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.
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
Rate this question :