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 ∠A meets BC
A
12(4^i+8^j+11^k)
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B
13(6^i+11^j+15^k)
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C
13(6^i+13^j+18^k)
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D
14(8^i+14^j+9^k)
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Solution
The correct option is C13(6^i+13^j+18^k) Given: →A=4^i+7^j+8^k,→B=2^i+3^j+4^k and →C=2^i+5^j+7^k
Suppose the bisector of angle A meets −−→BC at D.
Then −−→AD divides −−→BC in the ratio −−→AB:−−→AC ∴P.V.ofD=|−−→AC|(2^i+3^j+4^k)+|−−→AB|(2^i+5^j+7^k)|−−→AB|+|−−→AC| −−→AB=→B−→A=−2^i−4^j−4^k −−→AC=→C−→A=−2^i−2^j−^k |−−→AB|=√4+16+16=6 |−−→AC|=√4+4+1=3 ∴P.V.ofD=3(2^i+3^j+4^k)+6(2^i+5^j+7^k)6+3 ∴P.V.ofD=13(6^i+13^j+18^k)