Date | May 2017 | Marks available | 4 | Reference code | 17M.2.hl.TZ2.7 |
Level | HL only | Paper | 2 | Time zone | TZ2 |
Command term | Prove that | Question number | 7 | Adapted from | N/A |
Question
Given that a × b = b × c ≠ 0 prove that a + c = sb where s is a scalar.
Markscheme
METHOD 1
a × b = b × c
(a × b) − (b × c) = 0
(a × b) + (c × b) = 0 M1A1
(a + c) × b = 0 A1
(a + c) is parallel to b ⇒ a + c = sb R1AG
Note: Condone absence of arrows, underlining, or other otherwise “correct” vector notation throughout this question.
Note: Allow “is in the same direction to”, for the final R mark.
METHOD 2
a × b = b × c ⇒(a2b3−a3b2a3b1−a1b3a1b2−a2b1)=(b2c3−b3c2b3c1−b1c3b1c2−b2c1) M1A1
a2b3−a3b2=b2c3−b3c2⇒b3(a2+c2)=b2(a3+c3)
a3b1−a1b3=b3c1−b1c3⇒b1(a3+c3)=b3(a1+c1)
a1b2−a2b1=b1c2−b2c1⇒b2(a1+c1)=b1(a2+c2)
(a1+c1)b1=(a2+c2)b2=(a3+c3)b3=s A1
⇒a1+c1=sb1
⇒a2+c2=sb2
⇒a3+c3=sb3
⇒(a1a2a3)+(c1c2c3)=s(b1b2b3) A1
⇒ a + c = sb AG
[4 marks]