The plane ax + by = 0 is rotated about its line of intersection with the plane z = 0 through an angle `alpha`. Prove that the equation of the plane in its new position is ` ax + by +- (sqrt(a^2 + b^2) tan alpha)z = 0`

Explanation

Equation of the plane is `ax+by=0`……i
`:.` Equation of the plane after new position is
`(ax cos alpha)/(sqrt(a^(2)+b^(2)))+(by cos alpha)/(sqrt(b^(2)+a^(2)))+-z sin alpha =0`
`implies (ax)/(sqrt(a^(2)+b^(2)))+(by)/(sqrt(b^(2)+a^(2)))+-z tan alpha =0` [on dividing by `cos alpha`]
`implies ax+by+-z tan alpha sqrt(alpha^(2)+b^(2))=0` [on multiplying with `sqrt(a^(2)+b^(2))`]
Alternate Method
Given planes are `ax+by=0`.................i
and `z=0`.................ii
Therefore the equation of any plane passing through the line of intersection of planes i and ii may be taken as `ax+by+k=0` ....................iii
Then, direction cosines of a normal to the plane iii are `a/(sqrt(a^(2)+b^(2)+k^(2))),b/(sqrt(a^(2)+b^(2)+k^(2))),c/(sqrt(a^(2)+b^(2)+k^(2)))` and direction cosines of the normal to the plane i are `a/(sqrt(a^(2)+b^(2))),b/(sqrt(a^(2)+b^(2))),0`
Since the angle between the plane i and i is `alpha`,
`:. cos alpha=(a.a+b.b+k.0)/(sqrt(a^(2)+b^(2)+k^(2))sqrt(a^(2)+b^(2)))`
`= sqrt((a^(2)+b^(2))/(a^(2)+b^(2)+k^(2)))`
`implies k^(2)cos^(2)alpha=a^(2)(1-cos^(2)alpha)+b^(2)(1-cos^(2)alpha)`
`implies k^(2)=((a^(2)+b^(2)sin^(2)alpha))/(cos^(2) alpha)`
`=k=+-sqrt(a^(2)+b^(2)) tan alpha`
On putting this value in plane iii we get the equation of the planes as
`ax+by+zsqrt(a^(2)+b^(2)) tan alpha=0`