question_answer

The   vectors   \[\overrightarrow{AB}=3\hat{i}+5\hat{j}+4\hat{k}\]   and \[\overrightarrow{AC}=5\hat{i}-5\hat{j}+2\hat{k}\] are the sides of a triangle ABC. The length of the median through A is:

A) \[\sqrt{13}\]units

B) \[2\sqrt{5}\] units

C) 5 units

D) 10 units

Correct Answer: C

Solution :

[c] Let the given vectors be \[\overrightarrow{AB}=3\hat{i}+5\hat{j}+4\hat{k}\] and \[\overrightarrow{AC}=5\hat{i}-5\hat{j}+2\hat{k}\] Let AM be the median through A   \[\therefore \overrightarrow{AM}=\frac{1}{2}(\overrightarrow{AB}+\overrightarrow{AC})\]             \[=\frac{1}{2}[(3\hat{i}+5\hat{j}+4\hat{k})+(5\hat{i}-5\hat{j}+2\hat{k})]\]             \[=\frac{1}{2}(8\hat{i}+6\hat{k})=(4\hat{i}+3\hat{k})\] \[\therefore \] Length of the median \[AM=\sqrt{{{4}^{2}}+{{3}^{2}}}\]                                     \[=5\,units\]