If \( \overrightarrow{AB}=2\hat{\imath}+3\hat{\jmath}-\hat{k} \) and \( \overrightarrow{AC}=3\hat{\imath}-2\hat{\jmath}+4\hat{k} \) are the two sides of a triangle \( ABC \), then the length of the median through vertex \( A \) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{\sqrt{35}}{2} \) units
The median from A goes to D, the midpoint of BC, so
\( \displaystyle \overrightarrow{AD}=\frac{\overrightarrow{AB}+\overrightarrow{AC}}{2} \)
\( \displaystyle =\frac{5\hat{\imath}+\hat{\jmath}+3\hat{k}}{2} \)
\( \displaystyle |\overrightarrow{AD}|=\frac{\sqrt{5^2+1^2+3^2}}{2} \)
\( \displaystyle =\frac{\sqrt{35}}{2}\ \text{units} \)
Why the other options are incorrect
Option (1) \( \sqrt{33} \) units — Incorrect. The squared length of \( \overrightarrow{AB}+\overrightarrow{AC} \) is 25 + 1 + 9 = 35, not 33, and this length must still be halved.
Option (2) \( \sqrt{35} \) units — Incorrect. \( \sqrt{35} \) is the length of \( \overrightarrow{AB}+\overrightarrow{AC} \); it must still be halved.
Option (4) \( \displaystyle\frac{\sqrt{33}}{2} \) units — Incorrect. The numerator should be \( \sqrt{35} \), not \( \sqrt{33} \).