(a)     Show that the two planes

\[{\pi _1}:x + 2y - z = 1\]

\[{\pi _2}:x + z =  - 2\]

are perpendicular.

(b)     Find the equation of the plane \({\pi _3}\) that passes through the origin and is perpendicular to both \({\pi _1}\) and \({\pi _2}\).

(a)     for using normal vectors     (M1)

\(\left( {\begin{array}{*{20}{c}}
  1 \\
  2 \\
  { - 1}
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
  1 \\
  0 \\
  1
\end{array}} \right) = 1 - 1 = 0\)     M1A1

hence the two planes are perpendicular     AG

(b)     METHOD 1

EITHER

\(\left| {\begin{array}{*{20}{c}}
  i&j&k \\
  1&2&{ - 1} \\
  1&0&1
\end{array}} \right| = \) 2i – 2j– 2k     M1A1

OR

if \(\left( {\begin{array}{*{20}{c}}
  a \\
  b \\
  c
\end{array}} \right)\) is normal to \({\pi _3}\), then

\(a + 2b - c = 0\) and \(a + c = 0\)     M1

a solution is a = 1, b = –1, c = –1     A1

THEN

\({\pi _3}\) has equation \(x - y - z = d\)     (M1)

as it goes through the origin, d = 0 so \({\pi _3}\) has equation \(x - y - z = 0\)     A1

Note: The final (M1)A1 are independent of previous working.

METHOD 2

\(r = \left( {\begin{array}{*{20}{c}}
  0 \\
  0 \\
  0
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
  1 \\
  2 \\
  { - 1}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
  1 \\
  0 \\
  1
\end{array}} \right)\)     A1(A1)A1A1

[7 marks]

Although many candidates were successful in answering this question, a surprising number showed difficulties in working with normal vectors. In part (b) there were several candidates who found the cross product of the vectors but were unable to use it to write the equation of the plane.