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To find the parametric equations of a line that passes through a point \((-1, -2, 2)\) and is parallel to the vector \( \vec{v} = 3\hat{i} - 4\hat{j} + \hat{k} \), we can follow this method:

### Parametric Form of a Line:
The parametric equations of a line through a point \((x_0, y_0, z_0)\) and parallel to a direction vector \((a, b, c)\) are given by:

\[
x(t) = x_0 + at
\]
\[
y(t) = y_0 + bt
\]
\[
z(t) = z_0 + ct
\]

Here:
- The point \((-1, -2, 2)\) gives \(x_0 = -1\), \(y_0 = -2\), and \(z_0 = 2\).
- The direction vector \( \vec{v} = 3\hat{i} - 4\hat{j} + \hat{k} \) gives \(a = 3\), \(b = -4\), and \(c = 1\).

### Parametric Equations:
Using these values, the parametric equations for the line are:

\[
x(t) = -1 + 3t
\]
\[
y(t) = -2 - 4t
\]
\[
z(t) = 2 + t
\]

### Final Answer:
\[
x(t) = -1 + 3t
\]
\[
y(t) = -2 - 4t
\]
\[
z(t) = 2 + t
\]

Would you like further details or have any questions?

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Here are 5 questions to explore this topic further:
1. What are the general steps to find parametric equations for a line given a point and a direction vector?
2. How do we derive a vector from two points on a line?
3. Can you explain how to convert parametric equations to symmetric form?
4. How does changing the direction vector affect the parametric equations of a line?
5. What is the difference between a parametric equation and a vector equation of a line?

**Tip:** Always check if the direction vector is a scalar multiple of another vector to ensure lines are truly parallel or perpendicular.

Find parametric equations of the line through (-1, -2, 2) that is parallel to 3i - 4j + k.

This problem involves finding the parametric equations of a line that passes through the point (-1, -2, 2) and is parallel to the vector 3i - 4j + k. Explore the method to solve this, including the formulas for parametric equations in 3D space.

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