Q. 84.1( 7 Votes )

# 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.

Answer :

We know that solution of a plane passing through (x_{1},y_{1},z_{1}) is given as -

a(x – x_{1}) + b(y – y_{1}) + c(z – z_{1}) = 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 a_{1}x + b_{1}y + c_{1}z + d_{1} = 0 and a_{2}x + b_{2}y + c_{2}z + d_{2} = 0 are at right angles if,

a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 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

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