| 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∈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(13a−122a−231a−3) M1
=1(2(a−3)−(a−2))−3(2(a−3)−3(a−2))+(a−1)(2−6)
(or equivalent) A1
= 0 (therefore there is no unique solution) A1
[3 marks]
METHOD 2
(13a−122a−231a−3|11b):(13a−10−4−a0−8−2a|1−1b−3) M1A1
:(13a−10−4−a000|1−1b−1) (and 3 zeros imply no unique solution) A1
[3 marks]
METHOD 1
(13a−122a−231a−3|11b):(13a−10−4−a0−8−2a|1−1b−3) M1A1
:(13a−10−4−a000|1−1b−1) 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).
