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 to the plane in new position is

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

Answer 1

Answer: (D)

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

Given planes are
$ax+by=0$ .... (i)
and $z=0$ ... (ii)
Therefore, 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 DC's of a normal to the plane (iii) are
$\frac{a}{\sqrt{a^2+b^2+K^2}},\frac{b}{\sqrt{a^2+b^2+K^2}},\frac{K}{\sqrt{a^2+b^2+K^2}}$
The DC's 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 $\alpha$.
Thus $\cos \alpha =\frac{a\cdot a+b\cdot b+K\cdot 0}{\sqrt{a^2+b^2+K^2}\sqrt{a^2+b^2}}$
$=\sqrt{\frac{a^2+b^2}{a^2+b^2+K^2}}$
$\Rightarrow K^2{\cos }^2\alpha =a^2(1-{\cos }^2\alpha )+b^2(1-{\cos }^2)+b^2(1-{\cos }^2\alpha )$
$\Rightarrow K^2=\frac{(a^2+b^2){\sin }^2\alpha }{{\cos }^2\alpha }$
$\Rightarrow K=\pm \sqrt{a^2+b^2}\tan \alpha$
On putting the value of $K$ in Eq. (iii), we get the equation of plane as
$ax+by\pm z\sqrt{a^2+b^2}\tan \alpha =0$