Vector Algebra Question 188
Question: The perimeter of the triangle whose vertices have the position vectors $ (\mathbf{i}+\mathbf{j}+\mathbf{k}),(5\mathbf{i}+3\mathbf{j}-3\mathbf{k}) $ and $ (2\mathbf{i}+5\mathbf{j}+9\mathbf{k}), $ is given by
[MP PET 1993]
Options:
A) $ 15+\sqrt{157} $
B) $ 15-\sqrt{157} $
C) $ \sqrt{15}-\sqrt{157} $
D) $ \sqrt{15}+\sqrt{157} $
Show Answer
Answer:
Correct Answer: A
Solution:
- $ \mathbf{a}=4\mathbf{i}+2\mathbf{j}-4\mathbf{k}\Rightarrow |\mathbf{a}|=\sqrt{16+16+4}=6 $ $ \mathbf{b}=-3\mathbf{i}+2\mathbf{j}+12\mathbf{k}\Rightarrow |\mathbf{b}|=\sqrt{144+4+9}=\sqrt{157} $ $ \mathbf{c}=-\mathbf{i}-4\mathbf{j}-8\mathbf{k}\Rightarrow |\mathbf{c}|=\sqrt{64+16+1}=9 $ Hence perimeter is $ 15+\sqrt{157}. $
BETA
