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Date May 2017 Marks available 2 Reference code 17M.2.hl.TZ2.9
Level HL only Paper 2 Time zone TZ2
Command term Show that Question number 9 Adapted from N/A

Question

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

OA=2i + j – 2k

OB=2ij + 2k

OC= i + 3j + 3k

The plane Π2 contains the points O, A and B and the plane Π3 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 Π1, which passes through C and is perpendicular to OA.

[3]
c.

Show that the line (BC) lies in the plane Π1.

[2]
d.

Verify that 2j + k is perpendicular to the plane Π2.

[3]
e.

Find a vector perpendicular to the plane Π3.

[1]
f.

Find the acute angle between the planes Π2 and Π3.

[4]
g.

Markscheme

BC = (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

2μ=2λ

μ=1+4λ

2μ=2+λ     A1

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

solving first two equations gives λ=49, μ=79     (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 2x+y2z=1     A1

METHOD 2

plane is of the form 2x+y2z=d     (A1)

substituting (1, 3, 3) (to find gives 2+36=1)     (M1)

hence Cartesian form of plane is 2x+y2z=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+42

=0     A1

hence BC lies in Π1     AG

METHOD 2

substitute eqn of line into plane     M1

line r=(212)+λ(141). Plane π1:2x+y2z=1

2(2λ)+(1+4λ)2(2+λ)

=1     A1

hence BC lies in Π1     AG

 

Note:     Candidates may also just substitute 2ij+2k into the plane since they are told C lies on π1.

 

Note:     Do not award A1FT.

 

[2 marks]

d.

METHOD 1

applying scalar product to OA and OB     M1

(2j + k) (2i + j 2k) = 0     A1

(2j + k) (2i j + 2k) =0     A1

METHOD 2

attempt to find cross product of OA and OB     M1

plane Π2 has normal OA×OB = 8j 4k     A1

since 8j 4k = 4(2j + k), 2j + k is perpendicular to the plane Π2     R1

[3 marks]

e.

plane Π3 has normal OA×OC = 9i 8j + 5k     A1

[1 mark]

f.

attempt to use dot product of normal vectors     (M1)

cosθ=(2j+k)(9i8j+5k)|2j+k||9i8j+5k|     (M1)

=115170(=0.377)     (A1)

 

Note:     Accept 115170.   acute angle between planes =67.8(=1.18)     A1

 

[4 marks]

g.

Examiners report

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

Syllabus sections

Topic 4 - Core: Vectors » 4.6
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