A) \[\frac{\pi }{2}\]
B) \[\frac{\pi }{4}\]
C) \[\frac{\pi }{6}\]
D) None of these
Correct Answer: B
Solution :
Normal of plane \[{{P}_{1}}\] is \[{{\vec{n}}_{1}}=(2\hat{j}\,+3\hat{k})\times (4\hat{j}-3\hat{k})=-\,18\,\hat{i}\] Normal to plane \[{{P}_{2}}\] is \[=\,\,{{\vec{n}}_{2}}(\hat{j}\,-\hat{k})\,\times \,(3\hat{i}+3\hat{j})\,\,=\,\,3\hat{i}-3\hat{j}-3\,\hat{k}\] \[\therefore \,\,\vec{A}\,\,is\,parallel\,\,to\,\,\pm ({{\vec{n}}_{1}}\times {{\vec{n}}_{2}})=\pm (-\,54\hat{j}\,+54\,\hat{k})\] Now angle between \[\vec{A}\,\,and\,\,2\hat{i}+\hat{j}\,-2\hat{k}\] is given by cos \[\cos \theta =\pm \frac{(-54\hat{j}+54\hat{k}).(2\hat{i}+\hat{j}-2\hat{k})}{54\sqrt{2}.3}=\,\,\pm \,\,\frac{1}{\sqrt{2}}\] \[\therefore \,\,\,\theta =\frac{\pi }{4}\,\,or\,\,\frac{3\pi }{4}\]
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