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(B) The equation of the plane is given by $x + 2y - 2z + 5 = 0$. The perpendicular distance $P$ from a point $(x_1, y_1, z_1)$ to the plane $ax + by +

Vedclass · Accepted Answer

$\frac{5}{3}$

Vedclass · Answer

(B) The equation of the plane is given by $x + 2y - 2z + 5 = 0$. The perpendicular distance $P$ from a point $(x_1, y_1, z_1)$ to the plane $ax + by + cz + d = 0$ is given by the formula: $P = \left| \frac{ax_1 + by_1 + cz_1 + d}{\sqrt{a^2 + b^2 + c^2}} \right|$. Here,the point is the origin $(0, 0, 0)$,so $x_1 = 0, y_1 = 0, z_1 = 0$. The coefficients of the plane are $a = 1, b = 2, c = -2$,and $d = 5$. Substituting these values into the formula: $P = \left| \frac{1(0) + 2(0) - 2(0) + 5}{\sqrt{1^2 + 2^2 + (-2)^2}} \right|$. $P = \left| \frac{5}{\sqrt{1 + 4 + 4}} \right|$. $P = \left| \frac{5}{\sqrt{9}} \right|$. $P = \frac{5}{3}$ units.

The perpendicular distance from the origin to the plane $x + 2y - 2z + 5 = 0$ equals $.........$ units.

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