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

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo

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