If (2, -3, 6) is the foot of the perpendicula | vedclass

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Question

(A) Let the plane be $ax + by + cz = d$. The normal vector to the plane is the vector from the origin $(0, 0, 0)$ to the foot of the perpendicular $(2

Vedclass · Accepted Answer

$2x - 3y + 6z - 49 = 0$

Vedclass · Answer

(A) Let the plane be $ax + by + cz = d$. The normal vector to the plane is the vector from the origin $(0, 0, 0)$ to the foot of the perpendicular $(2, -3, 6)$.Thus,the normal vector $\vec{n} = (2, -3, 6)$.The equation of the plane is $2x - 3y + 6z = d$.Since the point $(2, -3, 6)$ lies on the plane,we substitute these coordinates into the equation:$2(2) - 3(-3) + 6(6) = d$$4 + 9 + 36 = d$$d = 49$.Therefore,the equation of the plane is $2x - 3y + 6z = 49$,which can be written as $2x - 3y + 6z - 49 = 0$.

If $(2, -3, 6)$ is the foot of the perpendicular drawn from the origin to a plane,then the equation of that plane is

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