A) \[\frac{1}{2\sqrt{2}}\]
B) \[\frac{1}{\sqrt{2}}\]
C) \[\frac{1}{4\sqrt{2}}\]
D) \[\frac{1}{3\sqrt{2}}\]
Correct Answer: D
Solution :
\[2x2y+3z2+\lambda (xy+z+1)=0\] \[x+2yz3+\mu (3xy+2z1)=0\] \[\Rightarrow \frac{2+\lambda }{1+3\mu }=\frac{-2-\lambda }{2-\mu }=\frac{3+\lambda }{-1+2\mu }=\frac{-2+\lambda }{-3-\mu }\] \[\Rightarrow \frac{1}{1+\mu }=\frac{5}{2+3\mu }\] \[\Rightarrow 5+5\mu =2+3\mu \] \[\Rightarrow \mu =\frac{-3}{2}\] \[2(x+2yz3)3(3xy+2z1)=0\] \[\Rightarrow -7x+7y8z3=0\] Distance \[=\frac{3}{\sqrt{162}}=\frac{3}{9\sqrt{2}}=\frac{1}{3\sqrt{2}}\]You need to login to perform this action.
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