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Date	May 2019	Marks available	7	Reference code	19M.1.SL.TZ1.S_6
Level	Standard Level	Paper	Paper 1	Time zone	Time zone 1
Command term	Find	Question number	S_6	Adapted from	N/A

Question

The magnitudes of two vectors, u and v, are 4 and $\sqrt{3}$ $\sqrt 3$ respectively. The angle between u and v is $\frac{π}{6}$ $\frac{\pi }{6}$ .

Let w = u − v. Find the magnitude of w.

Markscheme

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

METHOD 1 (cosine rule)

diagram including u, v and included angle of $\frac{π}{6}$ $\frac{\pi }{6}$ (M1)

sketch of triangle with w (does not need to be to scale) (A1)

choosing cosine rule (M1)

eg $a^{2} + b^{2} - 2 a b cos C$ ${a^2} + {b^2} - 2ab\,{\text{cos}}\,C$

correct substitution A1

eg $4^{2} + {(\sqrt{3})}^{2} - 2 (4) (\sqrt{3}) cos \frac{π}{6}$ ${4^2} + {\left( {\sqrt 3 } \right)^2} - 2\left( 4 \right)\left( {\sqrt 3 } \right){\text{cos}}\frac{\pi }{6}$

$cos \frac{π}{6} = \frac{\sqrt{3}}{2}$ ${\text{cos}}\frac{\pi }{6} = \frac{{\sqrt 3 }}{2}$ (seen anywhere) (A1)

correct working (A1)

eg 16 + 3 − 12

| w | = $\sqrt{7}$ $\sqrt 7$ A1 N2

METHOD 2 (scalar product)

valid approach, in terms of u and v (seen anywhere) (M1)

eg | w |² = (u − v)•(u − v), | w |² = u•u − 2u•v + v•v, | w |²= ${(u_{1} - v_{1})}^{2} + {(u_{2} - v_{2})}^{2}$ ${\left( {{u_1} - {v_1}} \right)^2} + {\left( {{u_2}\; - \;{v_2}} \right)^2}$ ,

| w | = $\sqrt{{(u_{1} - v_{1})}^{2} + {(u_{2} - v_{2})}^{2} + {(u_{3} - v_{3})}^{2}}$ $\sqrt {{{\left( {{u_1} - {v_1}} \right)}^2} + {{\left( {{u_2}\; - \;{v_2}} \right)}^2} + {{\left( {{u_3}\; - \;{v_3}} \right)}^2}}$

correct value for u•u (seen anywhere) (A1)

eg | u |² = 16, u•u = 16, ${u_{1}}_{1}^{2} + {u_{2}}_{2}^{2} = 16$ ${u_1}^2 + {u_2}^2 = 16$

correct value for v•v (seen anywhere) (A1)

eg | v |² = 16, v•v = 3, ${v_{1}}_{1}^{2} + {v_{2}}_{2}^{2} + {v_{3}}_{3}^{2} = 3$ ${v_1}^2 + {v_2}^2 + {v_3}^2 = 3$

$cos (\frac{π}{6}) = \frac{\sqrt{3}}{2}$ ${\text{cos}}\left( {\frac{\pi }{6}} \right) = \frac{{\sqrt 3 }}{2}$ (seen anywhere) (A1)

u•v $= 4 \times \sqrt{3} \times \frac{\sqrt{3}}{2}$ $= 4 \times \sqrt 3 \times \frac{{\sqrt 3 }}{2}$ (= 6) (seen anywhere) A1

correct substitution into u•u − 2u•v + v•v or ${u_{1}}_{1}^{2} + {u_{2}}_{2}^{2} + {v_{1}}_{1}^{2} + {v_{2}}_{2}^{2} - 2 (u_{1} v_{1} + u_{2} v_{2})$ ${u_1}^2 + {u_2}^2 + {v_1}^2 + {v_2}^2 - 2\left( {{u_1}{v_1} + {u_2}{v_2}} \right)$ (2 or 3 dimensions) (A1)

eg 16 − 2(6) + 3 (= 7)

| w | = $\sqrt{7}$ $\sqrt 7$ A1 N2

Examiners report

[N/A]

Syllabus sections

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

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