Find the coordinates of the foot of the perpe | vedclass

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Question

(A) Let the coordinates of the foot of the perpendicular $P$ from the origin $(0, 0, 0)$ to the plane $2x - 3y + 4z = 6$ be $(x_1, y_1, z_1)$.The dire

Vedclass · Accepted Answer

$\left(\frac{12}{29}, \frac{-18}{29}, \frac{24}{29}\right)$

Vedclass · Answer

(A) Let the coordinates of the foot of the perpendicular $P$ from the origin $(0, 0, 0)$ to the plane $2x - 3y + 4z = 6$ be $(x_1, y_1, z_1)$.The direction ratios of the normal to the plane are $(2, -3, 4)$.Since the line $OP$ is perpendicular to the plane,its direction ratios are proportional to the direction ratios of the normal. Thus,we can write:$x_1 = 2k, y_1 = -3k, z_1 = 4k$ for some constant $k$.Since the point $P(x_1, y_1, z_1)$ lies on the plane $2x - 3y + 4z = 6$,we substitute the coordinates:$2(2k) - 3(-3k) + 4(4k) = 6$$4k + 9k + 16k = 6$$29k = 6 \implies k = \frac{6}{29}$Substituting $k$ back into the expressions for $x_1, y_1, z_1$:$x_1 = 2 \left(\frac{6}{29}\right) = \frac{12}{29}$$y_1 = -3 \left(\frac{6}{29}\right) = \frac{-18}{29}$$z_1 = 4 \left(\frac{6}{29}\right) = \frac{24}{29}$Thus,the coordinates of the foot of the perpendicular are $\left(\frac{12}{29}, \frac{-18}{29}, \frac{24}{29}\right)$.

Find the coordinates of the foot of the perpendicular drawn from the origin to the plane $2x - 3y + 4z - 6 = 0$.

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