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JEE Main & Advanced Sample Paper JEE Main Sample Paper-11

  • question_answer
    Consider the lines represented parametrically as : \[{{\text{L}}_{\text{1}}}\text{:}\,\text{x}\,\text{=1-2}\,\text{t}\,\text{;}\,\text{y}\,\text{=}\,\text{t}\,\text{;}\,\text{z}\,\text{=}\,\text{-1+}\,\text{t}\] \[{{\text{L}}_{2}}\text{:}\,\text{x}\,\text{=4-s}\,\,\text{;}\,\text{y}\,\text{=5+4s}\,\text{;}\,\text{z}\,\text{=}\,\text{-2-s}\]

    A)  \[{{\cos }^{-1}}\left( \frac{1}{18} \right)\]                           

    B)  \[{{\cos }^{-1}}\left( \frac{1}{3\sqrt{6}} \right)\]

    C)  \[{{\cos }^{-1}}\left( \frac{1}{6\sqrt{3}} \right)\]              

    D) \[{{\cos }^{-1}}\left( \frac{1}{3\sqrt{2}} \right)\]               

    Correct Answer: C

    Solution :

     \[{{L}_{1}}:\frac{x-1}{-2}=\frac{y-0}{1}=\frac{z+1}{1};\] \[{{L}_{2}}:\frac{x-4}{1}=\frac{y-5}{4}=\frac{z+2}{-1}\] \[{{\vec{V}}_{1}}=-2\hat{i}+\hat{j}+\hat{k},{{\vec{V}}_{2}}=\hat{i}+4\hat{j}-\hat{k}\] \[\cos \theta =\left| \frac{-2+4-1}{\sqrt{6}.\sqrt{18}} \right|=\frac{1}{6\sqrt{3}}\Rightarrow \theta ={{\cos }^{-1}}\left( \frac{1}{6\sqrt{3}} \right)\]


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