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JEE Main & Advanced Mathematics Three Dimensional Geometry Question Bank Self Evaluation Test - Three Dimensional Geometry

  • question_answer
    \[{{L}_{1}}\] and \[{{L}_{2}}\] are two lines whose vector equations are \[{{L}_{1}}:\overset{\to }{\mathop{r}}\,=\lambda ((cos\,\,\theta +\sqrt{3})\hat{i}+(\sqrt{2}sin\,\,\theta )\hat{j}\]\[+(cos\theta -\sqrt{3})\hat{k}){{L}_{2}}:\overset{\to }{\mathop{r}}\,=\mu \left( a\hat{i}+b\hat{j}+c\hat{k} \right)\], 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

    A) \[\frac{\pi }{6}\]

    B) \[\frac{\pi }{4}\]

    C) \[\frac{\pi }{3}\]

    D) \[\frac{\pi }{2}\]

    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 \[\therefore \cos \alpha =\frac{2a\sqrt{3}}{a\sqrt{2}2\sqrt{2}}=\frac{\sqrt{3}}{2}\Rightarrow \alpha =\frac{\pi }{6}\]


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