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Date May 2017 Marks available 4 Reference code 17M.1.SL.TZ2.S_9
Level Standard Level Paper Paper 1 (without calculator) Time zone Time zone 2
Command term Find and Hence or otherwise Question number S_9 Adapted from N/A

Question

Note: In this question, distance is in metres and time is in seconds.

Two particles P 1 and P 2 start moving from a point A at the same time, along different straight lines.

After t seconds, the position of P 1 is given by r = ( 4 1 3 ) + t ( 1 2 1 ) .

Two seconds after leaving A, P 1 is at point B.

Two seconds after leaving A, P 2 is at point C, where AC = ( 3 0 4 ) .

Find the coordinates of A.

[2]
a.

Find  AB ;

[3]
b.i.

Find  | AB | .

[2]
b.ii.

Find cos B A ^ C .

[5]
c.

Hence or otherwise, find the distance between P 1 and P 2 two seconds after they leave A.

[4]
d.

Markscheme

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

recognizing t = 0 at A     (M1)

A is ( 4 ,   1 ,   3 )     A1     N2

[2 marks]

a.

METHOD 1

valid approach     (M1)

eg ( 4 1 3 ) + 2 ( 1 2 2 ) ,   ( 6 ,   3 ,   1 )

correct approach to find AB     (A1)

eg AO + OB ,  B A,  ( 6 3 1 ) ( 4 1 3 )

AB = ( 2 4 4 )     A1     N2

METHOD 2

recognizing AB is two times the direction vector     (M1)

correct working     (A1)

eg AB = 2 ( 1 2 2 )

AB = ( 2 4 4 )     A1     N2

[3 marks]

b.i.

correct substitution     (A1)

eg | AB | = 2 2 + 4 2 + 4 2 ,   4 + 16 + 16 ,   36

| AB | = 6     A1     N2

[2 marks]

b.ii.

METHOD 1 (vector approach)

valid approach involving AB and AC     (M1)

eg AB AC ,   BA AC AB × AC

finding scalar product and | AC |     (A1)(A1)

scalar product 2 ( 3 ) + 4 ( 0 ) 4 ( 4 )   ( = 10 )

| AC | = 3 2 + 0 2 + 4 2   ( = 5 )

substitution of their scalar product and magnitudes into cosine formula     (M1)

eg cos B A ^ C = 6 + 0 16 6 3 2 + 4 2

cos B A ^ C = 10 30 ( = 1 3 )     A1     N2

 

METHOD 2 (triangle approach)

valid approach involving cosine rule     (M1)

eg cos B A ^ C = A B 2 + A C 2 B C 2 2 × AB × AC

finding lengths AC and BC     (A1)(A1)

AC = 5 ,  BC = 9

substitution of their lengths into cosine formula     (M1)

eg cos B A ^ C = 5 2 + 6 2 9 2 2 × 5 × 6

cos B A ^ C = 20 60   ( = 1 3 )     A1     N2

[5 marks]

c.

Note:     Award relevant marks for working seen to find BC in part (c) (if cosine rule used in part (c)).

 

METHOD 1 (using cosine rule)

recognizing need to find BC     (M1)

choosing cosine rule     (M1)

eg c 2 = a 2 + b 2 2 a b cos C

correct substitution into RHS     A1

eg B C 2 = ( 6 ) 2 + ( 5 ) 2 2 ( 6 ) ( 5 ) ( 1 3 ) ,   36 + 25 + 20

distance is 9     A1     N2

 

METHOD 2 (finding magnitude of B C

recognizing need to find BC     (M1)

valid approach     (M1)

eg attempt to find O B or O C , O B = ( 6 3 1 ) or O C = ( 7 1 7 ) ,   B A + A C

correct working     A1

eg B C = ( 1 4 8 ) ,   C B = ( 1 4 8 ) ,   1 2 + 4 2 + 8 2 = 81

distance is 9     A1     N2

 

METHOD 3 (finding coordinates and using distance formula)

recognizing need to find BC     (M1)

valid approach     (M1)

eg attempt to find coordinates of B or C, B ( 6 ,   3 ,   1 ) or C ( 7 ,   1 ,   7 )

correct substitution into distance formula     A1

eg BC = ( 6 7 ) 2 + ( 3 ( 1 ) ) 2 + ( 1 7 ) 2 ,   1 2 + 4 2 + 8 2 = 81

distance is 9     A1     N2

[4 marks]

d.

Examiners report

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

Syllabus sections

Topic 3— Geometry and trigonometry » AHL 3.14—Vector equation of line
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Topic 3— Geometry and trigonometry

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