Final answer:
The vector and parametric equations for the line passing through the point (0, 13, -9) and parallel to the given line are found using the direction vector (2, -2, 9), resulting in the vector equation r(t) = (2t, 13 - 2t, -9 + 9t) and parametric equations x(t) = 2t, y(t) = 13 - 2t, z(t) = -9 + 9t.
Step-by-step explanation:
To find a vector equation and parametric equations for the line that passes through the point (0, 13, -9) and is parallel to the given line x = -1 + 2t, y = 6 - 2t, z = 3 + 9t, we need to determine the direction vector of the given line, which can be done by looking at the coefficients of t in the equations.
The given line has a direction vector d = (2, -2, 9), and since the new line is parallel to the given line, it will have the same direction vector. We can form the vector equation for the new line using the point it passes through (0, 13, -9) and the direction vector d.
The vector equation is: r(t) = r0 + t * d, where r0 = (0, 13, -9) and d = (2, -2, 9). Substituting the values, we get:
r(t) = (0, 13, -9) + t * (2, -2, 9)
So, the vector equation is:
r(t) = (0 + 2t, 13 - 2t, -9 + 9t)
The parametric equations are:
- x(t) = 0 + 2t
- y(t) = 13 - 2t
- z(t) = -9 + 9t