Are the planes x+y-2z+4=0 and 3x+3y-6z+5=0 intersecting

Answer

To determine whether the planes given by the equations \( x + y - 2z + 4 = 0 \) and \( 3x + 3y - 6z + 5 = 0 \) are intersecting, we can follow these steps:

### Step 1: Identify the normal vectors of the planes

The general equation of a plane can be expressed as \( Ax + By + Cz + D = 0 \), where the vector \( \vec{n} = (A, B, C) \) is the normal vector of the plane.

For the first plane \( x + y - 2z + 4 = 0 \):
- The coefficients are \( A = 1, B = 1, C = -2 \).
- Thus, the normal vector \( \vec{n_1} = (1, 1, -2) \).

For the second plane \( 3x + 3y - 6z + 5 = 0 \):
- The coefficients are \( A = 3, B = 3, C = -6 \).
- Thus, the normal vector \( \vec{n_2} = (3, 3, -6) \).

### Step 2: Check if the normal vectors are parallel

Two vectors \( \vec{n_1} \) and \( \vec{n_2} \) are parallel if there exists a scalar \( k \) such that:
\[ \vec{n_1} = k \cdot \vec{n_2} \]

We can check if:
\[ (1, 1, -2) = k \cdot (3, 3, -6) \]

This gives us the following equations:
1. \( 1 = 3k \)
2. \( 1 = 3k \)
3. \( -2 = -6k \)

From the first equation:
\[ k = \frac{1}{3} \]

From the third equation:
\[ -2 = -6k \implies k = \frac{1}{3} \]

Since both equations yield the same value for \( k \), we conclude that the normal vectors \( \vec{n_1} \) and \( \vec{n_2} \) are parallel.

### Step 3: Conclusion about the planes

Since the normal vectors are parallel, it implies that the planes themselves are parallel. Therefore, they will never intersect.

### Final Answer

No, the planes are parallel and do not intersect.

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