Q

# 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)

Views

$\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z}{1}=\lambda$

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

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)

Exams
Articles
Questions