The coordinates of the foot of the perpendicu | vedclass

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Question

(D) The coordinates of the foot of the perpendicular $(x, y, z)$ from a point $(x_1, y_1, z_1)$ to the plane $ax + by + cz + d = 0$ are given by the f

Vedclass · Accepted Answer

$\left(\frac{1}{5}, \frac{-1}{10}, \frac{1}{2}\right)$

Vedclass · Answer

(D) The coordinates of the foot of the perpendicular $(x, y, z)$ from a point $(x_1, y_1, z_1)$ to the plane $ax + by + cz + d = 0$ are given by the formula: $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = -\frac{ax_1 + by_1 + cz_1 + d}{a^2 + b^2 + c^2}$.Given the plane $2x - y + 5z - 3 = 0$ and the origin $(0, 0, 0)$,we have $x_1 = 0, y_1 = 0, z_1 = 0$.Substituting these values into the formula:$\frac{x - 0}{2} = \frac{y - 0}{-1} = \frac{z - 0}{5} = -\frac{2(0) - 1(0) + 5(0) - 3}{2^2 + (-1)^2 + 5^2}$.$\frac{x}{2} = \frac{y}{-1} = \frac{z}{5} = -\frac{-3}{4 + 1 + 25} = \frac{3}{30} = \frac{1}{10}$.Equating each part to $\frac{1}{10}$:$x = 2 	imes \frac{1}{10} = \frac{1}{5}$.$y = -1 	imes \frac{1}{10} = -\frac{1}{10}$.$z = 5 	imes \frac{1}{10} = \frac{1}{2}$.Thus,the coordinates are $\left(\frac{1}{5}, -\frac{1}{10}, \frac{1}{2}ight)$.

The coordinates of the foot of the perpendicular drawn from the origin to the plane $2x - y + 5z - 3 = 0$ are $ . . . . . . $.

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