| Date | May 2010 | Marks available | 4 | Reference code | 10M.2.hl.TZ2.7 |
| Level | HL only | Paper | 2 | Time zone | TZ2 |
| Command term | Find | Question number | 7 | Adapted from | N/A |
Question
The three planes
\(2x - 2y - z = 3\)
\(4x + 5y - 2z = - 3\)
\(3x + 4y - 3z = - 7\)
intersect at the point with coordinates (a, b, c).
Find the value of each of a, b and c.
The equations of three planes are
\(2x - 4y - 3z = 4\)
\( - x + 3y + 5z = - 2\)
\(3x - 5y - z = 6\).
Find a vector equation of the line of intersection of these three planes.
Markscheme
(a) use GDC or manual method to find a, b and c (M1)
obtain \(a = 2,{\text{ }}b = - 1,{\text{ }}c = 3\) (in any identifiable form) A1
[2 marks]
use GDC or manual method to solve second set of equations (M1)
obtain \(x = \frac{{4 - 11t}}{2};{\text{ }}y = \frac{{ - 7t}}{2};{\text{ }}z = t\) (or equivalent) (A1)
\(r = \left( {\begin{array}{*{20}{c}}
2 \\
0 \\
0
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
{ - 5.5} \\
{ - 3.5} \\
1
\end{array}} \right)\) (accept equivalent vector forms) M1A1
Note: Final A1 requires r = or equivalent.
[4 marks]
Examiners report
Generally well done.
Moderate success here. Some forgot that an equation must have an = sign.
