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A plane which is perpendicular to two planes 2x-2y+z=0 and x-y+2z=4 passes through ( 1,-2 ,1) . the distsnce of the plane from the point (1,2,2) is

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Solution:   The equation of  the plane through the point (1,-2,1)

       and  perpendicular to planes 

               2x-2y+z=0 hspace0.5cmand hspace0.5cmx-y+2z=4  is given by 

                             eginvmatrix x-1 & y+2 & z-1\ 2& -2 & 1\ 1 &-1 &2 endvmatrix=0Rightarrow hspace0.5cmx+y+1=0

   Its distance from the point    (1,2,2)   is    left | frac1+2+1sqrt2 
ight |=2sqrt2.      

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Deependra Verma

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