If 4^i+7^j+8^k,2^i+3^j+4^k and 2^i+5^j+7^k are the position vectors of the vertices A, B and C respectively of triangle ABC. The position vector of the point where the bisector of angle A meets BC is
A
23(−6^i−8^j−6^k)
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B
23(6^i+8^j+6^k)
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C
13(6^i+13^j+18^k)
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D
13(5^i+12^k)
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Solution
The correct option is C13(6^i+13^j+18^k) Suppose the bisector of angle A meets BC at D. Then, AD divides BC in the ratio AB : AC. So, PV of D =|−−→AB|(2^i+5^j+7^k)+|−−→AC|(2^i+3^j+4^k)|−−→AB|+−−→AC But, −−→AB=−2^i−4^j−4^k and −−→AC=−2^i−2^j−^k⇒|−−→AB|=6 and |−−→AC|=3 ∴ pv of D=6(2^i+5^j+7^k)+3(2^i+3^j+4^k)6+3=13(6^i+13^j+18^k)