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JEE Main & Advanced Mathematics Three Dimensional Geometry Question Bank Line and Plane

  • question_answer
    The line \[\frac{x+3}{3}=\frac{y-2}{-2}=\frac{z+1}{1}\] and the plane \[4x+5y+3z-5=0\] intersect at a point

    A)            (3, 1, ?2)

    B)            (3, ? 2, 1)

    C)            (2, ?1, 3)

    D)            (?1, ?2, ?3)

    Correct Answer: B

    Solution :

                       Line is \[\frac{x+3}{3}=\frac{y-2}{-2}=\frac{z+1}{1}=\lambda \,\](Let)            \[x=3\lambda -3;\,\,y=-2\lambda +2;\,\,z=\lambda -1\]line intersects plane, therefore,\[4(3\lambda -3)+5(-2\lambda +2)+3(\lambda -1)-5=0\]\[\,\,\Rightarrow \lambda =2\]. So, \[x=3;\,\,y=-2;\,\,z=1\].                    Trick : Since the point (3, ? 2, 1) satisfies both the equations.


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