Vector Algebra Question 130
Question: What is the vector equally inclined to the vectors $ \hat{i}+3\hat{j} $ and $ 3\hat{i}+\hat{j} $ ?
Options:
A) $ \hat{i}+\hat{j} $
B) $ 2\hat{i}-\hat{j} $
C) $ 2\hat{i}+\hat{j} $
D) None of theses
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] Let the required vector be $ \hat{i}+\hat{j} $ Since the vector $ \hat{i}+\hat{j} $ is equally inclined to the vectors $ \hat{i}+3\hat{j} $ and $ 3\hat{i}+\hat{j} $ therefore angle b/w $ \hat{i}+\hat{j} $ and $ \hat{i}+3\hat{j}={\theta_1} $ is equal to angle between $ \hat{i}+\hat{j} $ and $ \hat{i}+3\hat{j} $ $ ={{\cos }^{-1}}[ \frac{(1)(1)+(1)(3)}{\sqrt{{{(1)}^{2}}+{{(1)}^{2}}}\sqrt{{{(1)}^{2}}+{{(3)}^{2}}}} ] $ $ ={{\cos }^{-1}}[ \frac{1+3}{\sqrt{2}\sqrt{10}} ]={{\cos }^{-1}}[ \frac{4}{\sqrt{2}\sqrt{10}} ] $ $ ={{\cos }^{-1}}[ \frac{2}{\sqrt{5}} ] $ and angle between $ \hat{i}+\hat{j} $ and $ (3\hat{i}+\hat{j}) $ $ ={{\cos }^{-1}}| \frac{1+3}{\sqrt{10}\sqrt{2}} | $ $ ={{\cos }^{-1}}( \frac{4}{\sqrt{2}\sqrt{10}} )={{\cos }^{-1}}( \frac{2}{\sqrt{5}} ) $ Hence, required vector is $ \hat{i}+\hat{j} $
BETA
