The Cartesian equation of a line is 2x-3=3y+1 =5-6z. Find the vector equation of a line passing through (7,-5,0) and the parallel to the given line.

Answer

To find the vector equation of the line passing through the point (7, -5, 0) and parallel to the given line defined by the Cartesian equation \(2x - 3 = 3y + 1 = 5 - 6z\), we will follow these steps:

### Step 1: Identify the direction ratios of the given line
The given equation can be expressed in the form:
\[ \frac{2x - 3}{2} = \frac{3y + 1}{3} = \frac{5 - 6z}{-6} \]
From this, we can identify the direction ratios of the line as:
- For \(x\): The coefficient is \(2\)
- For \(y\): The coefficient is \(3\)
- For \(z\): The coefficient is \(-6\)

Thus, the direction ratios of the line are \(2, 3, -6\).

### Step 2: Write the direction vector
The direction vector \(\mathbf{b}\) corresponding to the direction ratios is:
\[ \mathbf{b} = \langle 2, 3, -6 \rangle \]

### Step 3: Write the position vector of the point
The position vector \(\mathbf{a}\) of the point (7, -5, 0) is:
\[ \mathbf{a} = 7\mathbf{i} - 5\mathbf{j} + 0\mathbf{k} = 7\mathbf{i} - 5\mathbf{j} \]

### Step 4: Write the vector equation of the line
The vector equation of a line can be expressed as:
\[ \mathbf{r} = \mathbf{a} + \lambda \mathbf{b} \]
Substituting the values we have:
\[ \mathbf{r} = (7\mathbf{i} - 5\mathbf{j}) + \lambda (2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}) \]

### Step 5: Simplify the vector equation
This can be simplified to:
\[ \mathbf{r} = (7 + 2\lambda)\mathbf{i} + (-5 + 3\lambda)\mathbf{j} + (-6\lambda)\mathbf{k} \]

### Final Vector Equation
Thus, the vector equation of the line passing through the point (7, -5, 0) and parallel to the given line is:
\[ \mathbf{r} = (7 + 2\lambda)\mathbf{i} + (-5 + 3\lambda)\mathbf{j} + (-6\lambda)\mathbf{k} \]