If the foot of the perpendicular drawn from ( | vedclass

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Question

(A) The foot of the perpendicular from the origin $(0,0,0)$ to the plane is given as $P(1,2,3)$.Since the line segment joining the origin $(0,0,0)$ an

Vedclass · Accepted Answer

$x+2y+3z=14$

Vedclass · Answer

(A) The foot of the perpendicular from the origin $(0,0,0)$ to the plane is given as $P(1,2,3)$.Since the line segment joining the origin $(0,0,0)$ and the foot of the perpendicular $(1,2,3)$ is normal to the plane,the direction ratios of the normal to the plane are $(1-0, 2-0, 3-0) = (1, 2, 3)$.The equation of a plane passing through a point $(x_1, y_1, z_1)$ with normal direction ratios $(a, b, c)$ is given by $a(x-x_1) + b(y-y_1) + c(z-z_1) = 0$.Substituting the point $(1, 2, 3)$ and normal vector $(1, 2, 3)$ into the equation:$1(x-1) + 2(y-2) + 3(z-3) = 0$$x - 1 + 2y - 4 + 3z - 9 = 0$$x + 2y + 3z - 14 = 0$$x + 2y + 3z = 14$.

If the foot of the perpendicular drawn from $(0,0,0)$ to a plane is $(1,2,3)$,then the equation of the plane is:

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