The distance between two parallel planes ax+b | vedclass

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(C) The given plane is $2x-y+2z+3=0$ $(1)$.The second plane is $4x-2y+4z+\lambda=0$,which can be written as $2x-y+2z+\frac{\lambda}{2}=0$ $(2)$.The di

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$13$

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(C) The given plane is $2x-y+2z+3=0$ $(1)$.The second plane is $4x-2y+4z+\lambda=0$,which can be written as $2x-y+2z+\frac{\lambda}{2}=0$ $(2)$.The distance between $(1)$ and $(2)$ is $\frac{|\frac{\lambda}{2}-3|}{\sqrt{2^2+(-1)^2+2^2}} = \frac{1}{3}$.$\frac{|\frac{\lambda}{2}-3|}{3} = \frac{1}{3} \Rightarrow |\frac{\lambda}{2}-3| = 1$.This gives $\frac{\lambda}{2}-3 = 1$ or $\frac{\lambda}{2}-3 = -1$.So,$\lambda = 8$ or $\lambda = 4$.The third plane is $2x-y+2z+\mu=0$ $(3)$.The distance between $(1)$ and $(3)$ is $\frac{|\mu-3|}{\sqrt{2^2+(-1)^2+2^2}} = \frac{2}{3}$.$\frac{|\mu-3|}{3} = \frac{2}{3} \Rightarrow |\mu-3| = 2$.This gives $\mu-3 = 2$ or $\mu-3 = -2$.So,$\mu = 5$ or $\mu = 1$.To find the maximum value of $\lambda+\mu$,we take $\lambda=8$ and $\mu=5$.Thus,$\lambda+\mu = 8+5 = 13$.

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