Question

The plane ax+by =0 is rotated through an angle $\alpha$ about its line of intersection with the plane z=0. Then the equation of the plane in the new position is

• A. $ax-by\pm z\sqrt{a^2+b^2}cot\alpha =0$
• B. $ax+by\pm z\sqrt{a^2+b^2}tan\alpha =0$
• C. $ax-by\pm z\sqrt{a^2+b^2}cot\alpha =0$
• D. $ax+by\pm z\sqrt{a^2+b^2}tan\alpha =0$

Answer 1

The correct option is D

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

Given planes are ax+by=0...........(i)
and z=0 .................(ii)
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:
$\frac{a}{\sqrt{a^2+b^2+k^2}}$,$\frac{b}{\sqrt{a^2+b^2+k^2}}$,$\frac{c}{\sqrt{a^2+b^2+k^2}}$
The 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 $\alpha$,
$\therefore cos\alpha =\frac{a.a+b.b+k.0}{\sqrt{a^2+b^2+k^2}\sqrt{a^2+b^2}}=\sqrt{\frac{a^2+b^2}{a^2+b^2+k^2}}\Rightarrow k^{2}co{s}^2\alpha =a^2(1-co{s}^2\alpha )+b^2(1-co{s}^2\alpha )\Rightarrow k^2=\frac{(a^2+b^2)si{n}^2\alpha }{co{s}^2\alpha }\Rightarrow k=\pm \sqrt{a^2+b^2}tan\alpha ,$
Hence, we get equation of plane as $ax+by\pm z\sqrt{a^2+b^2}tan\alpha =0$