In a triangle OAB, O is the origin, \(\overrightarrow{OA}=3\hat{\imath}+2\hat{\jmath}-\hat{k}\) and \(\overrightarrow{OB}=\hat{\imath}-4\hat{\jmath}+5\hat{k}\). The length of the median of the triangle through O is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 3 units
The median through O ends at the mid-point C of AB, whose position vector is the average of those of A and B:
\( \displaystyle \overrightarrow{OC} = \frac{(3\hat{\imath}+2\hat{\jmath}-\hat{k})+(\hat{\imath}-4\hat{\jmath}+5\hat{k})}{2} \)
\( \displaystyle = \frac{4\hat{\imath}-2\hat{\jmath}+4\hat{k}}{2} = 2\hat{\imath}-\hat{\jmath}+2\hat{k} \)
So the length of the median is
\(|\overrightarrow{OC}| = \sqrt{4+1+4} = 3\)
Why the other options are incorrect
Option (1) 6 units — Incorrect. It is \(|\overrightarrow{OA}+\overrightarrow{OB}|\); the sum must be halved to reach the mid-point.
Option (2) \(\sqrt{14}\) units — Incorrect. It is the length of the side OA, \(\sqrt{9+4+1}\).
Option (4) 9 units — Incorrect. It is \(|\overrightarrow{OC}|^2\), not \(|\overrightarrow{OC}|\).