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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