The distance of the plane 2x - y - 2z - 9 = 0 | vedclass

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(A) The distance of a plane $Ax + By + Cz + D = 0$ from a point $(x_1, y_1, z_1)$ is given by the formula: $d = \left| \frac{Ax_1 + By_1 + Cz_1 + D}{\

Vedclass · Accepted Answer

$3$

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

The distance of the plane $2x - y - 2z - 9 = 0$ from the origin is $d$ units.

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