The distance between the two parallel planes | vedclass

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(C) The given equations of the planes are $2x + y + 2z - 8 = 0$ and $4x + 2y + 4z + 5 = 0$. To make the coefficients of $x, y, z$ identical,multiply t

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$7$

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(C) The given equations of the planes are $2x + y + 2z - 8 = 0$ and $4x + 2y + 4z + 5 = 0$. To make the coefficients of $x, y, z$ identical,multiply the first equation by $2$: $4x + 2y + 4z - 16 = 0$ Now,the planes are $4x + 2y + 4z - 16 = 0$ and $4x + 2y + 4z + 5 = 0$. The distance $d$ between two parallel planes $Ax + By + Cz + D_1 = 0$ and $Ax + By + Cz + D_2 = 0$ is given by $d = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}$. Here,$A = 4, B = 2, C = 4, D_1 = -16, D_2 = 5$. $d = \frac{|-16 - 5|}{\sqrt{4^2 + 2^2 + 4^2}} = \frac{|-21|}{\sqrt{16 + 4 + 16}} = \frac{21}{\sqrt{36}} = \frac{21}{6} = \frac{7}{2}$.

The distance between the two parallel planes $2x + y + 2z = 8$ and $4x + 2y + 4z + 5 = 0$ is: (in $/2$)

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