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Date November 2019 Marks available 2 Reference code 19N.1.SL.TZ0.S_9
Level Standard Level Paper Paper 1 (without calculator) Time zone Time zone 0
Command term Find Question number S_9 Adapted from N/A

Question

The points A and B have position vectors  ( 2 4 4 ) and  ( 6 8 0 )  respectively.

Point C has position vector  ( 1 k 0 ) . Let O be the origin.

Find, in terms of k ,

OA OC .

[2]
a.i.

OB OC .

[1]
a.ii.

Given that  A O ^ C = B O ^ C , show that k = 7 .

[8]
b.

Calculate the area of triangle  AOC .

[6]
c.

Markscheme

correct substitution into either  OA OC or into  OB OC (in (ii))          (A1)     

eg       2 × ( 1 ) + 4 × k ,   6 × ( 1 ) + 8 × k

correct expression           A1   N1

eg       2 + 4 k ,   4 k + 2

[2 marks]

a.i.

correct expression           A1   N1

eg       8 k 6 ,   6 + 8 k

[1 mark]

a.ii.

finding magnitudes (seen anywhere)           A1A1

eg       ( 2 ) 2 + ( 4 ) 2 + ( 4 ) 2 ( = 6 ) ,   ( 6 ) 2 + ( 8 ) 2 + 0 2 ( = 10 )

correct substitution of their values into formula for angle  AOC            (A1)

eg       cos θ = 2 + 4 k ( 2 ) 2 + ( 4 ) 2 + ( 4 ) 2 | OC |

correct substitution of their values into formula for angle BOC            (A1)

eg       cos θ = 8 k 6 ( 6 ) 2 + ( 8 ) 2 + 0 2 | OC |

recognizing that  cos A O ^ C = cos B O ^ C   (seen anywhere)           (M1)

eg       2 + 4 k | OC | ( 2 ) 2 + ( 4 ) 2 + ( 4 ) 2 = 8 k 6 | OC | 6 2 + ( 8 ) 2 + 0 2 ,   2 + 4 k 6 1 + k 2 = 8 k 6 10 1 + k 2

correct working (without radicals)           (A2)

eg       10 ( 2 + 4 k ) = 6 ( 8 k 6 ) ,   11 k 2 79 k + 14 = 0

correct working clearly leading to the required answer           A1

eg      20+36=48k-40k,   56 = 8 k ,   k = 7   and   k = 2 11 ,   ( k 7 ) ( 11 k 2 ) = 0

k = 7            AG   N0

[8 marks]

b.

finding magnitude of  OC (seen anywhere)           A1

eg       ( 1 ) 2 + 7 2 + 0 2 ,   50

valid attempt to find  cos θ            (M1)

eg       cos θ = 2 + 28 6 ( 1 ) 2 + 7 2 + 0 2 ,   cos θ = 56 6 10 ( 1 ) 2 + 7 2 + 0 2 ,   ( 26 ) 2 = 6 2 + ( 50 ) 2 2 ( 6 ) 50 cos θ

finding cos θ            A1

eg       cos θ = 5 50 ( = 1 2 )

valid approach to find sin θ (seen anywhere)           (M1)

eg       θ = π 4 ,   sin θ = cos θ ,   sin θ = 1 25 50 ,   sin θ = 1 co s 2 θ ,   sin θ = 2 2

correct substitution of their values into  1 2 a b sin C            (A1)

eg      12×6×50×1-2550,   1 2 × 6 × 50 × 5 50

area is 15            A1   N3

[6 marks]

c.

Examiners report

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

Syllabus sections

Topic 3— Geometry and trigonometry » SL 3.2—2d and 3d trig, sine rule, cosine rule, area
Show 133 related questions
Topic 2—Functions » SL 2.10—Solving equations graphically and analytically
Topic 3— Geometry and trigonometry » AHL 3.12—Vector definitions
Topic 3— Geometry and trigonometry » AHL 3.13—Scalar (dot) product
Topic 3— Geometry and trigonometry » AHL 3.16—Vector product
Topic 2—Functions
Topic 3— Geometry and trigonometry

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