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

Explanation

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+kz=0" "`(iii)
The 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))) and (k)/(sqrt(a^(2)+b^(2)+k^(2)))`
The direction cosines of a normal to the plane (i) are
`" "(a)/(sqrt(a^(2)+b^(2))),(b)/(a^(2)+b^(2))) and 0`
Since the angle between the planes (i) and (ii) is `alpha`, we have
`" "cosalpha=(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)))`
or `" "k^(2)cos^(2)alpha=a^(2)(1-cos^(2)alpha)+b^(2)(1-cos^(2)alpha)`
or `" "k^(2)=((a^(2)+b^(2))sin^(2)alpha)/(cos^(2)alpha)or k=pmsqrt(a^(2)+b^(2))tanalpha`
Putting this in (iii), we get the equation of the plane as `ax+bypmzsqrt(a^(2)+b^(2))tanalpha=0`