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?

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.