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 . . .

Options

A. 4x+2y+5z=-13
B. 4x-2y-5z=45
C. 4x+2y-5z=37
D. 4x-2y+5z=-5

Explanation

To find the equation of the plane given that the foot of the perpendicular from the point (0, 0, 0) to the plane is (4, -2, -5), we can follow these steps:

### Step 1: Identify the foot of the perpendicular
The foot of the perpendicular from the point (0, 0, 0) to the plane is given as (4, -2, -5). This point lies on the plane.

### Step 2: Determine the normal vector
The normal vector to the plane can be derived from the coordinates of the foot of the perpendicular. The direction ratios of the normal vector (A, B, C) can be taken as the coordinates of the foot of the perpendicular, which are (4, -2, -5). Therefore, we have:
- A = 4
- B = -2
- C = -5

### Step 3: Use the point-normal form of the equation of the plane
The equation of a plane in point-normal form is given by:
\[ A(x - x_1) + B(y - y_1) + C(z - z_1) = 0 \]
where (x1, y1, z1) is a point on the plane, and (A, B, C) are the direction ratios of the normal vector.

Substituting the values:
- (x1, y1, z1) = (4, -2, -5)
- (A, B, C) = (4, -2, -5)

The equation becomes:
\[ 4(x - 4) - 2(y + 2) - 5(z + 5) = 0 \]

### Step 4: Expand the equation
Now, we expand the equation:
\[ 4x - 16 - 2y - 4 - 5z - 25 = 0 \]

### Step 5: Combine like terms
Combining the constant terms gives:
\[ 4x - 2y - 5z - 45 = 0 \]

### Step 6: Rearranging the equation
Rearranging the equation, we can write it as:
\[ 4x - 2y - 5z = 45 \]

Thus, the equation of the plane is:
\[ 4x - 2y - 5z = 45 \]