Given: Plane passes through A(2, 2, 1) and B(9, 3, 6). Plane is perpendicular to 2x + 6y + 6z = 1
To find: Equation of the plane
Formula Used: Equation of a plane is
a(x – x1) + b(y – y1) + c(z – z1) = 0
where a:b:c is the direction ratios of the normal to the plane.
(x1, y1, z1) is a point on the plane.
Explanation:
Let the equation of plane be a(x – x1) + b(y – y1) + c(z – z1) = 0
Since (2, 2, 1) is a point in the plane,
a(x – 2) + b(y – 2) + c(z – 1) = 0 … (1)
Since B(9, 3, 6) is another point on the plane,
a(9 – 2) + b(3 – 2) + c(6 – 1) = 0
7a + b + 5c = 0 … (1)
Since this plane is perpendicular to the plane 2x + 6y + 6z = 1, the direction ratios of the normal to the plane will also be perpendicular.
So, 2a + 6b + 6c = 0 ⇒ a + 3b + 3c = 0 … (2)
Solving (1) and (2),
a : b : c = 3 : 4 : -5
Substituting in (1),
3x – 6 + 4y – 8 – 5z + 5 = 0
3x + 4y – 5z – 9 = 0
Therefore the equation of the plane is 3x + 4y – 5z – 9 = 0
