If →a+→b+→c=0,|→a|=3,∣∣→b∣∣=5,|→c|=7, then the angle between →a and →b is
A
π/6
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B
2π/3
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C
5π/3
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D
π/3
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Solution
The correct option is Dπ/3 Given →a+→b+→c=0 →c=−(→a+→b) ⇒|→c|2=→c→c=(→a+→b).(→a+→b) ⇒|→c|2=|→a|2+∣∣→b∣∣2+2→a.→b ⇒|→c|2=|→a|2+∣∣→b∣∣2+2|→a|∣∣→b∣∣cosθ ⇒49=9+25+2×3×5cosθ ⇒15=30cosθ⇒cosθ=12⇒θ=π3