A) 3 unit
B) 1 unit
C) 0
D) None of the above
Correct Answer: D
Solution :
[d] Given planes are \[x-2y+z=1\] (i) And \[-3x+6y-3z=-2\] \[\equiv x-2y+z=\frac{3}{2}\] (ii) Since both planes are parallel and \[a=1,b=-2,c=1\] and \[{{d}_{1}}=-1,{{d}_{2}}=\frac{-2}{3}\] \[\therefore \] Distance \[=\left| \frac{{{d}_{2}}-{{d}_{1}}}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right|\] Distance \[=\left| \frac{1-\frac{2}{3}}{\sqrt{1+4+1}} \right|=\frac{1}{3\sqrt{6}}\]
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