Q. 74.7( 3 Votes )

# Find the equation of the plane passing through the origin and perpendicular to each of the planes x + 2y – z = 1 and 3x – 4y + z = 5.

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 (0,0,0), so the equation of plane is

a(x – 0) + b(y – 0) + c(z – 0) = 0

⇒ ax + by + cz = 0 …… (1)

Now, the required plane is also perpendicular to the planes,

x + 2y – z = 1 and 3x – 4y + z = 5

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 + 2b – c = 0 …… (b)

3a – 4b + c = 0 …… (c)

Solving (b) and (c) we get,

∴a = – 2λ, b = – 4λ, c = – 10λ

Putting values of a,b,c in equation (1) we get,

(– 2λ)x + (– 4λ)y + (– 10λ)z = 0

Dividing both sides by (– 2λ) we get

x + 2y + 5z = 0

So, the equation of the required planes is x + 2y + 5z = 0

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