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JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Morning)

  • question_answer
    If the line,\[\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{4}\]meets the plane, \[x+2y+3z=15\]at a point P, then the distance of P from the origin is [JEE Main 9-4-2019 Morning]

    A) \[\frac{9}{2}\]

    B) \[2\sqrt{5}\]

    C) \[\frac{\sqrt{5}}{2}\]                         

    D) \[\frac{7}{2}\]

    Correct Answer: A

    Solution :

    Any point on the given line can be \[(1+2\lambda ,-1+3\lambda ,2+4\lambda );\lambda \in R\] Put in plane \[1+2\lambda ,+(-2+6\lambda )+(6+12\lambda )=15\] \[20\lambda +5=15\] \[20\lambda =10\] \[\lambda =\frac{1}{2}\] \[\therefore \] Point\[\left( 2,\frac{1}{2},4 \right)\] Distance from origin \[=\sqrt{4+\frac{1}{4}+16}=\sqrt{\frac{16+1+64}{2}}=\sqrt{\frac{81}{2}}\]\[=\frac{9}{2}\]                    


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