Question

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 `"a"x + "b"y +- (sqrt("a"^2 + "b"^2) tan alpha)`z = 0.

Answer 1

Given planes are: 

ax + by = 0   ......(i)

z = 0  ......(ii)

Equation of any plane passing through the line of intersection of plane (i) and (ii) is

(ax + by) + kz = 0

⇒ ax + by + kz = 0   ......(iii)

Dividing both sides by `sqrt("a"^2 + "b"^2 + "k"^2)`, we get

`"a"/sqrt("a"^2 + "b"^2 + "k"^2) x + "b"/sqrt("a"^2 + "b"^2 + "k"^2) y + "k"/sqrt("a"^2 + "b"^2 + "k"^2) z` = 0

∴ Direction cosines of the normal to the plane are

`"a"/sqrt("a"^2 + "b"^2 + "k"^2)`

`"b"/sqrt("a"^2 + "b"^2 + "k"^2)`

`"k"/sqrt("a"^2 + "b"^2 + "k"^2)`

And the direction cosines of the plane (i) are

`"a"/sqrt("a"^2 + "b"^2), "b"/sqrt("a"^2 + "b"^2)`, 0

Since, a is the angle between the planes (i) and (iii), we get

`cos alpha = ("a"."a" + "b"."b" + "k".0)/(sqrt("a"^2 + "b"^2 + "k"^2)*sqrt("a"^2 + "b"^2)`

⇒ `cos alpha = ("a"^2 + "b"^2)/(sqrt("a"^2 + "b"^2 + "k"^2)*sqrt("a"^2 + "b"^2)`

⇒ `cos alpha = sqrt("a"^2 + "b"^2)/sqrt("a"^2 + "b"^2 + "k"^2)`

⇒ `cos^2alpha = ("a"^2 + "b"^2)/sqrt("a"^2 + "b"^2 + "k"^2)`

⇒ (a² + b² + k²) cos²α = a² + b²

⇒ a² cos²α + b² cos²α + k² cos²α = a² + b²

⇒ k² cos²α = a² – a² cos²α + b² – b² cos²α

⇒ k² cos²α = a²(1 – cos²α) + b²(1 – cos²α)

⇒ k² cos²α = a² sin²α + b² sin²α

⇒ k² cos²α = (a² + b²) sin²α

⇒ `"k"^2 = ("a"^2 + "b"^2) (sin^2 alpha)/(cos^2 alpha)`

⇒ k = `+- sqrt("a"^2 + "b"^2) . tan alpha`

Putting the value of k in equation (iii) we get

`"a"x + "b"y +- (sqrt("a"^2 + "b"^2) tan alpha)`z = 0 which is the required equation of plane.

Hence proved.