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JEE Main & Advanced JEE Main Solved Paper-2016

  • question_answer
    If the line,\[\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z+4}{3}\]lies in the plane, \[1x+my-z=9,\]then\[{{1}^{2}}{{+}^{2}}\] is equal to :- [JEE Main Solved Paper-2016 ]

    A) 2                                             

    B) 26

    C) 18                                          

    D) 5

    Correct Answer: A

    Solution :

                    Given line\[\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z+4}{3}\]and Given plane is \[\ell x+my-z=9\] Now, it is given that line lies on plane \[\therefore \]\[2\ell -m-3=0\Rightarrow 2\ell -m=3\]                    ...(1) Also, (3, ?2, ?4) lies on plane\[3\ell -2m=5\]...(2) Solving (1) and (2), we get \[\ell =1,m=-1\]\[\therefore \]\[{{\ell }^{2}}+{{m}^{2}}=2\]


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