Question

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

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

Answer 1

Answer: (A), (D)

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

Given:
$p_1:ax+by=0\cdots \left(i\right)$
$p_2:z=0\cdots \left(ii\right)$
Plane passing through the line of intersection of $p_1,p_2$
$p_1+k{p}_2=0$
$ax+by+kz=0\cdots \left(iii\right)$
DC's of normals of the plane $\left(iii\right)$
$(\frac{a}{\sqrt{a^2+b^2+k^2}},\frac{b}{\sqrt{a^2+b^2+k^2}},\frac{k}{\sqrt{a^2+b^2+k^2}})$
DC's of normals of the plane $\left(i\right)$
$(\frac{a}{\sqrt{a^2+b^2}},\frac{b}{\sqrt{a^2+b^2}},0)$
Since angle between plane $\left(i\right),\left(iii\right)$ is $\alpha$
$\cos \alpha =\frac{a\cdot a+b\cdot b+k\cdot 0}{\sqrt{a^2+b^2+k^2}\sqrt{a^2+b^2}}=\sqrt{\frac{a^2+b^2}{a^2+b^2+k^2}}$
Squaring both sides,
$\Rightarrow {\cos }^2\alpha =\frac{a^2+b^2}{a^2+b^2+k^2}$
$\Rightarrow (a^2+b^2){\cos }^2\alpha +k^2{\cos }^2\alpha =a^2+b^2$
$\Rightarrow k^2{\cos }^2\alpha =(a^2+b^2)(1-{\cos }^2\alpha )$
$\Rightarrow k^2=(a^2+b^2){\tan }^2\alpha$
$\Rightarrow k=\pm \sqrt{(a^2+b^2)}\tan \alpha$
put this in equation $\left(iii\right)$
$ax+by\pm z\sqrt{(a^2+b^2)}\tan \alpha =0$