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if a b c be three non-coplanar vectors and r be any arbitrary vector then (a×b)×(r×c)+(b×c)×(r×a)+(c×a)×(r×b) is equal to (1)0 (2)[a b c]r (3)2[a b c]r (4)3[a b c]r

if a b c be three non-coplanar vectors and r be any arbitrary vector then (a×b)×(r×c)+(b×c)×(r×a)+(c×a)×(r×b) is equal to (1)0 (2)[a b c]r (3)2[a b c]r (4)3[a b c]r

Grade:12

2 Answers

Arun
25763 Points
3 years ago
Dear Akshat
 
we have
 
(a x b) x (r x c)
= ((a x b ) . c) r – ((a x b) . r) c
 = [a b c] r – [a b r] c
similarly
(b x c) x (r x a) = [b c a] r – [b c r] c
 
and 
(c x a) x (r x b) = [c a b] r – [c a r] b
 
when we add all three
 
we get
3 [a b c] r  –  [a b c] r = 2 [a b c] r
hence option 3 is correct
 
reagrds
Arun (askIITians forum expert)
Kushagra Madhukar
askIITians Faculty 629 Points
one year ago
Dear student,
Please find the solution to your problem.
 
we have,
(a x b) x (r x c)
= ((a x b ) . c) r – ((a x b) . r) c
 = [a b c] r – [a b r] c
similarly
(b x c) x (r x a) = [b c a] r – [b c r] c
and 
(c x a) x (r x b) = [c a b] r – [c a r] b
when we add all three
we get
3 [a b c] r  –  [a b c] r = 2 [a b c] r
hence option 3 is correct
 
Thanks and regards,
Kushagra

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