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Question
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, then the position vector of the point where the bisector of ∠A meets BC is
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, then the position vector of the point where the bisector of ∠A meets BC is
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Solution
The correct option is C 13(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, P.V. of D is given by
|−−→AB|(2^i+5^j+7^k)+|−−→AC|(2^i+3^j+4^k)|−−→AB|+|−−→AC|
We know that,
−−→AB=−2^i−4^j−4^k|−−→AB|=6
−−→AC=−2^i−2^j−^k
|−−→AC|=3
Position vector of D is,
=18^i+39^j+54^k9=13(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, P.V. of D is given by
|−−→AB|(2^i+5^j+7^k)+|−−→AC|(2^i+3^j+4^k)|−−→AB|+|−−→AC|
We know that,
−−→AB=−2^i−4^j−4^k|−−→AB|=6
−−→AC=−2^i−2^j−^k
|−−→AC|=3
Position vector of D is,
=18^i+39^j+54^k9=13(6^i+13^j+18^k)
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