A) \[y-3z+6=0\]
B) \[3y-z+6=0\]
C) \[y+3z+6=0\]
D) \[3y-2z+6=0\]
Correct Answer: A
Solution :
The equation of the plane through the intersection of the plane \[x+y+z=1\] and \[2x+3y-z+4=0\] is \[(x+y+z-1)\,+\lambda (2x+3y-z+4)=0\] or \[(1+2\lambda )x+(1+3\lambda )y+(1-\lambda )z+4\lambda -1=0\] Since the plane parallel to x-axis, \[\therefore \] \[1+2\lambda =0\Rightarrow \lambda =-\frac{1}{2}\] Hence, the required equation will be \[y-3z+6=0\].
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