A) \[0\]
B) \[1\]
C) \[\sqrt{2}\]
D) \[2\sqrt{2}\]
Correct Answer: D
Solution :
The equation of a plane passing through\[(1,\,\,-2,\,\,1)\]is \[a(x-1)+b(y+2)+c(z-1)=0\] ... (i) It is perpendicular to \[2x-2y+z=0\] and\[x-y+2z=4\] \[\therefore \] \[2a-2b+c=0\]and\[a-b+2c=0\] \[\Rightarrow \] \[\frac{a}{-4+1}=\frac{-b}{4-1}=\frac{c}{-2+2}\] \[\Rightarrow \] \[\frac{a}{-3}=\frac{b}{-3}=\frac{c}{0}\] On substuting the values of \[a,\,\,b\] and \[c\] in Eq. (i), we get \[-3(x-1)-3(y+2)+0(z-1)=0\] \[\Rightarrow \] \[x+y+1=0\] Distance of this plane from\[(1,\,\,2,\,\,2)\] is given by \[d=\frac{1+2+1}{\sqrt{1+1}}=2\sqrt{2}\]You need to login to perform this action.
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