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Date November 2021 Marks available 1 Reference code 21N.1.AHL.TZ0.16
Level Additional Higher Level Paper Paper 1 Time zone Time zone 0
Command term Find Question number 16 Adapted from N/A

Question

A ship S is travelling with a constant velocity, v, measured in kilometres per hour, where

v=-1215.

At time t=0 the ship is at a point A(300, 100) relative to an origin O, where distances are measured in kilometres.

A lighthouse is located at a point (129, 283).

Find the position vector OS of the ship at time t hours.

[1]
a.

Find the value of t when the ship will be closest to the lighthouse.

[6]
b.

An alarm will sound if the ship travels within 20 kilometres of the lighthouse.

State whether the alarm will sound. Give a reason for your answer.

[2]
c.

Markscheme

OS=300100+t-1215             A1

 

[1 mark]

a.

attempt to find the vector from L to S           (M1)

LS=171-183+t-1215             A1


EITHER

LS=171-12t2+15t-1832           (M1)(A1)

minimize to find t on GDC           (M1)


OR

S closest when LS·-1215=0           (M1)

171-183+t-1215·-1215=0

-2052+144t-2745+225t=0           (M1)(A1)


OR

S closest when LS·-1215=0           (M1)

LS=5k4k

OS=129+5k283+4k           (A1)

129+5k283+4k=300-12t100+15t

Solving simultaneously            (M1)


THEN

t=13             A1

 

[6 marks]

b.

the alarm will sound            A1

LS=19.2<20            R1


Note: Do not award A1R0.

 

[2 marks]

c.

Examiners report

Part (a) was generally well answered. Following that there were several possible approaches to this question. It was clear that many candidates understood that the closest approach could be found using the scalar product to find orthogonal vectors. However, typical efforts involved attempting to find the value of t by assuming that OS and OL are perpendicular. Other methods incorrectly applied were equating OS and OL. Of course, this led to separate values of t for each component. The method of minimizing LS was not commonly employed.

a.

Part (a) was generally well answered. Following that there were several possible approaches to this question. It was clear that many candidates understood that the closest approach could be found using the scalar product to find orthogonal vectors. However, typical efforts involved attempting to find the value of t by assuming that OS and OL are perpendicular. Other methods incorrectly applied were equating OS and OL. Of course, this led to separate values of t for each component. The method of minimizing LS was not commonly employed.

b.

Part (a) was generally well answered. Following that there were several possible approaches to this question. It was clear that many candidates understood that the closest approach could be found using the scalar product to find orthogonal vectors. However, typical efforts involved attempting to find the value of t by assuming that OS and OL are perpendicular. Other methods incorrectly applied were equating OS and OL. Of course, this led to separate values of t for each component. The method of minimizing LS was not commonly employed.

c.

Syllabus sections

Topic 3—Geometry and trigonometry » AHL 3.10—Vector definitions
Topic 3—Geometry and trigonometry » AHL 3.11—Vector equation of a line in 2d and 3d
Topic 3—Geometry and trigonometry » AHL 3.12—Vector applications to kinematics
Topic 3—Geometry and trigonometry » AHL 3.13—Scalar and vector products
Topic 3—Geometry and trigonometry

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