Date | May 2008 | Marks available | 6 | Reference code | 08M.2.sl.TZ2.7 |
Level | SL only | Paper | 2 | Time zone | TZ2 |
Command term | Find | Question number | 7 | Adapted from | N/A |
Question
The line L1 is represented by \({{\boldsymbol{r}}_1} = \left( {\begin{array}{*{20}{c}}
2\\
5\\
3
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
1\\
2\\
3
\end{array}} \right)\) and the line L2 by \({{\boldsymbol{r}}_2} = \left( {\begin{array}{*{20}{c}}
3\\
{ - 3}\\
8
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
{ - 1}\\
3\\
{ - 4}
\end{array}} \right)\) .
The lines L1 and L2 intersect at point T. Find the coordinates of T.
Markscheme
evidence of equating vectors (M1)
e.g. \({L_1} = {L_2}\)
for any two correct equations A1A1
e.g. \(2 + s = 3 - t\) , \(5 + 2s = - 3 + 3t\) , \(3 + 3s = 8 - 4t\)
attempting to solve the equations (M1)
finding one correct parameter \((2 = - 1{\text{, }}t = 2)\) A1
the coordinates of T are \((1{\text{, }}3{\text{, }}0)\) A1 N3
[6 marks]
Examiners report
Those candidates prepared in this topic area answered the question particularly well, often only making some calculation error when solving the resulting system of equations. Curiously, a few candidates found correct values for s and t, but when substituting back into one of the vector equations, neglected to find the z-coordinate of T.