Distance between the two planes: 2x + 3y + 4z | vedclass

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(C) The equations of the planes are:$2x + 3y + 4z = 4$  $(1)$$4x + 6y + 8z = 12$Dividing the second equation by $2$,we get:$2x + 3y + 4z = 6$  $(2)$Si

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The equations of the planes are:$2x + 3y + 4z = 4$  $(1)$$4x + 6y + 8z = 12$Dividing the second equation by $2$,we get:$2x + 3y + 4z = 6$  $(2)$Since the coefficients of $x, y,$ and $z$ are the same in both equations,the planes are parallel.The distance $D$ between two parallel planes $ax + by + cz = d_1$ and $ax + by + cz = d_2$ is given by the formula:$D = \left| \frac{d_2 - d_1}{\sqrt{a^2 + b^2 + c^2}} \right|$Here,$a = 2, b = 3, c = 4, d_1 = 4,$ and $d_2 = 6$.Substituting these values into the formula:$D = \left| \frac{6 - 4}{\sqrt{2^2 + 3^2 + 4^2}} \right|$$D = \left| \frac{2}{\sqrt{4 + 9 + 16}} \right|$$D = \frac{2}{\sqrt{29}}$ units.Thus,the correct answer is $C$.

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

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