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Q.

Let r¯×a¯=b¯×a¯ and r¯.c¯=0 where a¯.c¯0 then a¯.c¯r¯×b¯+b¯.c¯a¯×b¯=

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a

0¯

b

c¯

c

a¯

d

b¯

answer is A.

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Detailed Solution

r×a=b×a      r·c=0 c×(r×a)=c×(b×a) (c·a) r-0=(c·a) b-(c·b) a r=b-c·bc·a a r×b=0-c·bc·a (a×b) (c·a) (r×a)+(c·b) (a×b)=0

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