Finding the Equation of the Plane

We are given a point P(-2, 1, 0) and the foot of the perpendicular drawn from P to a plane, which is Q(1, -2, 1). Our goal is to find the equation of the plane.

The line segment PQ is perpendicular to the plane. This is a key property when dealing with the foot of the perpendicular. This means the vector $\vec{PQ}$ is a normal vector to the plane. A normal vector is a vector that is perpendicular to every vector in the plane.

Calculating the Normal Vector

To find the normal vector $\vec{n}$, we calculate the vector $\vec{PQ}$ by subtracting the coordinates of the starting point P from the coordinates of the ending point Q:

$\vec{PQ} = Q - P = (1 - (-2), -2 - 1, 1 - 0)$

$\vec{PQ} = (1 + 2, -3, 1)$

$\vec{PQ} = (3, -3, 1)$

So, the normal vector to the plane is $\vec{n} = (3, -3, 1)$.

Using the Normal Vector to Write the Plane Equation

The general equation of a plane with a normal vector $\vec{n} = (A, B, C)$ is given by $Ax + By + Cz = D$, where D is a constant.

Using our calculated normal vector $\vec{n} = (3, -3, 1)$, the equation of the plane is partially determined as:

$3x - 3y + 1z = D$

or

$3x - 3y + z = D$

Now, we need to find the value of D. We know that the plane passes through the foot of the perpendicular Q(1, -2, 1), because the foot of the perpendicular is a point that lies on the plane itself.

Finding the Constant D

Since the point Q(1, -2, 1) lies on the plane, its coordinates must satisfy the equation of the plane. We substitute x=1, y=-2, and z=1 into the equation $3x - 3y + z = D$:

$3(1) - 3(-2) + (1) = D$

$3 + 6 + 1 = D$

$10 = D$

The Final Equation of the Plane

Substituting the value of D back into the equation, we get the final equation of the plane:

$3x - 3y + z = 10$

This problem is a fundamental application of vectors in 3D geometry to find the equation of the plane.