The distance between the two planes 2x + 3y + | vedclass

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(C) The equations of the planes are:$2x + 3y + 4z = 4$  ....$(i)$$4x + 6y + 8z = 12$Dividing the second equation by $2$,we get:$2x + 3y + 4z = 6$  ...

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$\frac{2}{\sqrt{29}}$ units

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(C) The equations of the planes are:$2x + 3y + 4z = 4$  ....$(i)$$4x + 6y + 8z = 12$Dividing the second equation by $2$,we get:$2x + 3y + 4z = 6$  ....$(ii)$Since the normal vectors $(2, 3, 4)$ are the same,the planes are parallel.The distance $D$ between two parallel planes $ax + by + cz = d_1$ and $ax + by + cz = d_2$ is given by the formula:$D = \left| \frac{d_2 - d_1}{\sqrt{a^2 + b^2 + c^2}} \right|$Substituting the values $a = 2, b = 3, c = 4, d_1 = 4$,and $d_2 = 6$:$D = \left| \frac{6 - 4}{\sqrt{2^2 + 3^2 + 4^2}} \right| = \left| \frac{2}{\sqrt{4 + 9 + 16}} \right| = \frac{2}{\sqrt{29}}$ units.

The distance between the two planes $2x + 3y + 4z = 4$ and $4x + 6y + 8z = 12$ is

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