In the space the equation `by+cz+d=0` represents a plane perpendicular to the plane

Options

A. YOZ
B. ZOX
C. XOY
D. `Z=k`

Explanation

To solve the problem, we need to determine which plane the equation \( by + cz + d = 0 \) is perpendicular to. We will follow these steps:

### Step 1: Identify the normal vector of the given plane
The equation of the plane is given as \( by + cz + d = 0 \). We can rewrite this in the standard form \( Ax + By + Cz + D = 0 \), where \( A = 0 \), \( B = b \), \( C = c \), and \( D = d \). The normal vector \( \vec{n_1} \) of this plane can be represented as:
\[ \vec{n_1} = (0, b, c) \]

### Step 2: Identify the normal vector of the yz-plane
The yz-plane is defined by the equation \( x = 0 \). The normal vector \( \vec{n_2} \) of the yz-plane is:
\[ \vec{n_2} = (1, 0, 0) \]

### Step 3: Calculate the dot product of the normal vectors
To check if the two planes are perpendicular, we need to calculate the dot product of the normal vectors \( \vec{n_1} \) and \( \vec{n_2} \):
\[ \vec{n_1} \cdot \vec{n_2} = (0, b, c) \cdot (1, 0, 0) = 0 \cdot 1 + b \cdot 0 + c \cdot 0 = 0 \]

### Step 4: Conclusion
Since the dot product is equal to 0, this means that the plane represented by the equation \( by + cz + d = 0 \) is perpendicular to the yz-plane.

Thus, the answer is that the plane \( by + cz + d = 0 \) is perpendicular to the yz-plane.