Find the equation of the plane passing through the points (2, 2, 1) and (9, 3, 6) and perpendicular to the plane 2x + 6y + 6z = 1.
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 (2,2,1), so the equation of the plane is
a(x – 2) + b(y – 2) + c(z – 1) = 0 …… (i)
Plane (i) is also passing through (9,3,6), so(9,3,6) must satisfy the equation of plane, so we have
a(9 – 2) + b(3 – 2) + c(6 – 1) = 0
⇒7a + b + 5c = 0 …… (ii)
Plane 2x + 6y + 6z = 1 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(2) + b(6) + c(6) = 0
⇒2a + 6b + 6c = 0 …… (iii)
Solving (ii) and (iii) we get,
∴a = – 24λ, b = – 32λ, c = – 40λ
Putting values of a,b,c in equation (i) we get,
(– 24λ)(x – 2) + (– 32λ)(y – 2) + (– 40λ)(z – 1) = 0
⇒ – 24λx + 48λ – 32λy + 64λ – 40λz + 40λ = 0
⇒ – 24λx – 32λy – 40λz + 152λ = 0
Dividing by – 8λ we get,
3x + 4y + 5z – 19 = 0
So, the required plane is 3x + 4y + 5z = 19