Find the vector equation of the line parallel to the line :
`(x -1)/(5) = (3 -y)/(2) = (z + 1)/(4)`

Answer

To find the vector equation of a line that is parallel to the given line, we can follow these steps:

### Step 1: Identify the direction ratios of the given line
The given line is represented in symmetric form as:
\[ \frac{x - 1}{5} = \frac{3 - y}{2} = \frac{z + 1}{4} \]
From this equation, we can identify the direction ratios of the line. The direction ratios are the coefficients of \(x\), \(y\), and \(z\) in the symmetric form.

**Direction ratios**: \(5, -2, 4\)

### Step 2: Write the vector form of the given line
The vector form of a line can be expressed as:
\[ \mathbf{r} = \mathbf{a} + \lambda \mathbf{b} \]
where \(\mathbf{a}\) is a position vector of a point on the line, \(\mathbf{b}\) is the direction vector, and \(\lambda\) is a scalar parameter.

From the symmetric form, we can take a point on the line. When \(x = 1\), \(y = 3\), and \(z = -1\), we have:
\[ \mathbf{a} = \begin{pmatrix} 1 \\ 3 \\ -1 \end{pmatrix} \]
And the direction vector \(\mathbf{b}\) corresponding to the direction ratios is:
\[ \mathbf{b} = \begin{pmatrix} 5 \\ -2 \\ 4 \end{pmatrix} \]

### Step 3: Write the vector equation of the line parallel to the given line
Since we need to find a line parallel to the given line, we can use the same direction vector \(\mathbf{b}\) but with a different point \(\mathbf{a}\). Let's denote the new point as \((\alpha, \beta, \gamma)\).

Thus, the vector equation of the line parallel to the given line is:
\[ \mathbf{r} = \begin{pmatrix} \alpha \\ \beta \\ \gamma \end{pmatrix} + \lambda \begin{pmatrix} 5 \\ -2 \\ 4 \end{pmatrix} \]

### Final Answer
The vector equation of the line parallel to the given line is:
\[ \mathbf{r} = \begin{pmatrix} \alpha \\ \beta \\ \gamma \end{pmatrix} + \lambda \begin{pmatrix} 5 \\ -2 \\ 4 \end{pmatrix} \]

---