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Question

Let a =a1i^+a2j^+a3k^, b =b1i^+b2j^+b3k^ and c =c1i^+c2j^+c3k^ be three non-zero vectors such that c is a unit vector perpendicular to both a and b . If the angle between a and b is π6, then a1a2a3b1b2b3c1c2c32is equal to

(a) 0
(b) 1
(c) 14 a 2 b 2

(d) 34 a 2 b 2

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Solution

(c) 14 a 2 b 2

We have a1a2a3b1b2b3c1c2c32=a×b.c2 By definition of scalar triple product= a×bccos0°2 a×b is parallel to vector c as c is perpendicular to both a and b= absinπ62 c=1 and angle between a and b is π6= a2b2122 = 14 a2b2

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