Consider the plane with equation \(4x - 2y - z = 1\) and the line given by the parametric equations

     \(x = 3 - 2\lambda \)

     \(y = (2k - 1) + \lambda \)

     \(z = - 1 + k\lambda .\)

Given that the line is perpendicular to the plane, find

(a)     the value of k;

(b)     the coordinates of the point of intersection of the line and the plane.

(a)     \(\boldsymbol{a} = \left( {\begin{array}{*{20}{c}}
  4 \\
  { - 2} \\
  { - 1}
\end{array}} \right) \bot \) to the plane     \(\boldsymbol{e} = \left( {\begin{array}{*{20}{c}}
  { - 2} \\
  1 \\
  k
\end{array}} \right)\) is parallel to the line     (A1)(A1)

Note: Award A1 for each correct vector written down, even if not identified.

line \( \bot \) plane \( \Rightarrow \) \(\boldsymbol{e}\) parallel to \(\boldsymbol{a}\)

since \(\left( {\begin{array}{*{20}{c}}
  4 \\
  { - 2} \\
  { - 1}
\end{array}} \right) = t\left( {\begin{array}{*{20}{c}}
  { - 2} \\
  1 \\
  k
\end{array}} \right) \Rightarrow k = \frac{1}{2}\)     (M1)A1

(b)     \(4(3 - 2\lambda ) - 2\lambda - \left( { - 1 + \frac{1}{2}\lambda } \right) = 1\)     (M1)(A1)

Note: FT their value of k as far as possible.

\(\lambda = \frac{8}{7}\)     A1

\({\text{P}}\left( {\frac{5}{7},\frac{8}{7}, - \frac{3}{7}} \right)\)     A1

[8 marks]

Solutions to this question were often disappointing. In (a), some candidates found the value of k, incorrectly, by taking the scalar product of the normal vector to the plane and the direction of the line. Such candidates benefitted partially from follow through in (b) but not fully because their line turned out to be parallel to the plane and did not intersect it.