Find both the vector equation and the parametric equations of the line through $$ (-1,1,8) $$ and $$ (2,-4,0) $$, where $$ t=0 $$ corresponds to the first given point. The vector equation is $$ \langle x, y, z\rangle=\langle $$ of 1

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To find the **vector equation** and the **parametric equations** of the line passing through the points $$ (-1, 1, 8) $$ and $$ (2, -4, 0) $$, with $$ t = 0 $$ corresponding to the first point $$ (-1, 1, 8) $$, follow these steps:

### 1. **Determine the Direction Vector**

The direction vector **$$ \mathbf{v} $$** of the line can be found by subtracting the coordinates of the initial point from the terminal point:

$$
\mathbf{v} = (2 - (-1),\; -4 - 1,\; 0 - 8) = (3, -5, -8)
$$

### 2. **Vector Equation of the Line**

The vector equation of a line in three-dimensional space is given by:

$$
\langle x, y, z \rangle = \langle x_0, y_0, z_0 \rangle + t \langle a, b, c \rangle
$$

Where:
- $$ \langle x_0, y_0, z_0 \rangle $$ is a point on the line (here, $$ (-1, 1, 8) $$),
- $$ \langle a, b, c \rangle $$ is the direction vector $$ \mathbf{v} $$,
- $$ t $$ is a parameter.

Substituting the known values:

$$
\boxed{ \langle x, y, z \rangle = \langle -1,\; 1,\; 8 \rangle + t \langle 3,\; -5,\; -8 \rangle }
$$

### 3. **Parametric Equations of the Line**

The vector equation can be broken down into parametric form by equating each component:

$$
\begin{cases}
x = x_0 + at \
y = y_0 + bt \
z = z_0 + ct \
\end{cases}
$$

Using the given point and direction vector:

$$
\begin{cases}
x = -1 + 3t \
y = 1 - 5t \
z = 8 - 8t \
\end{cases}
$$

So, the **parametric equations** are:

$$
\boxed{
\begin{aligned}
x &= -1 + 3t \
y &= 1 - 5t \
z &= 8 - 8t \
\end{aligned}
}
$$

### **Summary**

- **Vector Equation:**
  $$
  \langle x, y, z \rangle = \langle -1,\; 1,\; 8 \rangle + t \langle 3,\; -5,\; -8 \rangle
  $$
  
- **Parametric Equations:**
  $$
  \begin{aligned}
  x &= -1 + 3t \
  y &= 1 - 5t \
  z &= 8 - 8t \
  \end{aligned}
  $$

Here, $$ t = 0 $$ corresponds to the point $$ (-1, 1, 8) $$, and varying $$ t $$ moves along the line in the direction of the vector $$ \langle 3, -5, -8 \rangle $$.
The vector equation of the line is $$ \langle x, y, z \rangle = \langle -1, 1, 8 \rangle + t \langle 3, -5, -8 \rangle $$. The parametric equations are: $$ \begin{cases} x = -1 + 3t \ y = 1 - 5t \ z = 8 - 8t \ \end{cases} $$ where $$ t = 0 $$ corresponds to the point $$ (-1, 1, 8) $$.

Find both the vector equation and the parametric equations of the line through \( (-1,1,8) \) and \( (2,-4,0) \), where \( t=0 \) corresponds to the first given point. The vector equation is \( \langle x, y, z\rangle=\langle \) of 1

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