Vector Algebra Question 277
Question: If $ 4\mathbf{i}+7\mathbf{j}+8\mathbf{k},,2\mathbf{i}+3\mathbf{j}+4\mathbf{k}, $ and $ 2\mathbf{i}+5\mathbf{j}+7\mathbf{k} $ are the position vectors of the vertices A, B and C respectively of triangle ABC. The position vector of the point where the bisector of angle A meets BC is
[Pb. CET 2004]
Options:
A) $ \frac{1}{3},(6\mathbf{i}+13\mathbf{j}+18\mathbf{k}) $
B) $ \frac{2}{3},(6\mathbf{i}+12\mathbf{j}-8\mathbf{k}) $
C) $ \frac{1}{3},(-6\mathbf{i}-8\mathbf{j}-9\mathbf{k}) $
D) $ \frac{2}{3},(-6\mathbf{i}-12\mathbf{j}+8\mathbf{k}) $
Show Answer
Answer:
Correct Answer: A
Solution:
- Let the bisector of angle A meets BC at D, then AD divides BC in the ratio AB: AC Position vectors of D = $ \frac{|\overrightarrow{AB}|(2\mathbf{i}+5\mathbf{j}+7\mathbf{k})+|\overrightarrow{AC}|(2\mathbf{i}+3\mathbf{j}+4\mathbf{k})}{|\overrightarrow{AB}|+|\overrightarrow{AC}|} $ Here, $ |\overrightarrow{AB}|=|-2\mathbf{i}-4\mathbf{j}-4\mathbf{k}|=6 $ and $ |\overrightarrow{AC}|=|-2\mathbf{i}-2\mathbf{j}-\mathbf{k}|=3 $ Position vector of $ D=\frac{6(2\mathbf{i}+5\mathbf{j}+7\mathbf{k})+3(2\mathbf{i}+3\mathbf{j}+4\mathbf{k})}{6+3} $ $ =\frac{18\mathbf{i}+39\mathbf{j}+54\mathbf{k})}{9} $ $ =\frac{1}{3}(6\mathbf{i}+13\mathbf{j}+18\mathbf{k}) $ .
BETA
