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Question

If a, b, are three vectors such that each is inclined at an angle π/3 with the other two and |a|=1,|b|=2,|c|=3, then the scalar product of the vectors 2a+3b5c and 4a6b+10c is equal to

A
-334
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B
188
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C
-522
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D
-514
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Solution

The correct option is D -334
(2a+3b5c)(4a6b+10c)=8a50cab+20ac+12ba+30bc20ca+30cb=8a218b250c2+60bc=8(1)250(3)2+6022cos(π3)18(2)2=334

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