The equation `ax+by +c=0` represents a plane perpendicular to the

Options

A. xy-plane
B. yz-plane
C. zx-plane
D. none of these

Explanation

To solve the problem of determining which plane is perpendicular to the plane represented by the equation \( ax + by + c = 0 \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Normal Vector of the Given Plane:**
The equation of the plane is given as \( ax + by + c = 0 \). The normal vector \( \mathbf{n_1} \) of this plane can be derived from the coefficients of \( x \) and \( y \). Thus, we have:
\[ \mathbf{n_1} = a \mathbf{i} + b \mathbf{j} + 0 \mathbf{k} \]

2. **Understanding Perpendicular Planes:**
For a plane to be perpendicular to another, the dot product of their normal vectors must be zero. If we denote the normal vector of the required plane as \( \mathbf{n_2} \), then we need:
\[ \mathbf{n_1} \cdot \mathbf{n_2} = 0 \]

3. **Check Each Option:**
Let's evaluate the options provided to find which normal vector \( \mathbf{n_2} \) satisfies the condition \( \mathbf{n_1} \cdot \mathbf{n_2} = 0 \).

- **Option A:** \( \mathbf{n_2} = \mathbf{k} \)
\[ \mathbf{n_1} \cdot \mathbf{n_2} = (a \mathbf{i} + b \mathbf{j} + 0 \mathbf{k}) \cdot (0 \mathbf{i} + 0 \mathbf{j} + 1 \mathbf{k}) = 0 + 0 + 0 = 0 \]
Thus, this option is correct.

- **Option B:** \( \mathbf{n_2} = \mathbf{i} \)
\[ \mathbf{n_1} \cdot \mathbf{n_2} = (a \mathbf{i} + b \mathbf{j} + 0 \mathbf{k}) \cdot (1 \mathbf{i} + 0 \mathbf{j} + 0 \mathbf{k}) = a + 0 + 0 = a \]
Since \( a \neq 0 \), this option is incorrect.

- **Option C:** \( \mathbf{n_2} = \mathbf{j} \)
\[ \mathbf{n_1} \cdot \mathbf{n_2} = (a \mathbf{i} + b \mathbf{j} + 0 \mathbf{k}) \cdot (0 \mathbf{i} + 1 \mathbf{j} + 0 \mathbf{k}) = 0 + b + 0 = b \]
Since \( b \neq 0 \), this option is also incorrect.

4. **Conclusion:**
The only option that satisfies the condition for perpendicularity is Option A, which corresponds to the normal vector \( \mathbf{n_2} = \mathbf{k} \).

### Final Answer:
The equation \( ax + by + c = 0 \) represents a plane perpendicular to the plane with normal vector \( \mathbf{k} \).