RAJASTHAN ­ PET Rajasthan PET Solved Paper-2007

  • question_answer If the angle between the line \[\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}\]and the plane\[2x-y+\sqrt{\lambda z}+4=0\]is\[\theta \]such that\[\sin \theta =\frac{1}{3},\]then the value of\[\lambda \]is,

    A)  \[\frac{5}{3}\]

    B)  \[-\frac{3}{5}\]

    C)  \[\frac{3}{4}\]

    D)  \[-\frac{4}{3}\]

    Correct Answer: A

    Solution :

     For the line\[\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}\] \[{{a}_{1}}=1,{{b}_{1}}=2,{{c}_{1}}=2\] and for the plane\[2x-y+\sqrt{\lambda }z+4=0\] \[{{a}_{2}}=2,{{b}_{2}}=-1,{{c}_{2}}=\sqrt{\lambda }\] \[\therefore \]\[\sin \theta =\frac{1}{3}=\left| \frac{{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} \right|\] \[\Rightarrow \] \[\frac{1}{3}=\left| \frac{2-2+2\sqrt{\lambda }}{\sqrt{1+4+4}\sqrt{4+1+\lambda }} \right|\] \[\Rightarrow \] \[\frac{1}{3}=\frac{2\sqrt{\lambda }}{\sqrt{9}\sqrt{5+\lambda }}\] \[\Rightarrow \] \[\sqrt{5+\lambda }=2\sqrt{\lambda }\] \[\Rightarrow \] \[5+\lambda =4\lambda \] \[\Rightarrow \] \[3\lambda =5\] \[\Rightarrow \] \[\lambda =\frac{5}{3}\]

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