if 4i cap + 7j cap +8k cap ,2i cap + 3j cap + 4k cap and 2i cap + 5j cap + 7k cap are position vectors A,B,C of triangle ABC, the position vector of point where the bisector of angle A meets BC is ? Share with your friends Share 2 Vivek Rai answered this Dear student we must know the angle bisector theorem, which states that, An angle bisector of an angle of a triangle divides the opposite side in two segments that are proportional to the other two sides of the triangle.so, BD:DC::AB:ACnow it is given that , P.V of A =4i+7j+8kP.V of B=2i+3j+4kP.V of C= 2i+ 5j + 7k so, AB⇀=P.V of B-P.V of A=(2i+3j+4k)-(4i+7j+8k)=-2i-4j-4kAC⇀=P.V of C-P.V of A=(2i+ 5j + 7k )-(4i+7j+8k)=-2i-2j-k|AB⇀|=4+16+16=6|AC⇀|=4+4+1=3so as BD:DC::AB:AC,we have BDDC=63=21, so D divides the side BC in the ratio 2:1 internally if AD is the internal angle bisector. and 2:1 externally if AD is the external angle bisector. so postition vector of D can be obtained by using section formula P.V of D=2.P.V of C+1.P.V of B2+1 and 2.P.V of C-1.P.V of B2-1=2(2i+ 5j + 7k )+(2i+3j+4k)3and2(2i+ 5j + 7k )-(2i+3j+4k)1=2i+133j+6k and2i+7j+10k Hope this clears your doubt With regards 8 View Full Answer
