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Date	May Specimen paper	Marks available	2	Reference code	SPM.1.AHL.TZ0.11
Level	Additional Higher Level	Paper	Paper 1	Time zone	Time zone 0
Command term	Find	Question number	11	Adapted from	N/A

Question

A particle P moves with velocity v = $⎛ ⎜ ⎝ \begin{matrix} - 15 24 \end{matrix} ⎞ ⎟ ⎠$ $\left( {\begin{array}{*{20}{c}} { - 15} \\ 2 \\ 4 \end{array}} \right)$ in a magnetic field, B = $⎛ ⎜ ⎝ \begin{matrix} 0 d 1 \end{matrix} ⎞ ⎟ ⎠$ $\left( {\begin{array}{*{20}{c}} 0 \\ d \\ 1 \end{array}} \right)$ , $d \in \mathbb{R}$ .

Given that v is perpendicular to B, find the value of $d$ .

[2]

The force, F, produced by P moving in the magnetic field is given by the vector equation F = $a$ v × B, $a \in {\mathbb{R}^ + }$ .

Given that | F | = 14, find the value of $a$ .

[4]

Markscheme

$15 \times 0 + 2d + 4 = 0$ (M1)

$d = - 2$ A1

[2 marks]

$a\left( {\begin{array}{*{20}{c}} { - 15} \\ 2 \\ 4 \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 0 \\ { - 2} \\ 1 \end{array}} \right)$ (M1)

$= a\left( {\begin{array}{*{20}{c}} {10} \\ {15} \\ {30} \end{array}} \right)\left( { = 5a\left( {\begin{array}{*{20}{c}} 2 \\ 3 \\ 6 \end{array}} \right)} \right)$ A1

magnitude is $5a\sqrt {{2^2} + {3^2} + {6^2}} = 14$ M1

$a = \frac{{14}}{{35}}\,\left( { = 0.4} \right)$ A1

[4 marks]

Examiners report

[N/A]

Syllabus sections

Topic 3—Geometry and trigonometry » AHL 3.13—Scalar and vector products

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