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Date May 2019 Marks available 3 Reference code 19M.1.AHL.TZ1.H_11
Level Additional Higher Level Paper Paper 1 (without calculator) Time zone Time zone 1
Command term Verify Question number H_11 Adapted from N/A

Question

Two distinct lines, l 1 and l 2 , intersect at a point P . In addition to P , four distinct points are marked out on l 1 and three distinct points on l 2 . A mathematician decides to join some of these eight points to form polygons.

The line l 1 has vector equation r1 = ( 1 0 1 ) + λ ( 1 2 1 ) λ R  and the line l 2 has vector equation r2  = ( 1 0 2 ) + μ ( 5 6 2 ) μ R .

The point P has coordinates (4, 6, 4).

The point A has coordinates (3, 4, 3) and lies on l 1 .

The point B has coordinates (−1, 0, 2) and lies on l 2 .

Find how many sets of four points can be selected which can form the vertices of a quadrilateral.

[2]
a.i.

Find how many sets of three points can be selected which can form the vertices of a triangle.

[4]
a.ii.

Verify that P is the point of intersection of the two lines.

[3]
b.

Write down the value of λ corresponding to the point A .

[1]
c.

Write down PA and PB .

[2]
d.

Let C be the point on l 1 with coordinates (1, 0, 1) and D be the point on l 2 with parameter μ = 2 .

Find the area of the quadrilateral CDBA .

[8]
e.

Markscheme

appreciation that two points distinct from P need to be chosen from each line   M1

4 C 2 × 3 C 2

=18    A1

[2 marks]

a.i.

EITHER

consider cases for triangles including P or triangles not including P       M1

3 × 4 + 4 × 3 C 2 + 3 × 4 C 2      (A1)(A1)

Note: Award A1 for 1st term, A1 for 2nd & 3rd term.

OR

consider total number of ways to select 3 points and subtract those with 3 points on the same line      M1

8 C 3 5 C 3 4 C 3      (A1)(A1)

Note: Award A1 for 1st term, A1 for 2nd & 3rd term.

56−10−4

THEN

= 42    A1

[4 marks]

a.ii.

METHOD 1

substitution of (4, 6, 4) into both equations       (M1)

λ = 3 and  μ = 1        A1A1

(4, 6, 4)       AG

METHOD 2

attempting to solve two of the three parametric equations      M1

λ = 3 and  μ = 1        A1

check both of the above give (4, 6, 4)       M1AG

Note: If they have shown the curve intersects for all three coordinates they only need to check (4,6,4) with one of " λ " or " μ ".

[3 marks]

b.

λ = 2       A1

[1 mark]

c.

PA = ( 1 2 1 ) ,   PB = ( 5 6 2 )     A1A1

Note: Award A1A0 if both are given as coordinates.

[2 marks]

d.

METHOD 1

area triangle  ABP = 1 2 | PB × PA |     M1

( = 1 2 | ( 5 6 2 ) × ( 1 2 1 ) | ) = 1 2 | ( 2 3 4 ) |     A1

= 29 2     A1

EITHER

PC = 3 PA PD = 3 PB        (M1)

area triangle  PCD = 9 ×  area triangle ABP        (M1)A1

= 9 29 2     A1

OR

D  has coordinates (−11, −12, −2)    A1

area triangle  PCD = 1 2 | PD × PC | = 1 2 | ( 15 18 6 ) × ( 3 6 3 ) |     M1A1

Note: A1 is for the correct vectors in the correct formula.

= 9 29 2     A1

THEN

area of  CDBA = 9 29 2 29 2

= 4 29     A1

 

METHOD 2

D  has coordinates (−11, −12, −2)    A1

area  = 1 2 | CB × CA | + 1 2 | BC × BD |       M1

Note: Award M1 for use of correct formula on appropriate non-overlapping triangles.

Note: Different triangles or vectors could be used.

CB = ( 2 0 1 ) CA = ( 2 4 2 )     A1

CB × CA = ( 4 6 8 )     A1

BC = ( 2 0 1 ) BD = ( 10 12 4 )     A1

BC × BD = ( 12 18 24 )     A1

Note: Other vectors which might be used are  DA = ( 14 16 5 ) BA = ( 4 4 1 ) DC = ( 12 12 3 ) .

Note: Previous A1A1A1A1 are all dependent on the first M1.

valid attempt to find a value of  1 2 | a × b |       M1

Note: M1 independent of triangle chosen.

area  = 1 2 × 2 × 29 + 1 2 × 6 × 29

= 4 29     A1

Note: accept  1 2 116 + 1 2 1044 or equivalent.

 

[8 marks]

e.

Examiners report

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

Syllabus sections

Topic 1—Number and algebra » SL 1.9—Binomial theorem where n is an integer
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Topic 3— Geometry and trigonometry » AHL 3.16—Vector product
Topic 1—Number and algebra
Topic 3— Geometry and trigonometry

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