| Date | November 2017 | Marks available | 3 | Reference code | 17N.1.hl.TZ0.2 |
| Level | HL only | Paper | 1 | Time zone | TZ0 |
| Command term | Find | Question number | 2 | Adapted from | N/A |
Question
The points A and B are given by \({\text{A}}(0,{\text{ }}3,{\text{ }} - 6)\) and \({\text{B}}(6,{\text{ }} - 5,{\text{ }}11)\).
The plane Π is defined by the equation \(4x - 3y + 2z = 20\).
Find a vector equation of the line L passing through the points A and B.
Find the coordinates of the point of intersection of the line L with the plane Π.
Markscheme
\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 6 \\ { - 8} \\ {17} \end{array}} \right)\) (A1)
r = \(\left( {\begin{array}{*{20}{c}} 0 \\ 3 \\ { - 6} \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} 6 \\ { - 8} \\ {17} \end{array}} \right)\) or r = \(\left( {\begin{array}{*{20}{c}} 6 \\ { - 5} \\ {11} \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} 6 \\ { - 8} \\ {17} \end{array}} \right)\) M1A1
Note: Award M1A0 if r = is not seen (or equivalent).
[3 marks]
substitute line L in \(\Pi :4(6\lambda ) - 3(3 - 8\lambda ) + 2( - 6 + 17\lambda ) = 20\) M1
\(82\lambda = 41\)
\(\lambda = \frac{1}{2}\) (A1)
r = \(\left( {\begin{array}{*{20}{c}} 0 \\ 3 \\ { - 6} \end{array}} \right) + \frac{1}{2}\left( {\begin{array}{*{20}{c}} 6 \\ { - 8} \\ {17} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 3 \\ { - 1} \\ {\frac{5}{2}} \end{array}} \right)\)
so coordinate is \(\left( {3,{\text{ }} - 1,{\text{ }}\frac{5}{2}} \right)\) A1
Note: Accept coordinate expressed as position vector \(\left( {\begin{array}{*{20}{c}} 3 \\ { - 1} \\ {\frac{5}{2}} \end{array}} \right)\).
[3 marks]
