Three Dimensional Geometry Question 384
Question: The plane $ lx+my=0 $ is rotated an angle $ \alpha $ about its line of intersection with the plane $ z=0 $ , then the equation to the plane in its new position is
Options:
A) $ lx+my\pm z\sqrt{(l^{2}+m^{2})}\tan \alpha =0 $
B) $ lx-my\pm z\sqrt{(l^{2}+m^{2})}\tan \alpha =0 $
C) $ lx+my\pm z\sqrt{(l^{2}+m^{2})}\cos \alpha =0 $
D) $ lx-my\pm z\sqrt{(l^{2}+m^{2})}\cos \alpha =0 $
Show Answer
Answer:
Correct Answer: A
Solution:
Equation of any plane through the line of intersection of plane $ lx+my=0 $ and $ z=0 $ is $ lx+my+\lambda z=0 $ . Given that angle between plane is
$ ‘\alpha ’ $ so, angle between their normals is $ ‘\pi -\alpha ’ $ , also direction cosines of their normals are l, m, 0 and $ l,m,\lambda $
$ \tan ,(\pi -\alpha )=\pm \frac{\sqrt{\Sigma (m_1n_2-m_2n_1)}}{\Sigma l_1l_2} $
$ -\tan \alpha =\pm \frac{\lambda }{\sqrt{l^{2}+m^{2}}} $
Required equation is $ lx+my\pm z\sqrt{(l^{2}+m^{2})}\tan \alpha =0 $ .
BETA
