Three Dimensional Geometry Question 90
Question: $ L_1 $ and $ L_2 $ are two lines whose vector equations are $ L_1:\overset{\to }{\mathop{r}},=\lambda ((cos\theta +\sqrt{3})\hat{i}+(\sqrt{2}sin\theta )\hat{j} $ $ +(cos\theta -\sqrt{3})\hat{k})L_2:\overset{\to }{\mathop{r}},=\mu ( a\hat{i}+b\hat{j}+c\hat{k} ) $ , where $ \lambda $ and $ \mu $ are scalars and $ \alpha $ is the acute angle between $ L_1 $ and $ L_2 $ . If the angle $ ‘\alpha ’ $ is independent of $ \theta $ then the value of $ ‘\alpha ’ $ is
Options:
A) $ \frac{\pi }{6} $
B) $ \frac{\pi }{4} $
C) $ \frac{\pi }{3} $
D) $ \frac{\pi }{2} $
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Answer:
Correct Answer: A
Solution:
[a] Both the lines pass through origin. Line $ L_1 $ is parallel to the vector $ {{\overrightarrow{V}}_2}=a\hat{i}+b\hat{j}+c\hat{k} $
$ \therefore \cos \alpha =\frac{\overrightarrow{V_1.}\overrightarrow{V_2}}{|\overrightarrow{V_1}||\overrightarrow{V_2}|} $ $ =\frac{a(cso\theta +\sqrt{3})+(b\sqrt{2})sin\theta +c(cos\theta -\sqrt{3})}{\sqrt{a^{2}+b^{2}+c^{2}}\sqrt{{{(cos\theta +\sqrt{3})}^{2}}+2{{\sin }^{2}}\theta +{{(cos\theta -\sqrt{3})}^{2}}}} $ $ =\frac{(a+c)cos\theta +b\sqrt{3})+\sin \theta +(a-c-\sqrt{3})}{\sqrt{a^{2}+b^{2}+c^{2}}\sqrt{2+6}} $ In order that $ \cos \alpha $ is independent of $ \theta $ , we get $ a+c=0 $ and b=0
$ \therefore \cos \alpha =\frac{2a\sqrt{3}}{a\sqrt{2}2\sqrt{2}}=\frac{\sqrt{3}}{2}\Rightarrow \alpha =\frac{\pi }{6} $
BETA
