If G is the centroid of a ΔABC, then →GA+→GB+→GC is equal to
A
→0
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B
3→GA
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C
3→GB
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D
3→GC
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Solution
The correct option is A→0 We have →GB+→GC=(1+1)→GD=2→GD, where D is the midpoint of BC. Therefore, →GA+→GB+→GC=→GA+2→GD =→GA−→GA=0 (∵G divides AC in the ration 2:1,∴2→GD=−→GA)