Question

Find the distance of the point 3i-2j+k from the plane 3x+y-z+2=0 measured parallel to the line x-1/2=y+2/-3=z-1/1. Also find the foot of the perpendicular from the given point upon the given plane.

Answer 1

This is the required answer

Answer 2

Distance = $4\sqrt{14}$ units

and coordinate of foot of perpendicular =(-5,10,-3)

Step-by-step explanation:

Given equation of straight line is 3x + y - z +2= 0

The equation of straight line which passes through the point P( 3i -2j +k) and parallel to the given straight line is

$\frac{x-3}{2} =\frac{y+2}{-3} =\frac{z-1}{1}=r(say)$

Any point the straight line is Q(2r+3,-3r-2,r+1)

Let P be lie on the given plane . So Q will be satisfy the plane.

∴3(2r+3)+(-3r-2)-(r+1)+2=0

⇒r=-4

∴The coordinate of foot of perpendicular is (-5,10,-3)

The distance between PQ is $\sqrt{(3+5)^{2}+(-2-10)^2+(1+3)^{2} }$

                                                =$4\sqrt{14}$ units