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(A) Let the coordinates of the foot of the perpendicular $P$ from the origin $(0, 0, 0)$ to the plane be $(x_1, y_1, z_1)$.The equation of the plane i

Vedclass · Accepted Answer

$\left(\frac{24}{29}, \frac{36}{29}, \frac{48}{29}\right)$

Vedclass · Answer

(A) Let the coordinates of the foot of the perpendicular $P$ from the origin $(0, 0, 0)$ to the plane be $(x_1, y_1, z_1)$.The equation of the plane is $2x + 3y + 4z = 12$  --- $(1)$.The direction ratios of the normal to the plane are $(2, 3, 4)$.The magnitude of the normal vector is $\sqrt{2^2 + 3^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29}$.Dividing equation $(1)$ by $\sqrt{29}$,we get the normal form of the plane equation:$\frac{2}{\sqrt{29}}x + \frac{3}{\sqrt{29}}y + \frac{4}{\sqrt{29}}z = \frac{12}{\sqrt{29}}$.Comparing this with the standard normal form $lx + my + nz = d$,where $(l, m, n)$ are direction cosines and $d$ is the perpendicular distance from the origin,we have:$l = \frac{2}{\sqrt{29}}, m = \frac{3}{\sqrt{29}}, n = \frac{4}{\sqrt{29}}$ and $d = \frac{12}{\sqrt{29}}$.The coordinates of the foot of the perpendicular are given by $(ld, md, nd)$:$x_1 = \left(\frac{2}{\sqrt{29}}ight) 	imes \left(\frac{12}{\sqrt{29}}ight) = \frac{24}{29}$$y_1 = \left(\frac{3}{\sqrt{29}}ight) 	imes \left(\frac{12}{\sqrt{29}}ight) = \frac{36}{29}$$z_1 = \left(\frac{4}{\sqrt{29}}ight) 	imes \left(\frac{12}{\sqrt{29}}ight) = \frac{48}{29}$Thus,the coordinates are $\left(\frac{24}{29}, \frac{36}{29}, \frac{48}{29}ight)$.

Find the coordinates of the foot of the perpendicular drawn from the origin to the plane $2x + 3y + 4z - 12 = 0$.

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