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JEE Main & Advanced JEE Main Online Paper (Held On 08 April 2018)

  • question_answer
    If \[{{\text{L}}_{\text{1}}}\] is the line of intersection of the planes \[2x-2y+3z-2=0,\,\,x-y+z+1=0\] and \[{{L}_{2}}\] is the line of intersection of the planes \[x+2y-z-3=0,\text{ }3x-y+2z-1=0\], then the distance of the origin from the plane, containing the lines \[{{L}_{1}}\] and \[{{L}_{2}},\] is: [JEE Main Online 08-04-2018]

    A)  \[\frac{1}{2\sqrt{2}}\]                      

    B)  \[\frac{1}{\sqrt{2}}\]

    C)  \[\frac{1}{4\sqrt{2}}\]                      

    D)  \[\frac{1}{3\sqrt{2}}\]

    Correct Answer: D

    Solution :

    \[2x2y+3z2+\lambda (xy+z+1)=0\] \[x+2yz3+\mu (3xy+2z1)=0\] \[\Rightarrow \frac{2+\lambda }{1+3\mu }=\frac{-2-\lambda }{2-\mu }=\frac{3+\lambda }{-1+2\mu }=\frac{-2+\lambda }{-3-\mu }\] \[\Rightarrow \frac{1}{1+\mu }=\frac{5}{2+3\mu }\] \[\Rightarrow 5+5\mu =2+3\mu \] \[\Rightarrow \mu =\frac{-3}{2}\] \[2(x+2yz3)3(3xy+2z1)=0\] \[\Rightarrow -7x+7y8z3=0\] Distance \[=\frac{3}{\sqrt{162}}=\frac{3}{9\sqrt{2}}=\frac{1}{3\sqrt{2}}\]


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