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)

\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)

Most Viewed Questions

Preparation Products

JEE Main Rank Booster 2022

Booster and Kadha Video Lectures, Unlimited Full Mock Test, Adaptive Time Table, 24x7 Doubt Chat Support,.

₹ 6999/- ₹ 799/-
Buy Now
Knockout JEE Main 2022 (One Month)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Live Classes, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 13999/- ₹ 6999/-
Buy Now
Knockout JEE Main 2023 (One Month)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Live Classes, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 13999/- ₹ 6999/-
Buy Now
Knockout JEE Main (One Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Expert Mentorship, - Study Improvement Plan.

₹ 2999/- ₹ 1999/-
Buy Now
Knockout JEE Main (Twelve Months Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Expert Mentorship, - Study Improvement Plan.

₹ 23999/- ₹ 14999/-
Buy Now
Boost your Preparation for JEE Main 2021 with Personlized Coaching
 
Exams
Articles
Questions