Date | May 2018 | Marks available | 6 | Reference code | 18M.1.SL.TZ1.S_6 |
Level | Standard Level | Paper | Paper 1 | Time zone | Time zone 1 |
Command term | Find | Question number | S_6 | Adapted from | N/A |
Question
Six equilateral triangles, each with side length 3 cm, are arranged to form a hexagon.
This is shown in the following diagram.
The vectors p , q and r are shown on the diagram.
Find p•(p + q + r).
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1 (using |p| |2q| cosθ)
finding p + q + r (A1)
eg 2q,
| p + q + r | = 2 × 3 (= 6) (seen anywhere) A1
correct angle between p and q (seen anywhere) (A1)
π3π3 (accept 60°)
substitution of their values (M1)
eg 3 × 6 × cos(π3)(π3)
correct value for cos(π3)(π3) (seen anywhere) (A1)
eg 12,3×6×1212,3×6×12
p•(p + q + r) = 9 A1 N3
METHOD 2 (scalar product using distributive law)
correct expression for scalar distribution (A1)
eg p• p + p•q + p•r
three correct angles between the vector pairs (seen anywhere) (A2)
eg 0° between p and p, π3π3 between p and q, 2π32π3 between p and r
Note: Award A1 for only two correct angles.
substitution of their values (M1)
eg 3.3.cos0 +3.3.cosπ3π3 + 3.3.cos120
one correct value for cos0, cos(π3)(π3) or cos(2π3)(2π3) (seen anywhere) A1
eg 12,3×6×1212,3×6×12
p•(p + q + r) = 9 A1 N3
METHOD 3 (scalar product using relative position vectors)
valid attempt to find one component of p or r (M1)
eg sin 60 = x3x3, cos 60 = x3x3, one correct value 32,3√32,−3√3232,3√32,−3√32
one correct vector (two or three dimensions) (seen anywhere) A1
eg p=(323√32),q=(30),r=(32−3√320)p=⎛⎜ ⎜⎝323√32⎞⎟ ⎟⎠,q=(30),r=⎛⎜ ⎜ ⎜ ⎜ ⎜⎝32−3√320⎞⎟ ⎟ ⎟ ⎟ ⎟⎠
three correct vectors p + q + r = 2q (A1)
p + q + r = (60)(60) or (600)⎛⎝600⎞⎠ (seen anywhere, including scalar product) (A1)
correct working (A1)
eg (32×6)+(3√32×0),9+0+0(32×6)+(3√32×0),9+0+0
p•(p + q + r) = 9 A1 N3
[6 marks]