Find the equation of the plane passing through the points whose coordinates are (– 1, 1, 1) and (1, – 1, 1) and perpendicular to the plane x + 2y + 2z = 5.
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 (– 1,1,1), so the equation of plane is
a(x + 1) + b(y – 1) + c(z – 1) = 0 …… (i)
Plane (i) is also passing through (1, – 1,1), so(1, – 1,1) must satisfy the equation of plane, so we have
a(1 + 1) + b(– 1 – 1) + c(1 – 1) = 0
⇒2a – 2b = 0 …… (ii)
Plane x + 2y + 2z = 5 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(1) + b(2) + c(2) = 0
⇒a + 2b + 2c = 0 …… (iii)
Solving (ii) and (iii) we get,
∴a = – 4λ, b = – 4λ, c = 6λ
Putting values of a,b,c in equation (i) we get,
(– 4λ)(x + 1) + (– 4λ)(y – 1) + (6λ)(z – 1) = 0
⇒ – 4λx – 4λ – 4λy + 4λ + 6λz – 6λ = 0
⇒ – 4λx – 4λy + 6λz – 6λ = 0
Dividing by – 2λ we get,
2x + 2y – 3z + 3 = 0
So, the required plane is 2x + 2y – 3z + 3 = 0