question_answer

Find the equation of the plane through the point\[P(1,\,\,4,\,\,-2)\] and it is parallel to the plane \[-\,2x+y-3z=0.\]

Answer:

Given equation of plane is \[-\,2x+y-3z=0\]           ?(i) Equation of plane parallel to plane (i) is \[-\,2x+y-3z+\lambda =0\]                                 ?(ii) Since, the plane (ii) passes through the point\[P\,(1,\,\,4,\,\,-\,2)\] \[\therefore \]\[-\,2\,\,(1)+4-3\,\,(-\,2)+\lambda =0\] \[\Rightarrow -\,2+4+6+\lambda =0\] \[\Rightarrow \]   \[8+\lambda =0\Rightarrow -\,8\] On putting the value of \[\lambda \] in Eq. (ii), we get \[-\,2x+y-3z-8=0\] \[\Rightarrow \]   \[2x-y+3z+8=0\] which is the required equation of the plane.