If the foot of the perpendicular from `O(0,0,0)` to a plane is `P(1,2,2)`. Then the equation of the plane is

Options

A. `-x+2y+8z-9=0`
B. `x+2y+2z-9=0`
C. `x+y+z-5=0`
D. `x+2y-3z+1=0`

Explanation

To find the equation of the plane given that the foot of the perpendicular from the origin \( O(0,0,0) \) to the plane is \( P(1,2,2) \), we can follow these steps:

### Step 1: Identify the normal vector
The line segment \( OP \) (from the origin \( O \) to the point \( P \)) is perpendicular to the plane. Therefore, the vector \( OP \) can be considered as the normal vector \( \vec{n} \) of the plane.

\[ \vec{n} = P - O = (1 - 0, 2 - 0, 2 - 0) = (1, 2, 2) \]

### Step 2: Write the equation of the plane
The general equation of a plane can be expressed in the form:

\[ \vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n} \]

where \( \vec{r} \) is the position vector of any point on the plane, \( \vec{n} \) is the normal vector, and \( \vec{a} \) is the position vector of a known point on the plane.

Here, we have:
- Normal vector \( \vec{n} = (1, 2, 2) \)
- Point \( P(1, 2, 2) \) gives us \( \vec{a} = (1, 2, 2) \)

### Step 3: Calculate \( \vec{a} \cdot \vec{n} \)
Now we compute the dot product \( \vec{a} \cdot \vec{n} \):

\[ \vec{a} \cdot \vec{n} = (1, 2, 2) \cdot (1, 2, 2) = 1 \cdot 1 + 2 \cdot 2 + 2 \cdot 2 = 1 + 4 + 4 = 9 \]

### Step 4: Substitute into the plane equation
Now we substitute \( \vec{n} \) and \( \vec{a} \cdot \vec{n} \) into the plane equation:

\[ \vec{r} \cdot (1, 2, 2) = 9 \]

### Step 5: Express \( \vec{r} \) in terms of \( x, y, z \)
Let \( \vec{r} = (x, y, z) \). Then the equation becomes:

\[ (x, y, z) \cdot (1, 2, 2) = 9 \]

This expands to:

\[ 1 \cdot x + 2 \cdot y + 2 \cdot z = 9 \]

### Final Equation of the Plane
Thus, the equation of the plane is:

\[ x + 2y + 2z = 9 \]