JEE Main On 08 April 2018 Question 25
Question: Let $ \overset{\to }{\mathop{u}}, $ be a vector coplanar with the vectors $ \overset{\to }{\mathop{a}},=2\widehat{i}+3\widehat{j}-\widehat{k} $ and $ \overset{\to }{\mathop{b}},=\widehat{j}+\widehat{k}. $ If $ \overset{\to }{\mathop{u}}, $ is perpendicular to $ \overset{\to }{\mathop{a}}, $ and $ \overset{\to }{\mathop{u}},\cdot \overset{\to }{\mathop{b}},=24 $ , then $ {{| \overset{\to }{\mathop{u}}, |}^{2}} $ is equal to: [JEE Main Online 08-04-2018]
Options:
A) 256
B) 84
C) 336
D) 315
Show Answer
Answer:
Correct Answer: C
Solution:
$ \overset{\to }{\mathop{u}},=\lambda \overset{\to }{\mathop{a}},\times ( \overset{\to }{\mathop{a}},\times \overset{\to }{\mathop{b}}, ) $
$ =\lambda [ ( \overset{\to }{\mathop{a}},.\overset{\to }{\mathop{b}}, )\overset{\to }{\mathop{a}},-(\overset{\to }{\mathop{a}},.\overset{\to }{\mathop{a}},)\overset{\to }{\mathop{b}}, ] $
$ =\lambda [ 2\overset{\to }{\mathop{a}},-14\overset{\to }{\mathop{b}}, ] $
$ =2\lambda [ \overset{\to }{\mathop{a}},-7\overset{\to }{\mathop{b}}, ]$=$2\lambda ( 2\hat{i}-4 \widehat{j}-8\widehat{k} ) $
$ \overset{\to }{\mathop{u}},.\overset{\to }{\mathop{b}},=24 $ $ -12\times 2\lambda =24 $ $ \lambda =-1 $ $ \overset{\to }{\mathop{u}},=-4\widehat{i}+8\widehat{j}+16\widehat{k} $
$ {{| \overset{\to }{\mathop{u}}, |}^{2}}=16+64+256 $ $ {{| \overset{\to }{\mathop{u}}, |}^{2}}=336 $
