Q. 11

Mark against the correct answer in each of the following:

The equation of a plane through the point A(1, 0, -1) and perpendicular to the line is

A. 2x + 4y – 3z = 3

B. 2x – 4y + 3z = 5

C. 2x + 4y – 3z = 5

D. x + 3y + 7z = -6

Answer :

Given: Plane passes through the point A(1, 0, -1).


Plane is perpendicular to the line



To find: Equation of the plane.


Formula Used: Equation of a plane is ax + by + cz = d where (a, b, c) are the direction ratios of the normal to the plane.


Explanation:


Let the equation of the plane be


ax + by + cz = d … (1)


Substituting point A,


a – z = d


Since the given line is perpendicular to the plane, it is the normal.


Direction ratios of line is 2, 4, -3


Therefore, 2 + 3 = d


d = 5


So the direction ratios of perpendicular to plane is 2, 4, -3 and d = 5


Substituting in (1),


2x + 4y – 3z = 5


Therefore, equation of plane is 2x + 4y – 3z = 5

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