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.

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]

On the whole this question was well answered. Some candidates failed to find the complementary angle when using the formula with cosine.