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∈ℝ.
l1:r1=(32-1)+λ(2-22)
l2:r2=(204)+μ(1-11)
Show that l1 and l2 do not intersect.
Find the minimum distance between l1 and l2.
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]
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
=1√3|(1-11)×(-1-25)| (A1)
=1√3|(363)| A1
minimum distance is √18(=3√2) 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∈ℝ) 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(=3√2) 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∈ℝ) (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(=3√2) A1
[5 marks]