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

Vedclass · Accepted Answer

$6$

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(C) The perpendicular distance $d$ of a point $(x_1, y_1, z_1)$ from the plane $Ax + By + Cz + D = 0$ is given by the formula: $d = \left| \frac{Ax_1 + By_1 + Cz_1 + D}{\sqrt{A^2 + B^2 + C^2}} ight|$ Here,the point is the origin $(0, 0, 0)$ and the plane is $2x + y - 2z - 18 = 0$. Substituting the values: $d = \left| \frac{2(0) + 1(0) - 2(0) - 18}{\sqrt{2^2 + 1^2 + (-2)^2}} ight|$ $d = \left| \frac{-18}{\sqrt{4 + 1 + 4}} ight|$ $d = \left| \frac{-18}{\sqrt{9}} ight|$ $d = \left| \frac{-18}{3} ight|$ $d = |-6| = 6 	ext{ units}$

The perpendicular distance of the origin from the plane $2x + y - 2z - 18 = 0$ is (in $\text{ units}$)

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