Find the value of \(k\).

Given that c = a + 2b, find c.

evidence of scalar product     M1

eg\(\,\,\,\,\,\)a \( \bullet \) b, \(4(k + 3) + 2k\)

recognizing scalar product must be zero     (M1)

eg\(\,\,\,\,\,\)a \( \bullet \) b \( = 0,{\text{ }}4k + 12 + 2k = 0\)

correct working (must involve combining terms)     (A1)

eg  \(6k + 12,\,\,\,6k =  - 12\)

\(\,\,\,\,\,\)\(k =  - 2\)     A1     N2

[4 marks]

attempt to substitute their value of \(k\) (seen anywhere)     (M1)

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

correct working     (A1)

eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 2 \\ { - 4} \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} {4 + 2k + 6} \\ {2 + 2k} \end{array}} \right)\)

c = \(\left( {\begin{array}{*{20}{c}} 6 \\ { - 2} \end{array}} \right)\)     A1     N2

[3 marks]