Vector Algebra Question 71
Question: 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:
Options:
A) $ \sqrt{13} $ units
B) $ 2\sqrt{5} $ units
C) 5 units
D) 10 units
Show Answer
Answer:
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 $
BETA
