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Date November 2018 Marks available 3 Reference code 18N.1.AHL.TZ0.H_9
Level Additional Higher Level Paper Paper 1 (without calculator) Time zone Time zone 0
Command term Find Question number H_9 Adapted from N/A

Question

Consider a triangle OAB such that O has coordinates (0, 0, 0), A has coordinates (0, 1, 2) and B has coordinates (2 b , 0, b − 1) where b < 0.

Let M be the midpoint of the line segment [OB].

Find, in terms of b , a Cartesian equation of the plane Π containing this triangle.

[5]
a.

Find, in terms of b , the equation of the line L which passes through M and is perpendicular to the plane П.

[3]
b.

Show that L does not intersect the y -axis for any negative value of b .

 

[7]
c.

Markscheme

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

METHOD 1

n = ( 0 1 2 ) × ( 2 b 0 b 1 )       (M1)

= ( b 1 4 b 2 b )       (M1)A1

(0, 0, 0) on Π so  ( b 1 ) x + 4 b y 2 b z = 0       (M1)A1

 

METHOD 2

using equation of the form  p x + q y + r z = 0       (M1)

(0, 1, 2) on Π ⇒  q + 2 r = 0

(2 b , 0, b − 1) on Π ⇒  2 b p + r ( b 1 ) = 0       (M1)A1

Note: Award (M1)A1 for both equations seen.

solve for  p q and  r       (M1)

( b 1 ) x + 4 b y 2 b z = 0       A1

 

[5 marks]

a.

M has coordinates  ( b , 0 , b 1 2 )       (A1)

r ( b 0 b 1 2 ) + λ ( b 1 4 b 2 b )       M1A1

Note: Award M1A0 if r = (or equivalent) is not seen.

Note: Allow equivalent forms such as  x b b 1 = y 4 b = 2 z b + 1 4 b .

 

[3 marks]

b.

METHOD 1

x = z = 0       (M1)

Note: Award M1 for either x = 0 or z = 0 or both.

b + λ ( b 1 ) = 0 and  b 1 2 2 λ b = 0       A1

attempt to eliminate λ        M1

b b 1 = b 1 4 b       (A1)

4 b 2 = ( b 1 ) 2       A1

EITHER

consideration of the signs of LHS and RHS       (M1)

the LHS is negative and the RHS must be positive (or equivalent statement)       R1

OR

4 b 2 = b 2 2 b + 1

5 b 2 2 b + 1 = 0

Δ = ( 2 ) 2 4 × 5 × 1 = 16 ( < 0 )      M1

no real solutions       R1

THEN

so no point of intersection       AG

 

METHOD 2

x = z = 0       (M1)

Note: Award M1 for either  x = 0 or  z = 0 or both.

b + λ ( b 1 ) = 0 and  b 1 2 2 λ b = 0       A1

attempt to eliminate b       M1

λ 1 + λ = 1 1 4 λ       (A1)

4 λ 2 = 1 ( λ 2 = 1 4 )       A1

consideration of the signs of LHS and RHS       (M1)

there are no real solutions (or equivalent statement)       R1

so no point of intersection       AG

 

[7 marks]

c.

Examiners report

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

Syllabus sections

Topic 3— Geometry and trigonometry » AHL 3.14—Vector equation of line
Show 98 related questions
Topic 3— Geometry and trigonometry » AHL 3.17—Vector equations of a plane
Topic 3— Geometry and trigonometry » AHL 3.18—Intersections of lines & planes
Topic 3— Geometry and trigonometry

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