Find the vector equation of the following planes whose Cartesian equations are `x+2y+3z+5=0`

Answer

To find the vector equation of the plane given by the Cartesian equation \( x + 2y + 3z + 5 = 0 \), we will follow these steps:

### Step 1: Identify the coefficients
The given equation can be rewritten in the standard form \( Ax + By + Cz + D = 0 \). Here, we have:
- \( A = 1 \)
- \( B = 2 \)
- \( C = 3 \)
- \( D = 5 \)

### Step 2: Write the normal vector
The normal vector \( \mathbf{n} \) to the plane can be formed using the coefficients \( A, B, \) and \( C \):
\[ \mathbf{n} = A \mathbf{i} + B \mathbf{j} + C \mathbf{k} = 1 \mathbf{i} + 2 \mathbf{j} + 3 \mathbf{k} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k} \]

### Step 3: Write the vector form of the plane
The vector equation of a plane can be expressed as:
\[ \mathbf{r} \cdot \mathbf{n} = D \]
where \( \mathbf{r} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} \) is the position vector of any point on the plane.

### Step 4: Substitute the values
We need to express \( D \) in terms of the normal vector. Since the original equation is \( x + 2y + 3z + 5 = 0 \), we can rewrite it as:
\[ x + 2y + 3z = -5 \]
Thus, \( D = -5 \).

### Step 5: Write the final vector equation
Substituting the normal vector and \( D \) into the equation gives:
\[ \mathbf{r} \cdot (1 \mathbf{i} + 2 \mathbf{j} + 3 \mathbf{k}) = -5 \]
This can be written as:
\[ \mathbf{r} \cdot \mathbf{n} = -5 \]
or explicitly:
\[ x + 2y + 3z = -5 \]

### Final Vector Equation
The vector equation of the plane is:
\[ \mathbf{r} \cdot (1 \mathbf{i} + 2 \mathbf{j} + 3 \mathbf{k}) = -5 \]