| Date | May 2008 | Marks available | 6 | Reference code | 08M.2.hl.TZ1.3 |
| Level | HL only | Paper | 2 | Time zone | TZ1 |
| Command term | Find | Question number | 3 | Adapted from | N/A |
Question
A ray of light coming from the point (−1, 3, 2) is travelling in the direction of vector \(\left( {\begin{array}{*{20}{c}}
4 \\
1 \\
{ - 2}
\end{array}} \right)\) and meets the plane \(\pi :x + 3y + 2z - 24 = 0\) .
Find the angle that the ray of light makes with the plane.
Markscheme
The normal vector to the plane is \(\left( {\begin{array}{*{20}{c}}
1 \\
3 \\
2
\end{array}} \right)\) . (A1)
EITHER
\(\theta \) is the angle between the line and the normal to the plane.
\(\cos \theta = \frac{{\left( {\begin{array}{*{20}{c}}
4 \\
1 \\
{ - 2}
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
1 \\
3 \\
2
\end{array}} \right)}}{{\sqrt {14} \sqrt {21} }} = \frac{3}{{\sqrt {14} \sqrt {21} }} = \left( {\frac{3}{{7\sqrt 6 }}} \right)\) (M1)A1A1
\( \Rightarrow \theta = 79.9^\circ {\text{ }}( = 1.394…)\) A1
The required angle is 10.1° (= 0.176) A1
OR
\(\phi \) is the angle between the line and the plane.
\(\sin \phi = \frac{{\left( {\begin{array}{*{20}{c}}
4 \\
1 \\
{ - 2}
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
1 \\
3 \\
2
\end{array}} \right)}}{{\sqrt {14} \sqrt {21} }} = \frac{3}{{\sqrt {14} \sqrt {21} }}\) (M1)A1A1
\(\phi \) = 10.1° (= 0.176) A2
[6 marks]
Examiners report
On the whole this question was well answered. Some candidates failed to find the complementary angle when using the formula with cosine.
