The coordinates of the foot of the perpendicu | vedclass

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Question

(A) Let the foot of the perpendicular from the origin $(0, 0, 0)$ to the plane be $P(2, 4, -3)$.Since $OP$ is the normal to the plane,the direction ra

Vedclass · Accepted Answer

$2x + 4y - 3z = 29$

Vedclass · Answer

(A) Let the foot of the perpendicular from the origin $(0, 0, 0)$ to the plane be $P(2, 4, -3)$.Since $OP$ is the normal to the plane,the direction ratios of the normal are $(2 - 0, 4 - 0, -3 - 0) = (2, 4, -3)$.The equation of a plane passing through a point $(x_1, y_1, z_1)$ with normal vector $(a, b, c)$ is given by $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$.Substituting the point $P(2, 4, -3)$ and the normal vector $(2, 4, -3)$,we get:$2(x - 2) + 4(y - 4) - 3(z - (-3)) = 0$$2(x - 2) + 4(y - 4) - 3(z + 3) = 0$$2x - 4 + 4y - 16 - 3z - 9 = 0$$2x + 4y - 3z - 29 = 0$$2x + 4y - 3z = 29$.

The coordinates of the foot of the perpendicular drawn from the origin to a plane is $(2, 4, -3)$. The equation of the plane is

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