If |→a|=3,∣∣→b∣∣=1,|→c|=4 and →a+→b+→c=→0, then the value of →a⋅→b+→b⋅→c+→c⋅→a is
A
26
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B
−13
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C
−26
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D
13
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Solution
The correct option is B−13 Now, (→a+→b+→c)2=(→a+→b+→c)⋅(→a+→b+→c) 0=⇒|→a|2+∣∣→b∣∣2+|→c|2+2(→a⋅→b+→b⋅→c+→c⋅→a) 0=⇒(3)2+(1)2+(4)2+2(→a⋅→b+→b⋅→c+→c⋅→a) ⇒(→a⋅→b+→b⋅→c+→c⋅→a)=−262=−13