JEE Main & Advanced AIEEE Paper (Held On 11 May 2011)

  • question_answer
    The distance of the point (1, -5, 9) from the plane x - y + z = 5 measured along a straight line x = y = z is :     AIEEE  Solved  Paper (Held On 11 May  2011)

    A) \[10\sqrt{3}\]

    B) \[5\sqrt{3}\]

    C) \[3\sqrt{10}\]

    D) \[3\sqrt{5}\]

    Correct Answer: A

    Solution :

                    Line through P(1, - 5, 9) parallel to x = y = z is \[\frac{x-1}{1}=\frac{y+5}{1}=\frac{z-9}{1}=\lambda (say)\] \[Q(x=1+\lambda ,y=-5+\lambda ,z=9+\lambda )\] Given plane \[x-y+z=5\] \[\therefore \]\[1+\lambda +5-\lambda +9+\lambda =5\] \[\Rightarrow \]\[\lambda =-10\] \[\therefore \]\[Q(-9,-15,-1)\] \[\therefore \]\[PQ=\sqrt{{{(1+9)}^{2}}+{{(15-5)}^{2}}+{{(9+1)}^{2}}}\] \[=\sqrt{300}=10\sqrt{3}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner