Q Find the equation of a line lieson a planeplane 2x-3y + 4z=-8,perp

1.	Q Find the equation of a line lieson a planeplane 2x-3y + 4z=-8,perpendicular to a linex-1/3=y-1/2=z-2/3 and passes throughthe points (1,2,-1),
Answer» ong>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.

Q Find the equation of a line lieson a planeplane 2x-3y + 4z=-8,perpendicular to a linex-1/3=y-1/2=z-2/3 and passes throughthe points (1,2,-1),

Answer»

ong>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.

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