The foot of the perpendicular drawn from the origin to a plane is (2, -3, -4). Find the equation of the plane.

Answer

To find the equation of the plane given that the foot of the perpendicular drawn from the origin to the plane is (2, -3, -4), we can follow these steps:

### Step 1: Identify the point and the normal vector
The foot of the perpendicular from the origin (0, 0, 0) to the plane is given as the point \( P(2, -3, -4) \). This point lies on the plane. The vector from the origin to this point is the position vector \( \vec{OP} = 2\hat{i} - 3\hat{j} - 4\hat{k} \).

### Step 2: Determine the normal vector
The normal vector \( \vec{n} \) to the plane can be taken as the vector \( \vec{OP} \) itself, which is:
\[ \vec{n} = \langle 2, -3, -4 \rangle \]

### Step 3: Use the point-normal form of the plane equation
The equation of a plane in point-normal form can be expressed as:
\[ \vec{n} \cdot (\vec{r} - \vec{r_0}) = 0 \]
where \( \vec{r} = \langle x, y, z \rangle \) is the position vector of any point on the plane, and \( \vec{r_0} = \langle 2, -3, -4 \rangle \) is the position vector of the point on the plane.

### Step 4: Substitute the values
Substituting the normal vector and the point into the equation:
\[ \langle 2, -3, -4 \rangle \cdot \langle x - 2, y + 3, z + 4 \rangle = 0 \]

### Step 5: Expand the dot product
Calculating the dot product:
\[ 2(x - 2) - 3(y + 3) - 4(z + 4) = 0 \]

### Step 6: Simplify the equation
Expanding this gives:
\[ 2x - 4 - 3y - 9 - 4z - 16 = 0 \]
Combining like terms:
\[ 2x - 3y - 4z - 29 = 0 \]

### Step 7: Rearranging to standard form
Rearranging this equation gives us:
\[ 2x + 3y + 4z = 29 \]

### Final Equation of the Plane
Thus, the equation of the plane is:
\[ 2x + 3y + 4z = 29 \]