Find the equation of the plane passing through the point (1, – 1, 2) and (2, – 2, 2) and which is perpendicular to the plane 6x – 2y + 2z = 9.
We know that solution of a plane passing through (x1,y1,z1) is given as -
a(x – x1) + b(y – y1) + c(z – z1) = 0
The required plane passes through (1, – 1, 2), so the equation of plane is
a(x – 1) + b(y + 1) + c(z – 2) = 0 …… (i)
Plane (i) is also passing through (2, – 2, 2), so(2, – 2, 2) must satisfy the equation of plane, so we have
a(2 – 1) + b(– 2 + 1) + c(2 – 2) = 0
⇒a – b = 0 …… (ii)
Plane 6x – 2y + 2z = 9 is perpendicular to the required plane
We know that planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 are at right angles if,
a1a2 + b1b2 + c1c2 = 0 …… (a)
Using (a) we have,
a(6) + b(– 2) + c(2) = 0
⇒6a – 2b + 2c = 0 …… (iii)
Solving (ii) and (iii) we get,
∴a = – 2λ, b = – 2λ, c = 4λ
Putting values of a,b,c in equation (i) we get,
(– 2λ)(x – 1) + (– 2λ)(y + 1) + (4λ)(z – 2) = 0
⇒ – 2λx + 2λ – 2λy – 2λ + 4λz – 8λ = 0
⇒ – 2λx – 2λy + 4λz – 8λ = 0
Dividing by – 2λ we get,
x + y – 2z + 4 = 0
So, the required plane is x + y – 2z + 4 = 0