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The position vectors of the vertices A, B and C of triangle ABC are 4î + 7ĵ + 8k̂, 2î + 3ĵ + 4k̂ and 2î + 5ĵ + 7k̂ respectively. The position vector of the point where the internal bisector of angle A intersects BC is

1. (1/3)(6î + 13ĵ + 18k̂)
2. (2/3)(6î + 12ĵ − 8k̂)
3. (1/3)(−6î − 8ĵ − 9k̂)
4. (2/3)(−6î − 12ĵ + 8k̂)

Correct answer: (1/3)(6î + 13ĵ + 18k̂)

Solution

AB=6, AC=3, so the bisector from A meets BC at D dividing it in ratio AB:AC=2:1 (BD:DC). D=(2*C+1*B)/3=(2,13/3,6)=(1/3)(6i+13j+18k).

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