The three vectors \(\boldsymbol{a}\), \(\boldsymbol{b}\) and \(\boldsymbol{c}\) are given by\[{\boldsymbol{a}} = \left( {\begin{array}{*{20}{c}}
  {2y} \\
  { - 3x} \\
  {2x}
\end{array}} \right),{\text{ }}{\boldsymbol{b}}{\text{ }} = \left( {\begin{array}{*{20}{c}}
  {4x} \\
  y \\
  {3 - x}
\end{array}} \right),{\text{ }}{\boldsymbol{c}}{\text{ }} = \left( {\begin{array}{*{20}{c}}
  4 \\
  { - 7} \\
  6
\end{array}} \right){\text{ where }}x,y \in \mathbb{R}{\text{ }}{\text{.}}\]

(a)     If a + 2b − c = 0, find the value of x and of y.

(b)     Find the exact value of \(|\)a + 2b\(|\).

(a)     \(2y + 8x = 4\)     M1

\( - 3x + 2y = - 7\)     A1

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

Note: Award M1 for attempt at components, A1 for two correct equations.

   No penalty for not checking the third equation.

solving : x = 1, y = –2     A1

(b)     \(|\)a + 2b\(| = \left| {\left( {\begin{array}{*{20}{c}}
  { - 4} \\
  { - 3} \\
  2
\end{array}} \right) + 2\left( {\begin{array}{*{20}{c}}
  4 \\
  { - 2} \\
  2
\end{array}} \right)} \right|\)

\( = \left| {\left( {\begin{array}{*{20}{c}}
  4 \\
  { - 7} \\
  6
\end{array}} \right)} \right|\)

\( \Rightarrow |\)a + 2b\(|\) \( = \sqrt {{4^2} + {{( - 7)}^2} + {6^2}} \)     (M1)

\( = \sqrt {101} \)     A1

[5 marks]

The majority of candidates understood what was required in part (a) of this question and gained the correct answer. Most candidates were able to do part (b) but few realised that they did not have to calculate \(|\)a + 2b\(|\) as this is \(|\)c\(|\). Many candidates lost time on this question.