Q. 94.8( 4 Votes )

# Find the equation

Answer :

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

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