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प्रश्न
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 ax + by `+- (sqrt(a^2 + b^2) tan alpha)z ` = 0
उत्तर
Given planes are: ax + by = 0 ......(i) and z = 0 ......(ii)
Now, the equation of any plan 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
So, direction consines 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`
As, α is the angle between the planes (i) and (iii), we get
⇒ cos α = `(a.a + b.b k.0)/(sqrt(a^2 + b^2 + k^2)* sqrt(a^2 + b^2))`
cos α = `(a^2 + b^2)/(sqrt(a^2 + b^2 + k^2) * sqrt(a^2 + b^2))`
cos α = `(a^2 + b^2)/sqrt(a^2 + b^2 + k^2)`
cos2α = `(a^2 + b^2)/sqrt(a^2 + b^2 + k^2)`
`(a^2 + b^2 + k^2) cos^2alpha = a^2 + b^2`
`a^2 cos^2 alpha + b^2 cos^2alpha + k^2 cos^2 alpha = a^2 + b^2`
`k^2 cos^2 alpha = a^2 - a^2 cos^2 alpha 6 b^2 - b^2 cos^2alpha`
`k^2 cos^2 alpha = alpha^2(1 - cos^2alpha)(1 - cos^2alpha)`
`k^2cos^2alpha = a^2 sin^2alpha + b^2 sin^2alpha`
`k^2 cos^2alpha = (a^2 + b^2) sin^2alpha`
⇒ = `k^2 = (a^2 + b^2) (sin^2alpha)/(cos^2alpha)`
⇒ `k = +- sqrt(a^2 + b^2) * tan alpha`
Putting the value of k in equation (iii) we get
`ax + by +- (sqrt(a^2 + b^2) * tan alpha)z` = 0 which is the required equation of plane.
Hence proved.
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