Three Dimensional Geometry Question 367
Question: The angle between the pair of planes represented by equation $ 2x^{2}-2y^{2}+4z^{2}+6xz+2yz+3xy=0 $ is
Options:
A) $ {{\cos }^{-1}}( \frac{1}{3} ) $
B) $ {{\cos }^{-1}}( \frac{4}{21} ) $
C) $ {{\cos }^{-1}}( \frac{4}{9} ) $
D) $ {{\cos }^{-1}}( \frac{7}{\sqrt{84}} ) $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ 2x^{2}-2y^{2}+4z^{2}+(6xz+2yz+3xy)=0 $ or $ 2x^{2}+x(6z+3y)-2y^{2}+4z^{2}+2yz=0 $ $ x=\frac{-(6z+3y)\pm \sqrt{36z^{2}+9y^{2}+36yz-8(-2y^{2}+4z^{2}+2yz)}}{4} $ $ x=\frac{-(6z+3y)\pm \sqrt{{{(2z+5y)}^{2}}}}{4} $
$ \Rightarrow x=\frac{-(6z+3y)\pm (2z+5y)}{4} $ or $ 2x-y+2z=0,x+2y+2z=0 $
$ \therefore $ Angle between planes is $ {{\cos }^{-1}}( \frac{4}{9} ). $
BETA
