Question 24 (OR 1st Question) For three non-zero vectors a , b and c , prove that [ (a - b) (b - c) (c - a)] = 0
[ (a - b) (b - c) (c - a)]
= (a - b) (b - c) x (c - a)
= (a - b) [ (b x c) - (b x a) - (c x c) + (c x a)
(c x c) = [c] [c] sin 0° n
= 0
= (a - b). [(b x c) - (b x a) - 0 + (c x a)]
= (a - b). [ (b x c) - (b x a) - 0 + (c x a)]
= a [ (b x c) - (b x a) + (c x a) -b [(b x c) - (b x a) + (c x a)
= a . (b x c) -a . (b x a) + a (c x a) - b (b x c) b (b x a) - b (c x a)
= [a, b, c] - [a, b, a] + [a, c, a] - [b, b, c] + [b, b, a] - [ b, c, a]
now
[ b, b, c] = [c, b, b]
= c (b x b)
as (b x b) = 0
= c .0
= 0
similarly,
[a, b, a] = 0
[b, b, a] = 0
[a,c, a] = 0
= [a, b, c] - 0 + 0 - 0 + 0 - [ b, c, a]
= [a, b, c] - [ b, c, a]
as [a, b, c] = [c, a, b ] = - [b, c, a]
= [a, b, c] - [a, b, c]
= 0

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.