Find the vector and the cartesian equations of the line that passes through the point (3, -2 , - 5), (3, -2, 6) .

Answer

To find the vector and Cartesian equations of the line that passes through the points \( A(3, -2, -5) \) and \( B(3, -2, 6) \), we can follow these steps:

### Step 1: Identify the Points
We have two points:
- Point A: \( (3, -2, -5) \)
- Point B: \( (3, -2, 6) \)

### Step 2: Find the Direction Vector
The direction vector \( \vec{d} \) can be found by subtracting the coordinates of point A from point B:
\[ \vec{d} = B - A = (3 - 3, -2 - (-2), 6 - (-5)) = (0, 0, 11) \]
So, the direction vector \( \vec{d} = (0, 0, 11) \).

### Step 3: Write the Vector Equation of the Line
The vector equation of the line can be expressed as:
\[ \vec{r} = \vec{a} + \lambda \vec{d} \]
where \( \vec{a} \) is the position vector of point A, and \( \lambda \) is a scalar parameter. The position vector of point A is:
\[ \vec{a} = 3\hat{i} - 2\hat{j} - 5\hat{k} \]
Thus, the vector equation becomes:
\[ \vec{r} = (3\hat{i} - 2\hat{j} - 5\hat{k}) + \lambda(0\hat{i} + 0\hat{j} + 11\hat{k}) \]
This simplifies to:
\[ \vec{r} = 3\hat{i} - 2\hat{j} + (-5 + 11\lambda)\hat{k} \]

### Step 4: Write the Cartesian Equation of the Line
To convert the vector equation to Cartesian form, we can express it in terms of \( x, y, z \):
1. The \( x \)-coordinate remains constant at \( 3 \):
\[ x - 3 = 0 \]
2. The \( y \)-coordinate also remains constant at \( -2 \):
\[ y + 2 = 0 \]
3. The \( z \)-coordinate varies with \( \lambda \):
\[ z + 5 = 11\lambda \implies \lambda = \frac{z + 5}{11} \]

Thus, the Cartesian equations of the line can be summarized as:
\[ \frac{x - 3}{0} = \frac{y + 2}{0} = \frac{z + 5}{11} \]

### Final Answer
- **Vector Equation**:
\[ \vec{r} = 3\hat{i} - 2\hat{j} + (-5 + 11\lambda)\hat{k} \]
- **Cartesian Equation**:
\[ \frac{x - 3}{0} = \frac{y + 2}{0} = \frac{z + 5}{11} \]