JEE Main On 16 April 2018 Question 16
Question: Let $ \vec{a}=\hat{i}+\hat{j}+\hat{k},\vec{c}=\hat{j}-\hat{k} $ and a vector $ \vec{b} $ be such that $ \vec{a}\times \vec{b}=\vec{c} $ and $ \vec{a}.\vec{b}=3. $ Then $ |\vec{b}| $ equals?
Options:
A) $ \sqrt{\frac{11}{3}} $
B) $ \sqrt{\frac{11}{3}} $
C) $ \frac{11}{\sqrt{3}} $
D) $ \frac{11}{3} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \vec{a}\times \vec{b}\times \vec{c} $
$ \Rightarrow $ $ |\vec{a}||\vec{b}|\sin \theta =|\vec{c}| $
$ \Rightarrow $ $ |\vec{a}||\vec{b}|\sin \theta =\sqrt{2} $ ?[1] $ \vec{a}.\vec{b}=3 $
$ \Rightarrow $ $ |\vec{a}||\vec{b}|\cos \theta =3 $ ?[2] Dividing [1] by [2], we get $ \tan \theta =\frac{\sqrt{2}}{3} $
$ \Rightarrow $ $ \sin \theta =\frac{\sqrt{2}}{\sqrt{11}} $ Substituting value of $ \sin \theta $ in [1], we get
$ \Rightarrow $ $ |\vec{a}||\vec{b}||\sin \theta |=\sqrt{2} $
$ \Rightarrow $ $ \sqrt{3}|\vec{b}|\frac{\sqrt{2}}{\sqrt{11}}=\sqrt{2} $ $ |\vec{b}|=\frac{\sqrt{11}}{\sqrt{3}} $ Hence, answer is option A.
