User interface language: English | Español

Date May 2022 Marks available 5 Reference code 22M.2.AHL.TZ2.11
Level Additional Higher Level Paper Paper 2 Time zone Time zone 2
Command term Find Question number 11 Adapted from N/A

Question

Two airplanes, A and B, have position vectors with respect to an origin O given respectively by

rA=19-11+t-624

rB=1012+t42-2

where t represents the time in minutes and 0t2.5.

Entries in each column vector give the displacement east of O, the displacement north of O and the distance above sea level, all measured in kilometres.

The two airplanes’ lines of flight cross at point P.

Find the three-figure bearing on which airplane B is travelling.

[2]
a.

Show that airplane A travels at a greater speed than airplane B.

[2]
b.

Find the acute angle between the two airplanes’ lines of flight. Give your answer in degrees.

[4]
c.

Find the coordinates of P.

[5]
d.i.

Determine the length of time between the first airplane arriving at P and the second airplane arriving at P.

[2]
d.ii.

Let D(t) represent the distance between airplane A and airplane B for 0t2.5.

Find the minimum value of D(t).

[5]
e.

Markscheme

let ϕ be the required angle (bearing)


EITHER

ϕ=90°-arctan12 =arctan2          (M1)


Note: Award M1 for a labelled sketch.


OR

cosϕ=01·421×20 =0.4472,=15          (M1)

ϕ=arccos0.4472


THEN

063°          A1


Note: Do not accept 063.6° or 63.4° or 1.10c.

 

[2 marks]

a.

METHOD 1

let bA be the speed of A and let bB be the speed of B

attempts to find the speed of one of A or B          (M1)

bA=-62+22+42  or  bB=42+22+-22


Note: Award M0 for bA=192+-12+12 and bB=12+02+122.


bA=7.48 =56 (km min-1) and bB=4.89 =24 (km min-1)          A1

bA>bB so A travels at a greater speed than B          AG

 

METHOD 2

attempts to use speed=distancetime

speedA=rAt2-rAt1t2-t1  and  speedB=rBt2-rBt1t2-t1          (M1)

for example:

speedA=rA1-rA01  and speedB=rB1-rB01

speedA=-62+22+421  and speedB=42+22+221

speedA=7.48214  and speedB=4.8924          A1

speedA>speedB so A travels at a greater speed than B          AG

 

[2 marks]

b.

attempts to use the angle between two direction vectors formula         (M1)

cosθ=-64+22+4-2-62+22+4242+22+-22         (A1)

cosθ=-0.7637 =-784  or  θ=arccos-0.7637 =2.4399

attempts to find the acute angle 180°-θ using their value of θ         (M1)

=40.2°         A1

 

[4 marks]

c.

for example, sets rAt1=rBt2 and forms at least two equations         (M1)

19-6t1=1+4t2

-1+2t1=2t2

1+4t1=12-2t2


Note: Award M0 for equations involving t only.


EITHER

attempts to solve the system of equations for one of t1 or t2         (M1)

t1=2  or  t2=32         A1


OR

attempts to solve the system of equations for t1 and t2         (M1)

t1=2  or  t2=32         A1


THEN

substitutes their t1 or t2 value into the corresponding rA or rB         (M1)

P7,3,9         A1


Note: Accept OP=739. Accept 7 km east of O3 km north of O and 9 km above sea level.

 

[5 marks]

d.i.

attempts to find the value of t1-t2           (M1)

t1-t2=2-32

0.5 minutes (30 seconds)         A1

 

[2 marks]

d.ii.

EITHER

attempts to find rB-rA           (M1)

rB-rA=-18111+t100-6

attempts to find their D(t)           (M1)

D(t)=10t-182+1+11-6t2         A1


OR

attempts to find rA-rB           (M1)

rA-rB=18-1-11+t-1006

attempts to find their D(t)           (M1)

D(t)=18-10t2+-12+-11+6t2         A1

 

Note: Award M0M0A0 for expressions using two different time parameters.


THEN

either attempts to find the local minimum point of D(t) or attempts to find the value of t such that D'(t)=0  (or equivalent)           (M1)

t=1.8088 =12368

D(t)=1.01459

minimum value of D(t) is 1.01 =119034 (km)         A1


Note: Award M0 for attempts at the shortest distance between two lines.

 

[5 marks]

e.

Examiners report

General comment about this question: many candidates were not exposed to this setting of vectors question and were rather lost.

Part (a) Probably the least answered question on the whole paper. Many candidates left it blank, others tried using 3D vectors. Out of those who calculated the angle correctly, only a small percentage were able to provide the correct true bearing as a 3-digit figure.

Part (b) Well done by many candidates who used the direction vectors to calculate and compare the speeds. A number of candidates tried to use the average rate of change but mostly unsuccessfully.

Part (c) Most candidates used the correct vectors and the formula to obtain the obtuse angle. Then only some read the question properly to give the acute angle in degrees, as requested.

Part (d) Well done by many candidates who used two different parameters. They were able to solve and obtain two values for time, the difference in minutes and the correct point of intersection. A number of candidates only had one parameter, thus scoring no marks in part (d) (i). The frequent error in part (d)(ii) was providing incorrect units.

Part (e) Many correct answers were seen with an efficient way of setting the question and using their GDC to obtain the answer, graphically or numerically. Some gave time only instead of actually giving the minimal distance. A number of candidates tried to find the distance between two skew lines ignoring the fact that the lines intersect.

a.
[N/A]
b.
[N/A]
c.
[N/A]
d.i.
[N/A]
d.ii.
[N/A]
e.

Syllabus sections

Topic 5 —Calculus » SL 5.8—Testing for max and min, optimisation. Points of inflexion
Show 202 related questions
Topic 3— Geometry and trigonometry » AHL 3.13—Scalar (dot) product
Topic 3— Geometry and trigonometry
Topic 5 —Calculus

View options