JEE Main Solved Paper 2017 Question 12
Question: Let $ \vec{a}=2\hat{i}+\hat{j}-2\hat{k} $ and $ \vec{b}=\hat{i}+\hat{j}. $ Let $ \vec{c} $ be a vector such that $ |\vec{c}-\vec{a}|=3,| (\vec{a}\times \vec{b})\times \vec{c} |=3 $ and the angle between $ \vec{c} $ and $ \vec{a}\times \vec{b} $ be $ 30^{o}. $ Then $ \vec{a}.\vec{c} $ is equal to: [JEE Main Solved Paper-2017]
Options:
A) $ \frac{1}{8} $
B) $ \frac{25}{8} $
C) 2
D) 5
Show Answer
Answer:
Correct Answer: C
Solution:
- $ \vec{a}=2\hat{i}+\hat{j}-2\hat{k},,\vec{b}=\hat{i}+\hat{j} $ and $ |\vec{a}|=3 $
$ \therefore $ $ \vec{a}\times \vec{b}=2\hat{i}-2\hat{j}+\hat{k} $ $ |\vec{a}\times \vec{b}|=3 $ Now: $ (\vec{a}\times \vec{b})\times \vec{c}=|\vec{a}\times \vec{b}||\vec{c}|\sin 30\hat{n} $ $ |(\vec{a}\times \vec{b})\times \vec{c}|=3.|\vec{c}|.\frac{1}{2} $ $ 3=3|\vec{c}|.\frac{1}{2} $
$ \therefore $ $ |\vec{c}|,=2 $ Now: $ |\vec{c}-\vec{a}|=3 $ $ c^{2}+a^{2}-2\vec{c}.\vec{a}=9 $ $ 4+9-2\vec{a}.\vec{c}=9 $ $ \vec{a}.\vec{c}=2 $
