Vector Algebra Question 147
Question: The vectors $ \overrightarrow{AB}=3\mathbf{i}+5\mathbf{j}+4\mathbf{k} $ and $ \overrightarrow{AC}=5\mathbf{i}-5\mathbf{j}+2\mathbf{k} $ are the sides of a triangle ABC. The length of the median through A is
[UPSEAT 2004]
Options:
A) $ \sqrt{13} $ unit
B) $ \theta $ unit
C) 5 unit
D) 10 unit
Show Answer
Answer:
Correct Answer: C
Solution:
- $ P.V. $ of $ \overrightarrow{AD}=\frac{(3+5)\mathbf{i}+(5-5)\mathbf{j}+(4+2)\mathbf{k}}{2} $
$ \overrightarrow{AD}=\frac{8\mathbf{i}+6\mathbf{k}}{2}=4\mathbf{i}+3\mathbf{k} $
\ Length of median $ =|\overrightarrow{AD}|=\sqrt{16+9}=5 $ unit.
BETA
