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I need help with - Three Dimensional Geometry - JEE Main-7

A perpendicular is drawn from a point on the line \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z}{1} 

to the plane x+y+z=3 such that the foot of the perpendicular Q also 

lies on the plane x-y+z=3. Then the co-ordinates of Q are : 

  • Option 1)

    (1, 0, 2)

  • Option 2)

    (2, 0 , 1)

  • Option 3)

    ( - 1, 0 , 4)

     

  • Option 4)

    (4, 0 , -1)

 
Answers (1)
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\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z}{1}=\lambda

P(2\lambda+1,-\lambda-1,\lambda)

So,

\\foot\;of\;perpendicular\;Q\\\frac{x-(2\lambda+1)}{1}=\frac{y+(\lambda+1)}{1}=\frac{z-\lambda }{1}=\frac{-(2\lambda+1-\lambda-1+\lambda -3) }{3}

\frac{x-(2\lambda+1)}{1}=\frac{y+\lambda+1}{1}=\frac{z-\lambda }{1}=-\frac{(2\lambda -3) }{3}

=> x=2\lambda +1+(-\frac{2\lambda-3}{3})=\frac{4\lambda +6}{3}

=> y=-\lambda -1-(\frac{2\lambda-3}{3})=\frac{-5\lambda}{3}

=> z=\lambda -(\frac{2\lambda-3}{3})=\frac{\lambda+3}{3}

\therefore  point is (\frac{4\lambda+6}{3},\frac{-5\lambda}{3},\frac{\lambda+3}{3})

It lies on equation x-y+z=3

                            \frac{4\lambda+6}{3}-\frac{-5\lambda}{3}+\frac{\lambda+3}{3}=3

                     =>\lambda =0

So, point Q (\frac{6}{3},0,\frac{3}{3})

                Q ( 2 , 0 , 1 )

So, option (2) is correct.


Option 1)

(1, 0, 2)

Option 2)

(2, 0 , 1)

Option 3)

( - 1, 0 , 4)

 

Option 4)

(4, 0 , -1)

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