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Date May 2021 Marks available 3 Reference code 21M.1.AHL.TZ2.8
Level Additional Higher Level Paper Paper 1 (without calculator) Time zone Time zone 2
Command term Show that Question number 8 Adapted from N/A

Question

The lines l1 and l2 have the following vector equations where λ, μ.

l1:r1=(32-1)+λ(2-22)

l2:r2=(204)+μ(1-11)

Show that l1 and l2 do not intersect.

[3]
a.

Find the minimum distance between l1 and l2.

[5]
b.

Markscheme

METHOD 1

setting at least two components of l1 and l2 equal           M1

3+2λ=2+μ      (1)

2-2λ=-μ        (2)

-1+2λ=4+μ  (3)

attempt to solve two of the equations eg. adding (1) and (2)           M1

gives a contradiction (no solution), eg 5=2           R1

so l1 and l2 do not intersect                   AG

 

Note: For an error within the equations award M0M1R0.
Note: The contradiction must be correct to award the R1.

 

METHOD 2

l1 and l2 are parallel, so l1 and l2 are either identical or distinct.           R1

Attempt to subtract two position vectors from each line,

e.g. (32-1)-(204)(=(12-5))           M1

(32-1)k(1-11)               A1

 

[3 marks]

a.

METHOD 1

l1 and l2 are parallel (as (2-22) is a multiple of (1-11))

let A be (3,2,-1) on l1 and let B be (2,0,4) on l2

Attempt to find vector AB(=(-1-25))            (M1)

Distance required is |v×AB||v|              M1

=13|(1-11)×(-1-25)|            (A1)

=13|(363)|                    A1

minimum distance is 18(=32)                    A1

 

METHOD 2

l1 and l2 are parallel (as (2-22) is a multiple of (1-11))

let A be a fixed point on l1 eg (3,2,-1) and let B be a general point on l2 (2+μ,-μ,4+μ)

attempt to find vector AB            (M1)

AB=(-1-25)+μ(1-11) (μ)                   A1

|AB|=(-1+μ)2+(-2-μ)2+(5+μ)2 (=3μ2+12μ+30)              M1


EITHER

null                   A1


OR

|AB|=3(μ+2)2+18 to obtain μ=-2                   A1


THEN

minimum distance is 18(=32)                   A1

 

METHOD 3

let A be (3,2,-1) on l1 and let B be (2+μ,-μ,4+μ) on l2              (M1)

(or let A be (2,0,4) on l2 and let B be (3+2λ,2-2λ,-1+2λ) on l1)

AB=(-1-25)+μ(1-11) (μ)  (or AB=(2λ+1-2λ+22λ-5))                 A1

(μ-1-μ-2μ+5)·(1-11)=0  (or (2λ+1-2λ+22λ-5)·(1-11)=0)              M1

μ=-2  or  λ=1                   A1

minimum distance is 18(=32)                   A1

 

[5 marks]

b.

Examiners report

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

Syllabus sections

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