17M.1.SL.TZ2.S_2

pestleMathematicsAASLPaper 117M· ahl-3-13-scalar-(dot)-productsource ↗

The vectors a = $\left( {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right)$ and b = $\left( {\begin{array}{*{20}{c}} {k + 3} \\ k \end{array}} \right)$ are perpendicular to each other.

Find the value of $k$ .

[4]

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

[3]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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]

Examiners’ report

[N/A]