To solve the problem, we will determine both the vector equation and the Cartesian equation of the line passing through the points $(-2, 0, 3)$ and $(3, 5, -2)$.

1. Vector Equation of the Line

The general form of the vector equation of a line is:

$\mathbf{r} = \mathbf{a} + t \mathbf{b}$

Where:

• $\mathbf{a}$ is a position vector of a point on the line (we can use $(-2, 0, 3)$),
• $\mathbf{b}$ is the direction vector of the line, which is determined by the difference between the two points, i.e., $\mathbf{b} = (3 - (-2), 5 - 0, -2 - 3)$.

Calculate $\mathbf{b}$:

$\mathbf{b} = (5, 5, -5)$

Thus, the vector equation of the line is: $\mathbf{r} = (-2, 0, 3) + t (5, 5, -5), \quad t \in \mathbb{R}$

In expanded form: $\mathbf{r} = (-2 + 5t, 5t, 3 - 5t)$

2. Cartesian Equation of the Line

To find the Cartesian equation, we eliminate the parameter $t$ from the vector equation. From the parametric equations: $x = -2 + 5t, \quad y = 5t, \quad z = 3 - 5t$

Solving for $t$ in terms of $x, y, z$:

• From $x = -2 + 5t$: $t = \frac{x + 2}{5}$,
• From $y = 5t$: $t = \frac{y}{5}$,
• From $z = 3 - 5t$: $t = \frac{3 - z}{5}$.

Equating these expressions for $t$: $\frac{x + 2}{5} = \frac{y}{5} = \frac{3 - z}{5}$

Remove the denominator to simplify: $x + 2 = y = 3 - z$

Rearrange to write the Cartesian equations: $x + 2 = y, \quad y + z = 3$

Final Answer:

1. Vector Equation: $\mathbf{r} = (-2, 0, 3) + t (5, 5, -5), \quad t \in \mathbb{R}$

2. Cartesian Equations: $x + 2 = y, \quad y + z = 3$

Would you like further details or explanations on any part of the solution? Here are some related questions:

1. How is the direction vector calculated from two points?
2. What does the parameter $t$ represent in the vector equation?
3. How can we verify that the Cartesian and vector equations describe the same line?
4. What are some real-world applications of these equations in 3D geometry?
5. How do you convert between vector and Cartesian forms systematically?

Tip: Always cross-check your direction vector to ensure it is accurate, as it forms the backbone of all line equations.