If (2,3,-1) is the foot of the perpendicular | vedclass

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(B) The foot of the perpendicular is $P(2,3,-1)$ and the point from which the perpendicular is drawn is $A(4,2,1)$.Since the line segment $AP$ is perp

Vedclass · Accepted Answer

$2x - y + 2z + 1 = 0$

Vedclass · Answer

(B) The foot of the perpendicular is $P(2,3,-1)$ and the point from which the perpendicular is drawn is $A(4,2,1)$.Since the line segment $AP$ is perpendicular to the plane,the direction ratios of the normal to the plane are the same as the direction ratios of the line $AP$.The direction ratios of the normal are $(4-2, 2-3, 1-(-1)) = (2, -1, 2)$.Thus,the equation of the plane is of the form $2x - y + 2z + d = 0$.Since the plane passes through the point $(2,3,-1)$,we substitute these coordinates into the equation:$2(2) - (3) + 2(-1) + d = 0$$4 - 3 - 2 + d = 0$$-1 + d = 0$$d = 1$.Therefore,the equation of the plane is $2x - y + 2z + 1 = 0$.

If $(2,3,-1)$ is the foot of the perpendicular from $(4,2,1)$ to a plane,then the equation of the plane is

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