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Question

AB=3i+j−k and AC=i−j+3k. If the point P on the line segment BC is equidistant from AB and AC, then AP is

A
2ik
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B
i2k
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C
2i+k
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D
None of these
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Solution

The correct option is C 2i+k
A point equidistant from AB and AC is on the bisector of the angle BAC.
A vector along the internal bisector of the angle BAC
=AB|AB|+AC|AC|
=3i+jk9+1+1+ij+3k1+1+9=111(4i+2k)AP=t(2i+k)BP=APAB=t(2i+k)(3i+jk)=(2t3)ij+(t+1)k
Also BC=ACAB=(ij+3k)(3i+jk)=2i2j+4k
But BP=sBC
(2t3)ij+(t+1)k=s(2i2j+4k)2t3=2s,1=2s,t+1=4s
s=12 and t=1
AP=2i+k

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