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(B) The foot of the perpendicular drawn from the origin $O(0, 0, 0)$ to the plane is given as $P(4, -2, -5)$.Since the line segment $OP$ is perpendicu

Vedclass · Accepted Answer

$4x - 2y - 5z = 45$

Vedclass · Answer

(B) The foot of the perpendicular drawn from the origin $O(0, 0, 0)$ to the plane is given as $P(4, -2, -5)$.Since the line segment $OP$ is perpendicular to the plane,the vector $\vec{OP}$ acts as the normal vector $\vec{n}$ to the plane.Thus,the direction ratios of the normal are $(4 - 0, -2 - 0, -5 - 0) = (4, -2, -5)$.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 values,we get $4(x - 4) - 2(y + 2) - 5(z + 5) = 0$.Expanding this,we have $4x - 16 - 2y - 4 - 5z - 25 = 0$.Simplifying,we get $4x - 2y - 5z - 45 = 0$,or $4x - 2y - 5z = 45$.

If the foot of the perpendicular drawn from the point $(0, 0, 0)$ to the plane is $(4, -2, -5)$,then the equation of the plane is $......$

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