InterviewSolution
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InterviewSolution
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Let `vec(A)D` be the angle bisector of `angle A" of " Delta ABC ` such that `vec(A)D=alpha vec(A)B+beta vec(A)C,` thenA. `alpha= (|vec(AB)|)/(|vec(AB)|+|vec(AC)|),beta=(|vec(AC)|)/(|vec(AB)|+|vec(AC)|)`B. `alpha= (|vec(AB)|+|vec(AC)|)/(|vec(AB)|),beta=(|vec(AB)|+|vec(AC)|)/(|vec(AC)|)`C. `alpha= (|vec(AC)|)/(|vec(AB)|+|vec(AC)|),beta=(|vec(AB)|)/(|vec(AB)|+|vec(AC)|)`D. `alpha= (|vec(AB)|)/(|vec(AC)|),beta=(|vec(AC)|)/(|vec(AB)|)` |
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Answer» Correct Answer - C Clearly, AD divides BC in the ratio AB : AC. `therefore vec(AD)=(|vec(AB)|vec(AC)+|vec(AC)|vec(AB))/(|vec(AB)|+|vec(AC)|)` `rArrvec(AD)=alpha vec(AB)+beta vec(AC),` where `alpha= (|vec(AC)|)/(|vec(AB)|+|vec(AC)|) and beta=(|vec(AB)|)/(|vec(AB)|+|vec(AC)|)` |
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