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Date May 2018 Marks available 4 Reference code 18M.2.AHL.TZ1.H_11
Level Additional Higher Level Paper Paper 2 Time zone Time zone 1
Command term Show that and Write down Question number H_11 Adapted from N/A

Question

Two submarines A and B have their routes planned so that their positions at time t hours, 0 ≤ t < 20 , would be defined by the position vectors rA =(241)+t(110.15)=241+t110.15 and rB =(03.22)+t(0.51.20.1)=03.22+t0.51.20.1 relative to a fixed point on the surface of the ocean (all lengths are in kilometres).

To avoid the collision submarine B adjusts its velocity so that its position vector is now given by

rB =(03.22)+t(0.451.080.09)=03.22+t0.451.080.09.

Show that the two submarines would collide at a point P and write down the coordinates of P.

[4]
a.

Show that submarine B travels in the same direction as originally planned.

[1]
b.i.

Find the value of t when submarine B passes through P.

[2]
b.ii.

Find an expression for the distance between the two submarines in terms of t.

[5]
c.i.

Find the value of t when the two submarines are closest together.

[2]
c.ii.

Find the distance between the two submarines at this time.

[1]
c.iii.

Markscheme

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

rA rB        (M1)

2 − t = − 0.5t ⇒ t = 4       A1

checking t = 4 satisfies 4 + t = 3.2 + 1.2t and − 1 − 0.15t = − 2 + 0.1t      R1

P(−2, 8, −1.6)      A1

Note: Do not award final A1 if answer given as column vector.

[4 marks]

a.

0.9×(0.51.20.1)=(0.451.080.09)0.9×0.51.20.1=0.451.080.09     A1

Note: Accept use of cross product equalling zero.

hence in the same direction      AG

[1 mark]

b.i.

(0.45t3.2+1.08t2+0.09t)=(281.6)0.45t3.2+1.08t2+0.09t=281.6      M1

Note: The M1 can be awarded for any one of the resultant equations.

t=409=4.44t=409=4.44     A1

[2 marks]

b.ii.

rA − rB(2t4+t10.15t)(0.45t3.2+1.08t2+0.09t)2t4+t10.15t0.45t3.2+1.08t2+0.09t      (M1)(A1)

=(20.55t0.80.08t10.24t)=20.55t0.80.08t10.24t     (A1)

Note: Accept rA − rB.

distance D=(20.55t)2+(0.80.08t)2+(10.24t)2      M1A1

(=8.642.688t+0.317t2)

[5 marks]

c.i.

minimum when dDdt=0      (M1)

t = 3.83      A1

[2 marks]

c.ii.

0.511 (km)      A1

[1 mark]

c.iii.

Examiners report

[N/A]
a.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
c.i.
[N/A]
c.ii.
[N/A]
c.iii.

Syllabus sections

Topic 3— Geometry and trigonometry » AHL 3.12—Vector definitions
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Topic 3— Geometry and trigonometry » AHL 3.15—Classification of lines
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