The plane ax + by = 0 is rotated about its line of intersection with the plane z = 0 through an angle α. Prove that the equation of the plane in its new position is z = 0.

Given: The plane ax + by = 0 is rotated about its line of intersection with the plane z = 0 through an angle α.

To Prove:

Equation of the plane in its new position is given by

Proof: We have two planes given to us,

ax + by = 0 …(i)

z = 0 …(ii)

We know that,

Equation of plane passing through the line of intersection of planes (i) and (ii) is given by

ax + by + λz = 0 …(iii)

where, λ ∈ ℝ

The angle between the new plane and plane (i) is given to be α.

Since, angle between two planes is equal to the angle between their normal.

So,

Direction ratio of normal to ax + by = 0 or ax + by + 0z = 0 is (a, b, 0).

And,

Direction ratio of normal to ax + by + λz = 0 is (a, b, λ).

Also, we know that

Angle between normal vectors of two planes and is given as,

Substituting values of these vectors, we get

Multiplying √(a² + b²) by numerator and denominator on R.H.S.,

Squaring on both sides,

⇒ (a² + b² + λ²) cos² α = a² + b²

⇒ a² cos² α + b² cos² α + λ² cos² α = a² + b²

⇒ λ² cos² α = a² + b² – a² cos² α – b² cos² α

⇒ λ² cos² α = a² – a² cos² α + b² – b² cos² α

⇒ λ² cos² α = a²(1 – cos² α) + b²(1 – cos² α)

⇒ λ² cos² α = (a² + b²)(1 – cos² α)

⇒ λ² cos² α = (a² + b²) sin² α [∵, sin² α + cos² α = 1]

Since, .

⇒ λ² = (a² + b²) tan² α

Substitute the value of λ in equation (iii) to find the equation of the plane,

ax + by + λz = 0

Hence, proved.