Question

If the plane ax+by=0 is rotated about its line of intersection with the plane z=0 through an angle $\alpha$,then prove that the equation of the plane in its new position is $ax+by\pm \left(\sqrt{a^2+b^2}tan\alpha \right)z=0.$

Answer 1

Equation of the plane is ax+by=0 ...(i)

$\therefore$ Equation of the plane after new position is

$\frac{axcos\alpha }{\sqrt{a^2+b^2}}+\frac{bycos\alpha }{\sqrt{b^2+a^2}}\pm zsin\alpha =0\Rightarrow \frac{ax}{\sqrt{a^2+b^2}}+\frac{by}{\sqrt{b^2+a^2}}\pm zsin\alpha =0\left[ondividingbycos\alpha \right]\Rightarrow ax+by\pm ztan\alpha \sqrt{a^2+b^2}=0\left[onmultiplyingwith\sqrt{a^2+b^2}\right]\text{Alternate Method}Given,planesareax+by=0...\left(i\right)andz=0...\left(ii\right)$

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 (i) are $\frac{a}{\sqrt{a^2+b^2}},\frac{b}{\sqrt{a^2+b^2}},0$

Since,the angle between the planes (i) and (ii) is $\alpha$,

$\therefore cos\alpha =\frac{a.a+b.b+k.0}{\sqrt{a^2+b^2+k^2}\sqrt{a^2+b^2}}=\sqrt{\frac{a^2+b^2}{a^2+b^2+k^2}}\Rightarrow k^{2}co{s}^2\alpha =a^2(1-co{s}^2\alpha )+b^2(1-co{s}^2\alpha )\Rightarrow k^2=\frac{(a^2+b^2)si{n}^2\alpha }{co{s}^2\alpha }k=\pm \sqrt{a^2+b^2}tan\alpha$

On putting this value in plane (iii),we get the equation of the plane as

$ax+by\pm ztan\alpha \sqrt{a^2+b^2}=0$