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Date May 2017 Marks available 3 Reference code 17M.2.AHL.TZ2.H_9
Level Additional Higher Level Paper Paper 2 Time zone Time zone 2
Command term Find Question number H_9 Adapted from N/A

Question

The points A, B and C have the following position vectors with respect to an origin O.

O A = 2 i + j – 2k

O B = 2 ij + 2k

O C =  i + 3j + 3k

The plane Π 2 contains the points O, A and B and the plane Π 3 contains the points O, A and C.

Find the vector equation of the line (BC).

[3]
a.

Determine whether or not the lines (OA) and (BC) intersect.

[6]
b.

Find the Cartesian equation of the plane Π 1 , which passes through C and is perpendicular to O A .

[3]
c.

Show that the line (BC) lies in the plane Π 1 .

[2]
d.

Verify that 2j + k is perpendicular to the plane Π 2 .

[3]
e.

Find a vector perpendicular to the plane Π 3 .

[1]
f.

Find the acute angle between the planes Π 2 and Π 3 .

[4]
g.

Markscheme

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

B C = (i + 3j + 3k) (2i  j + 2k) = i + 4j + k    (A1)

r = (2i j + 2k) + λ ( i + 4j + k)

(or r = (i + 3j + 3k) + λ ( i + 4j + k)     (M1)A1

 

Note:     Do not award A1 unless r = or equivalent correct notation seen.

 

[3 marks]

a.

attempt to write in parametric form using two different parameters AND equate     M1

2 μ = 2 λ

μ = 1 + 4 λ

2 μ = 2 + λ      A1

attempt to solve first pair of simultaneous equations for two parameters     M1

solving first two equations gives λ = 4 9 ,   μ = 7 9      (A1)

substitution of these two values in third equation     (M1)

since the values do not fit, the lines do not intersect     R1

 

Note:     Candidates may note that adding the first and third equations immediately leads to a contradiction and hence they can immediately deduce that the lines do not intersect.

 

[6 marks]

b.

METHOD 1

plane is of the form r (2i + j 2k) = d     (A1)

d = (i + 3j + 3k) (2i + j 2k) = 1     (M1)

hence Cartesian form of plane is 2 x + y 2 z = 1      A1

METHOD 2

plane is of the form 2 x + y 2 z = d      (A1)

substituting ( 1 ,   3 ,   3 ) (to find gives 2 + 3 6 = 1 )     (M1)

hence Cartesian form of plane is 2 x + y 2 z = 1      A1

[3 marks]

c.

METHOD 1

attempt scalar product of direction vector BC with normal to plane     M1

( i + 4j + k) (2i + j 2k) = 2 + 4 2

= 0      A1

hence BC lies in Π 1      AG

METHOD 2

substitute eqn of line into plane     M1

line  r = ( 2 1 2 ) + λ ( 1 4 1 ) .  Plane  π 1 : 2 x + y 2 z = 1

2 ( 2 λ ) + ( 1 + 4 λ ) 2 ( 2 + λ )

= 1      A1

hence BC lies in Π 1      AG

 

Note:     Candidates may also just substitute 2 i j + 2 k into the plane since they are told C lies on π 1 .

 

Note:     Do not award A1FT.

 

[2 marks]

d.

METHOD 1

applying scalar product to O A and O B      M1

(2j + k) (2i + j 2k) = 0     A1

(2j + k) (2i j + 2k) =0     A1

METHOD 2

attempt to find cross product of O A and O B      M1

plane Π 2 has normal OA × OB = 8j 4k     A1

since 8j 4k = 4(2j + k), 2j + k is perpendicular to the plane Π 2      R1

[3 marks]

e.

plane Π 3 has normal OA × OC = 9i 8j + 5k     A1

[1 mark]

f.

attempt to use dot product of normal vectors     (M1)

cos θ = ( 2 j + k ) ( 9 i 8 j + 5 k ) | 2 j + k | | 9 i 8 j + 5 k |      (M1)

= 11 5 170 ( = 0.377 )      (A1)

 

Note:     Accept 11 5 170 .   acute angle between planes = 67.8 ( = 1.18 )      A1

 

[4 marks]

g.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.
[N/A]
d.
[N/A]
e.
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f.
[N/A]
g.

Syllabus sections

Topic 3— Geometry and trigonometry » AHL 3.13—Scalar (dot) product
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Topic 3— Geometry and trigonometry

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