \right)\) M1A1
\(:\left( {\begin{array}{*{20}{c}}
1&3&{a - 1}\\
0&{ - 4}&{ - a}\\
0&0&0
\end{array}} \right|\left. {\begin{array}{*{20}{c}}
1\\
{ - 1}\\
{b - 1}
\end{array}} \right)\) (and 3 zeros imply no unique solution) A1
[3 marks]
METHOD 1
\(\left( {\begin{array}{*{20}{c}}
1&3&{a - 1}\\
2&2&{a - 2}\\
3&1&{a - 3}
\end{array}} \right|\left. {\begin{array}{*{20}{c}}
1\\
1\\
b
\end{array}} \right):\left( {\begin{array}{*{20}{c}}
1&3&{a - 1}\\
0&{ - 4}&{ - a}\\
0&{ - 8}&{ - 2a}
\end{array}} \right|\left. {\begin{array}{*{20}{c}}
1\\
{ - 1}\\
{b - 3}
\end{array}} \right)\) M1A1
\(:\left( {\begin{array}{*{20}{c}}
1&3&{a - 1}\\
0&{ - 4}&{ - a}\\
0&0&0
\end{array}} \right|\left. {\begin{array}{*{20}{c}}
1\\
{ - 1}\\
{b - 1}
\end{array}} \right)\) A1
b = 1 A1 N2
Note: Award M1 for an attempt to use row operations.
[4 marks]
METHOD 2
b = 1 A4
Note: Award A4 only if “ b −1 ” seen in (a).
[4 marks]
Examiners report
The best candidates used row reduction correctly in part a) and were hence able to deduce b = 1 in part b) for an easy final 4 marks. The determinant method was often usefully employed in part a).
The best candidates used row reduction correctly in part a) and were hence able to deduce b = 1 in part b) for an easy final 4 marks. The determinant method was often usefully employed in part a).
