Find the equation of the plane that contains the point (1, – 1, 2) and is perpendicular to each of the planes 2x + 3y–2z = 5and x + 2y – 3z = 8.
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
⇒ ax + by + cz = a – b + 2c …… (1)
Now, the required plane is also perpendicular to the planes,
2x + 3y – 2z = 5 and x + 2y – 3z = 8
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,
2a + 3b – 2c = 0 …… (b)
a + 2b – 3c = 0 …… (c)
Solving (b) and (c) we get,
∴a = – 5λ, b = 4λ, c = λ
Putting values of a,b,c in equation (1) we get,
(– 5λ)x + (4λ)y + (λ)z = – 5λ – 4λ + 2λ
⇒ – 5λx + 4λy + λz = – 7λ
Dividing both sides by (– λ) we get
5x – 4y – z = 7
So, the equation of the required planes is 5x - 4y – z = 7