A plane \(\pi \) has vector equation r = (−2i + 3j − 2k) + \(\lambda \)(2i + 3j + 2k) + \(\mu \)(6i − 3j + 2k).

(a)     Show that the Cartesian equation of the plane \(\pi \) is 3x + 2y − 6z = 12.

(b)     The plane \(\pi \) meets the x, y and z axes at A, B and C respectively. Find the coordinates of A, B and C.

(c)     Find the volume of the pyramid OABC.

(d)     Find the angle between the plane \(\pi \) and the x-axis.

(e)     Hence, or otherwise, find the distance from the origin to the plane \(\pi \).

(f)     Using your answers from (c) and (e), find the area of the triangle ABC.

(a)     EITHER

normal to plane given by

\(\left| {\begin{array}{*{20}{c}}
  i&j&k \\
  2&3&2 \\
  6&{ - 3}&2
\end{array}} \right|\)     M1A1

= 12i + 8j – 24k     A1

equation of \(\pi \) is \(3x + 2y - 6z = d\)     (M1)

as goes through (–2, 3, –2) so d = 12     M1A1

\(\pi :3x + 2y - 6z = 12\)     AG

OR

\(x = - 2 + 2\lambda + 6\mu \)

\(y = 3 + 3\lambda - 3\mu \)

\(z = - 2 + 2\lambda + 2\mu \)

eliminating \(\mu \)

\(x + 2y = 4 + 8\lambda \)

\(2y + 3z = 12\lambda \)     M1A1A1

eliminating \(\lambda \)

\(3(x + 2y) - 2(2y + 3z) = 12\)     M1A1A1

\(\pi :3x + 2y - 6z = 12\)     AG

[6 marks]

 

(b)     therefore A(4, 0, 0), B(0, 6, 0) and C(0, 0, 2)     A1A1A1

Note: Award A1A1A0 if position vectors given instead of coordinates.

 

[3 marks]

 

(c)     area of base \({\text{OAB}} = \frac{1}{2} \times 4 \times 6 = 12\)     M1

\(V = \frac{1}{3} \times 12 \times 2 = 8\)     M1A1

[3 marks]

 

(d)     \(\left( {\begin{array}{*{20}{c}}
  3 \\
  2 \\
  { - 6}
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
  1 \\
  0 \\
  0
\end{array}} \right) = 3 = 7 \times 1 \times \cos \phi \)     M1A1

\(\phi = \arccos \frac{3}{7}\)

so \(\theta = 90 - \arccos \frac{3}{7} = 25.4^\circ \,\,\,\,\,\)(accept 0.443 radians)     M1A1

[4 marks]

 

(e)     \(d = 4\sin \theta = \frac{{12}}{7}\,\,\,\,\,( = 1.71)\)     (M1)A1

[2 marks]

 

(f)     \(8 = \frac{1}{3} \times \frac{{12}}{7} \times {\text{area}} \Rightarrow {\text{area}} = 14\)     M1A1

Note: If answer to part (f) is found in an earlier part, award M1A1, regardless of the fact that it has not come from their answers to part (c) and part (e).

 

[2 marks]

Total [20 marks]

The question was generally well answered, although there were many students who failed to recognise that the volume was most logically found using a base as one of the coordinate planes.