Question

Q
Find the equation of a line lies
on a plane
plane 2x-3y + 4z=-8,
perpendicular to a line
x-1/3=y-1/2=z-2/3 and passes through
the points (1,2,-1),​

Answer 1

Answer:

No such line can exist because the point

(1,2,−1) does not lie on the plane 2x−3y+4z=8

, therefore, the line cannot lie on the plane.

Step-by-step explanation:

An additional reason why no such line can exist is that the direction of the line is

→l=3ˆi+2ˆj+3ˆk

; this means that the only planes that can have lines that intersect the given line perpendicularly must be of the form 3x+2y+3z=c where c is any real number. All other planes cannot have lines that intersect the given line perpendicularly.