Vector Algebra Question 211
Question: If $ 4\hat{i}+7\hat{j}+8\hat{k},2\hat{i}+3\hat{j}+4\hat{k} $ and $ 2\hat{i}+5\hat{j}+7\hat{k} $ are the position vectors of the vertices A, B and C, respectively, of triangle ABC, then the position vector of the point where the bisector of angle A meets BC is
Options:
A) $ \frac{2}{3}(-6\hat{i}-8\hat{j}-6\hat{k}) $
B) $ \frac{2}{3}(6\hat{i}+8\hat{j}+6\hat{k}) $
C) $ \frac{1}{3}(6\hat{i}+8\hat{j},+18\hat{k}) $
D) $ \frac{1}{3}(5\hat{j}+12\hat{k}) $
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Answer:
Correct Answer: C
Solution:
Suppose the bisector of angle A meets BC at D. Then AD divides BC in the ratio AB: AC. So, P.V. of D is given by $ \frac{|\overrightarrow{AB}|(2\hat{i}+5\hat{j}+7\hat{k})+|\overrightarrow{AC}|(2\hat{i}+3\hat{j}+4\hat{k})}{|\overrightarrow{AB}|+|\overrightarrow{AC}|} $ But $ \overrightarrow{AB}=-2\hat{i}-4\hat{j}-4\hat{k} $ And $ \overrightarrow{AC}=-2\hat{i}-2\hat{j}-\hat{k} $ $ \Rightarrow \overrightarrow{| AB |}=6 $ and $ \overrightarrow{| AC |}=3 $ Therefore, P.V. of D is given by $ \frac{6(2\hat{i}+5\hat{j}+7\hat{k})+3(2\hat{i}+3\hat{j}+4\hat{k})}{6+3} $ $ =\frac{1}{3}(6\hat{i}+13\hat{j}+18\hat{k}) $
BETA
