The coordinates of the foot of the perpendicu | vedclass

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Question

(B) The coordinates of the foot of the perpendicular $(x, y, z)$ from the origin $(0, 0, 0)$ to the plane $ax + by + cz = d$ are given by the formula:

Vedclass · Accepted Answer

$(\frac{18}{7}, \frac{54}{7}, \frac{-27}{7})$

Vedclass · Answer

(B) The coordinates of the foot of the perpendicular $(x, y, z)$ from the origin $(0, 0, 0)$ to the plane $ax + by + cz = d$ are given by the formula: $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = \frac{-(ax_1 + by_1 + cz_1 - d)}{a^2 + b^2 + c^2}$ Here,$(x_1, y_1, z_1) = (0, 0, 0)$,$a = 2$,$b = 6$,$c = -3$,and $d = 63$. Substituting these values: $\frac{x-0}{2} = \frac{y-0}{6} = \frac{z-0}{-3} = \frac{-(2(0) + 6(0) - 3(0) - 63)}{2^2 + 6^2 + (-3)^2}$ $\frac{x}{2} = \frac{y}{6} = \frac{z}{-3} = \frac{-(-63)}{4 + 36 + 9}$ $\frac{x}{2} = \frac{y}{6} = \frac{z}{-3} = \frac{63}{49} = \frac{9}{7}$ Now,solving for $x, y, z$: $x = 2 	imes \frac{9}{7} = \frac{18}{7}$ $y = 6 	imes \frac{9}{7} = \frac{54}{7}$ $z = -3 	imes \frac{9}{7} = \frac{-27}{7}$ Thus,the coordinates are $(\frac{18}{7}, \frac{54}{7}, \frac{-27}{7})$.

The coordinates of the foot of the perpendicular drawn from the origin to the plane $2x + 6y - 3z = 63$ are

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