If the perpendicular distance from (1, 2, 4) | vedclass

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(B) The formula for the perpendicular distance $d$ from a point $(x_1, y_1, z_1)$ to a plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|Ax_1 + By_

Vedclass · Accepted Answer

$7$

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(B) The formula for the perpendicular distance $d$ from a point $(x_1, y_1, z_1)$ to a plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$.Given the point $(1, 2, 4)$ and the plane $2x + 2y - z + k = 0$,we have $A = 2, B = 2, C = -1, D = k$.The distance is $3 = \frac{|2(1) + 2(2) - 1(4) + k|}{\sqrt{2^2 + 2^2 + (-1)^2}}$.$3 = \frac{|2 + 4 - 4 + k|}{\sqrt{4 + 4 + 1}}$.$3 = \frac{|2 + k|}{\sqrt{9}}$.$3 = \frac{|2 + k|}{3}$.$|2 + k| = 9$.This implies $2 + k = 9$ or $2 + k = -9$.So,$k = 7$ or $k = -11$.Comparing with the given options,$k = 7$ is the correct value.

If the perpendicular distance from $(1, 2, 4)$ to the plane $2x + 2y - z + k = 0$ is $3$,then $k =$

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