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) \[\frac{2}{3}\,(-6\mathbf{i}-12\mathbf{j}+8\mathbf{k})\]
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 \[D=\frac{6(2\mathbf{i}+5\mathbf{j}+7\mathbf{k})+3(2\mathbf{i}+3\mathbf{j}+4\mathbf{k})}{6+3}\] \[=\frac{18\mathbf{i}+39\mathbf{j}+54\mathbf{k})}{9}\]\[=\frac{1}{3}(6\mathbf{i}+13\mathbf{j}+18\mathbf{k})\].You need to login to perform this action.
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