JEE Main On 16 April 2018 Question 3
Question: Let $ \vec{A}=(\hat{i}+\hat{j}) $ and $ \vec{B}=(\hat{i}-\hat{j}). $ The magnitude of a coplanar vector $ \vec{C} $ such that $ \vec{A}.\vec{C}=\vec{B}.\vec{C}=\vec{A}.\vec{B}, $ is given by [JEE Main 16-4-2018]
Options:
A) $ \sqrt{\frac{5}{9}} $
B) $ \sqrt{\frac{10}{9}} $
C) $ \sqrt{\frac{20}{9}} $
D) $ \sqrt{\frac{9}{12}} $
Show Answer
Answer:
Correct Answer: A
Solution:
Let the vector $ \vec{C} $ be $ x\hat{i}+y\hat{j} $
$ \vec{A}.\vec{C}=x+y=1 $ and $ \vec{B}.\vec{C}=2x-y=1 $
Solving these two equations simultaneously
$ x=2/3 $ and $ y=1/3 $
Hence $ |\vec{C}|=\sqrt{\frac{4}{9}+\frac{1}{9}}=\sqrt{\frac{5}{9}} $
