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Date May 2018 Marks available 6 Reference code 18M.1.SL.TZ1.S_9
Level Standard Level Paper Paper 1 Time zone Time zone 1
Command term Find Question number S_9 Adapted from N/A

Question

Point A has coordinates (−4, −12, 1) and point B has coordinates (2, −4, −4).

The line L passes through A and B.

Show that  AB = ( 6 8 5 )

[1]
a.

Find a vector equation for L.

[2]
b.i.

Point C (k , 12 , −k) is on L. Show that k = 14.

[4]
b.ii.

Find OB AB .

[2]
c.i.

Write down the value of angle OBA.

[1]
c.ii.

Point D is also on L and has coordinates (8, 4, −9).

Find the area of triangle OCD.

[6]
d.

Markscheme

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

correct approach       A1

eg    AO + OB , B A ( 2 4 4 ) ( 4 12 1 )

AB = ( 6 8 5 )      AG  N0

[1 mark]

a.

any correct equation in the form r = a + tb (any parameter for t)      A2 N2

where a is  ( 2 4 4 ) or  ( 4 12 1 ) and b is a scalar multiple of  ( 6 8 5 )

eg  r  = ( 4 12 1 ) + t ( 6 8 5 ) , ( x , y , z ) = ( 2 , 4 , 4 ) + t ( 6 , 8 , 5 ) ,   = ( 4 + 6 t 12 + 8 t 1 5 t )

Note: Award A1 for the form a + tb, A1 for the form L = a + tb, A0 for the form rb + ta.

[2 marks]

b.i.

METHOD 1 (solving for t)

valid approach       (M1)

eg    ( k 12 k ) = ( 2 4 4 ) + t ( 6 8 5 ) , ( k 12 k ) = ( 4 12 1 ) + t ( 6 8 5 )

one correct equation       A1

eg −4 + 8t = 12, −12 + 8t = 12

correct value for t       (A1)

eg   t = 2 or 3

correct substitution      A1

eg  2 + 6(2), −4 + 6(3), −[1 + 3(−5)]

k = 14      AG N0

 

METHOD 2 (solving simultaneously)

valid approach      (M1)

eg   ( k 12 k ) = ( 2 4 4 ) + t ( 6 8 5 ) , ( k 12 k ) = ( 4 12 1 ) + t ( 6 8 5 )

two correct equations in        A1

eg   k = −4 + 6t, −k = 1 −5t

EITHER (eliminating k)

correct value for t       (A1)

eg    t = 2 or 3

correct substitution      A1

eg  2 + 6(2), −4 + 6(3)

OR (eliminating t)

correct equation(s)      (A1)

eg   5k + 20 = 30t and −6k − 6 = 30t, −k = 1 − 5 ( k + 4 6 )

correct working clearly leading to k = 14      A1

eg  k + 14 = 0, −6k = 6 −5k − 20, 5k = −20 + 6(1 + k)

THEN

k = 14       AG N0

[4 marks]

 

b.ii.

correct substitution into scalar product       A1

eg   (2)(6) − (4)(8) − (4)(−5), 12 − 32 + 20

OB AB = 0      A1 N0

[2 marks]

 

c.i.

O B A = π 2 , 90 ( accept 3 π 2 , 270 )       A1 N1

[1 marks]

c.ii.

METHOD 1 ( 1 2  × height × CD)

recognizing that OB is altitude of triangle with base CD (seen anywhere)      M1

eg    1 2 × | OB | × | CD | , OB CD ,  sketch showing right angle at B

CD = ( 6 8 5 ) or  DC = ( 6 8 5 )   (seen anywhere)       (A1)

correct magnitudes (seen anywhere)      (A1)(A1)

| OB | = ( 2 ) 2 + ( 4 ) 2 + ( 4 ) 2 = ( 36 )

| CD | = ( 6 ) 2 + ( 8 ) 2 + ( 5 ) 2 = ( 125 )

correct substitution into  1 2 b h       A1

eg      1 2 × 6 × 125  

area  = 3 125 , 15 5       A1 N3

 

METHOD 2 (subtracting triangles)

recognizing that OB is altitude of either ΔOBD or ΔOBC(seen anywhere)       M1

eg   1 2 × | OB | × | BD | , OB BC ,  sketch of triangle showing right angle at B

one correct vector  BD or  DB or  BC or  CB  (seen anywhere)       (A1)

eg    BD = ( 6 8 5 ) CB = ( 12 16 10 )

| OB | = ( 2 ) 2 + ( 4 ) 2 + ( 4 ) 2 = ( 36 )  (seen anywhere)       (A1)

one correct magnitude of a base (seen anywhere)        (A1)

| BD | = ( 6 ) 2 + ( 8 ) 2 + ( 5 ) 2 = ( 125 ) , | BC | = 144 + 256 + 100 = ( 500 )

correct working       A1

eg   1 2 × 6 × 500 1 2 × 6 × 5 5 , 1 2 × 6 × 500 × sin 90 1 2 × 6 × 5 5 × sin 90

area  = 3 125 , 15 5       A1 N3

 

METHOD 3 (using 1 2 ab sin C with ΔOCD)

two correct side lengths (seen anywhere)      (A1)(A1)

| OD | = ( 8 ) 2 + ( 4 ) 2 + ( 9 ) 2 = ( 161 ) , | CD | = ( 6 ) 2 + ( 8 ) 2 + ( 5 ) 2 = ( 125 ) ,   | OC | = ( 14 ) 2 + ( 12 ) 2 + ( 14 ) 2 = ( 536 )

attempt to find cosine ratio (seen anywhere)       M1
eg   536 286 2 161 125 , OD DC | O D | | D C |

correct working for sine ratio       A1

eg    ( 125 ) 2 161 × 125 + si n 2 D = 1

correct substitution into  1 2 a b sin C        A1

eg   0.5 × 161 × 125 × 6 161

area  = 3 125 , 15 5       A1 N3

[6 marks]

d.

Examiners report

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