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Date	May 2022	Marks available	5	Reference code	22M.2.AHL.TZ1.7
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 1
Command term	Find	Question number	7	Adapted from	N/A

Question

Consider the vectors $a$ and $b$ such that $a = (\begin{matrix} 12 \\ - 5 \end{matrix})$ and $|b| = 15$ .

Consider the vector $p$ such that $p = a + b$ .

Consider the vector $q$ such that $q = (\begin{matrix} x \\ y \end{matrix})$ , where $x, y \in ℝ^{+}$ .

Find the possible range of values for $|a + b|$ .

[2]

Given that $|a + b|$ is a minimum, find $p$ .

[2]

Find $q$ such that $| q | = | b |$ and $q$ is perpendicular to $a$ .

[5]

Markscheme

$|a| = \sqrt{12^{2} + {(- 5)}^{2}} (= 13)$ (A1)

$2 \leq |a + b| \leq 28$ (accept min $2$ and max $28$ ) A1

Note: Award (A1)A0 for $2$ and $28$ seen with no indication that they are the endpoints of an interval.

[2 marks]

recognition that $p$ or $b$ is a negative multiple of $a$ (M1)

$p = - 2 \hat{a}$ OR $b = - \frac{15}{13} a (= - \frac{15}{13} (\begin{matrix} 12 \\ - 5 \end{matrix}))$

$p = - \frac{2}{13} (\begin{matrix} 12 \\ - 5 \end{matrix}) (= (\begin{matrix} - 1.85 \\ 0.769 \end{matrix}))$ A1

[2 marks]

METHOD 1

$q$ is perpendicular to $(\begin{matrix} 12 \\ - 5 \end{matrix})$

$\Rightarrow q$ is in the direction $(\begin{matrix} 5 \\ 12 \end{matrix})$ (M1)

$q = k (\begin{matrix} 5 \\ 12 \end{matrix})$ (A1)

$(|q| =) \sqrt{{(5 k)}^{2} + {(12 k)}^{2}} = 15$ (M1)

$k = \frac{15}{13}$ (A1)

$q = \frac{15}{13} (\begin{matrix} 5 \\ 12 \end{matrix}) (= (\begin{matrix} \frac{75}{13} \\ \frac{180}{13} \end{matrix}) = (\begin{matrix} 5.77 \\ 13.8 \end{matrix}))$ A1

METHOD 2

$q$ is perpendicular to $(\begin{matrix} 12 \\ - 5 \end{matrix})$

attempt to set scalar product $q . a = 0$ OR product of gradients $= - 1$ (M1)

$12 x - 5 y = 0$ (A1)

$(|q| =) \sqrt{x^{2} + y^{2}} = 15$

attempt to solve simultaneously to find a quadratic in $x$ or $y$ (M1)

$x^{2} + {(\frac{12 x}{5})}^{2} = 15^{2}$ OR ${(\frac{5 y}{12})}^{2} + y^{2} = 15^{2}$

$q = (\begin{matrix} \frac{75}{13} \\ \frac{180}{13} \end{matrix}) (= (\begin{matrix} 5.77 \\ 13.8 \end{matrix}))$ A1A1

Note: Award A1 independently for each value. Accept values given as $x = \frac{75}{13}$ and $y = \frac{180}{13}$ or equivalent.

[5 marks]

Examiners report

[N/A]

Syllabus sections

Topic 3— Geometry and trigonometry » AHL 3.13—Scalar (dot) product

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