A) \[{{\tan }^{-1}}\left( \frac{2x\sqrt{1-{{x}^{2}}}}{1-2{{x}^{2}}} \right)\]
B) \[\frac{1}{8}\]
C) \[\frac{1}{4}\]
D) \[\frac{1}{2}\]
Correct Answer: A
Solution :
The equation of the plane through the intersection of the planes \[x+y+z=1\] and \[2x+3y-z+4=0\] is \[(x+y+z-1)+\lambda (2z+3y-z+4)=0\]or \[(2\lambda +1)x+(3\lambda +1)y+(1-\lambda )z+4\lambda -1=0\]..(i) It is parallel to X-axis. i.e. \[\frac{x}{1}=\frac{y}{0}=\frac{z}{0}\] \[\therefore \]\[1(2\lambda +1)+0\times (3\lambda +1)+0(1-\lambda )=0\] \[\Rightarrow \]\[\lambda =-\frac{1}{2}\] On substituting \[\lambda =-\frac{1}{2}\] in Eq. (i), we get \[\left( 2\times -\frac{1}{2}+1 \right)x+\left( 3\times \frac{-1}{2}+1 \right)y+\left( 1+\frac{1}{2} \right)\] \[z+4\times \left( 1-\frac{1}{2} \right)-1=0\] \[\Rightarrow \]\[\frac{-1}{2}y+\frac{3}{2}z-3=0\]\[\Rightarrow \]\[y-3z+6=0\] which is the equation of the required plane.
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