If the distance between the planes 4x - 2y - | vedclass

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(C) The given equations of the planes are $4x - 2y - 4z + 1 = 0$ and $4x - 2y - 4z + d = 0$.Since the coefficients of $x, y,$ and $z$ are proportional

Vedclass · Accepted Answer

$-41$ or $43$

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(C) The given equations of the planes are $4x - 2y - 4z + 1 = 0$ and $4x - 2y - 4z + d = 0$.Since the coefficients of $x, y,$ and $z$ are proportional,the planes are parallel.The distance $D$ between two parallel planes $Ax + By + Cz + D_1 = 0$ and $Ax + By + Cz + D_2 = 0$ is given by the formula $D = \frac{|D_2 - D_1|}{\sqrt{A^2 + B^2 + C^2}}$.Here,$A = 4, B = -2, C = -4, D_1 = 1,$ and $D_2 = d$.The distance is given as $7$.So,$7 = \frac{|d - 1|}{\sqrt{4^2 + (-2)^2 + (-4)^2}}$.$7 = \frac{|d - 1|}{\sqrt{16 + 4 + 16}}$.$7 = \frac{|d - 1|}{\sqrt{36}}$.$7 = \frac{|d - 1|}{6}$.$|d - 1| = 42$.This implies $d - 1 = 42$ or $d - 1 = -42$.If $d - 1 = 42$,then $d = 43$.If $d - 1 = -42$,then $d = -41$.Thus,$d = 43$ or $d = -41$.

If the distance between the planes $4x - 2y - 4z + 1 = 0$ and $4x - 2y - 4z + d = 0$ is $7$,then the value of $d$ is:

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