A) \[y-3z-6=0\]
B) \[y-3z+6=0\]
C) \[y-z-1=0\]
D) \[y-z+1=0\]
Correct Answer: B
Solution :
Equation of plane passing through intersection of given planes is, \[(x+y+z-1)+\lambda \,(2x+3y-z+4)=0\] ?..(i) Plane (i) is parallel to x-axis, then \[\,(1+2\lambda )\,1=0\,\] Þ \[\lambda =-\frac{1}{2}\] Put the value of \[\lambda \] in (i), we get \[y-3z+6=0\], which is the required plane.You need to login to perform this action.
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