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In ΔABC,−−→AB=^i+3^j−2^k, −−→AC=3^i−^j−2^k. If the bisector of ∠BAC meets BC at D and G is the centroid of ΔABC, then |−−→GD|= |
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Answer» In ΔABC,−−→AB=^i+3^j−2^k, −−→AC=3^i−^j−2^k. |
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