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Date November 2018 Marks available 5 Reference code 18N.2.SL.TZ0.S_8
Level Standard Level Paper Paper 2 Time zone Time zone 0
Command term Find Question number S_8 Adapted from N/A

Question

Consider the points A(−3, 4, 2) and B(8, −1, 5).

A line L has vector equation r = ( 2 0 5 ) + t ( 1 2 2 ) . The point C (5, y , 1) lies on line L.

Find | AB | .

[2]
a.ii.

Find the value of y .

[3]
b.i.

Show that AC = ( 8 10 1 ) .

[2]
b.ii.

Find the angle between AB and AC .

[5]
c.

Find the area of triangle ABC.

[2]
d.

Markscheme

Note: There may be slight differences in answers, depending on which combination of unrounded values and previous correct 3 sf values the candidates carry through in subsequent parts. Accept answers that are consistent with their working.

 

correct substitution into formula      (A1)

eg    11 2 + ( 5 ) 2 + 3 2

12.4498 

| AB | = 155   (exact), 12.4     A1 N2

 

[2 marks]

a.ii.

Note: There may be slight differences in answers, depending on which combination of unrounded values and previous correct 3 sf values the candidates carry through in subsequent parts. Accept answers that are consistent with their working.

 

valid approach to find t       (M1)

eg      ( 5 y 1 ) = ( 2 0 5 ) + t ( 1 2 2 ) ,   5 = 2 + t ,    1 = 5 + 2 t

t = 3      (seen anywhere)      (A1)

attempt to substitute their parameter into the vector equation      (M1)

eg      ( 5 y 1 ) = ( 2 0 5 ) + 3 ( 1 2 2 ) ,   3 ( 2 )

y = 6      A1 N2

 

[3 marks]

b.i.

Note: There may be slight differences in answers, depending on which combination of unrounded values and previous correct 3 sf values the candidates carry through in subsequent parts. Accept answers that are consistent with their working.

 

correct approach      A1

eg      ( 5 6 1 ) ( 3 4 2 ) ,  AO + OC,   c a

AC = ( 8 10 1 )      AG N0

Note: Do not award A1 in part (ii) unless answer in part (i) is correct and does not result from working backwards.

 

[2 marks]

b.ii.

Note: There may be slight differences in answers, depending on which combination of unrounded values and previous correct 3 sf values the candidates carry through in subsequent parts. Accept answers that are consistent with their working.

 

finding scalar product and magnitude      (A1)(A1)

scalar product = 11 × 8 + −5 × −10 + 3 × −1  (=135)

| AC | = 8 2 + ( 10 ) 2 + ( 1 ) 2 ( = 165 , 12.8452 )

evidence of substitution into formula      (M1)

eg  cos θ = 11 × 8 + 5 × 10 + 3 × 1 | AB | × 8 2 + ( 10 ) 2 + ( 1 ) 2 , cos θ = AB AC 155 × 8 2 + ( 10 ) 2 + ( 1 ) 2

correct substitution      (A1)

eg     cos θ = 11 × 8 + 5 × 10 + 3 × 1 155 × 8 2 + ( 10 ) 2 + ( 1 ) 2 ,    cos θ = 135 159.921 ,

cos θ = 0.844162

0.565795,  32.4177°

A ^ = 0.566,  32.4°     A1 N3

 

[5 marks]

c.

Note: There may be slight differences in answers, depending on which combination of unrounded values and previous correct 3 sf values the candidates carry through in subsequent parts. Accept answers that are consistent with their working.

 

correct substitution into area formula      (A1)

eg    1 2 × 155 × 165 × sin ( 0.566 ) ,   1 2 × 155 × 165 × sin ( 32.4 )

42.8660

area = 42.9      A1 N2

 

[2 marks]

d.

Examiners report

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

Syllabus sections

Topic 3—Geometry and trigonometry » AHL 3.11—Vector equation of a line in 2d and 3d
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Topic 3—Geometry and trigonometry

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