19M.1.AHL.TZ1.H_1

pestleMathematicsAAHLPaper 119M· ahl-3-13-scalar-(dot)-productsource ↗

Let a = $\left( {\begin{array}{*{20}{c}} 2 \\ k \\ { - 1} \end{array}} \right)$ and b = $\left( {\begin{array}{*{20}{c}} { - 3} \\ {k + 2} \\ k \end{array}} \right)$ , $k \in \mathbb{R}$ .

Given that a and b are perpendicular, find the possible values of $k$ .

Markscheme / solution

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

a • b = $\left( {\begin{array}{*{20}{c}} 2 \\ k \\ { - 1} \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} { - 3} \\ {k + 2} \\ k \end{array}} \right)$

$= - 6 + k\left( {k + 2} \right) - k$ A1

a • b = 0 (M1)

${k^2} + k - 6 = 0$

attempt at solving their quadratic equation (M1)

$\left( {k + 3} \right)\left( {k - 2} \right) = 0$

$k = - 3{\text{,}}\,\,2$ A1

Note: Attempt at solving using |a||b| = |a × b| will be M1A0A0A0 if neither answer found M1(A1)A1A0
for one correct answer and M1(A1)A1A1 for two correct answers.

[4 marks]

Examiners’ report

[N/A]