To determine which of the given lines is parallel to the line described by the equations:

$x = 4t - 2, \quad y = 8t - 9, \quad z = -4t + 13$

We first need to extract the direction vector of this line. A parametric form of the line is given by:

$\mathbf{r}(t) = (x(t), y(t), z(t)) = (4t - 2, 8t - 9, -4t + 13)$

The direction vector for this line is the vector formed by the coefficients of $t$ in each equation. So, the direction vector is:

$\mathbf{v} = (4, 8, -4)$

Now, we analyze each of the given lines:

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a) $x = 4t - 6, \, y = 8t + 2, \, z = t$

• The direction vector for this line is obtained from the coefficients of $t$: $\mathbf{v_a} = (4, 8, 1)$
• Since this direction vector is not proportional to $(4, 8, -4)$, the lines are not parallel.

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b) $x - 2 = \frac{y + 1}{2} = -z - 1$

This is a symmetric form. To convert it to parametric form, we solve for $x$, $y$, and $z$:

• Let $\lambda$ be the common parameter: $x = 2 + \lambda, \quad y = -1 + 2\lambda, \quad z = -1 - \lambda$ So, the parametric equations are: $x = 2 + \lambda, \quad y = -1 + 2\lambda, \quad z = -1 - \lambda$ The direction vector for this line is: $\mathbf{v_b} = (1, 2, -1)$
• This direction vector is not proportional to $(4, 8, -4)$, so the lines are not parallel.

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c) $\frac{x-4}{-2} = \frac{y-8}{-9} = \frac{z-4}{13}$

This is another symmetric form. To convert it to parametric form, we set each fraction equal to a parameter $\lambda$:

• $\frac{x - 4}{-2} = \lambda$, so $x = -2\lambda + 4$
• $\frac{y - 8}{-9} = \lambda$, so $y = -9\lambda + 8$
• $\frac{z - 4}{13} = \lambda$, so $z = 13\lambda + 4$

Thus, the parametric equations are: $x = -2\lambda + 4, \quad y = -9\lambda + 8, \quad z = 13\lambda + 4$

The direction vector for this line is: $\mathbf{v_c} = (-2, -9, 13)$

Since the direction vector $(-2, -9, 13)$ is proportional to $(4, 8, -4)$ (specifically, it is $-\frac{1}{2}$ times $(4, 8, -4)$), these lines are parallel.

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d) $x = -4t + 1, \, y = -8t, \, z = 4t + 4$

• The direction vector for this line is obtained from the coefficients of $t$: $\mathbf{v_d} = (-4, -8, 4)$
• Since this direction vector is proportional to $(4, 8, -4)$ (specifically, it is $-1$ times $(4, 8, -4)$), the lines are parallel.

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Conclusion:

The lines that are parallel to the given line are c and d.

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Would you like more details on the calculations, or do you have any questions about this? Here are some related questions:

1. How can you determine if two vectors are parallel?
2. What is the process for converting symmetric equations to parametric form?
3. What does it mean for two lines to be proportional in terms of direction vectors?
4. How do you find the direction vector from parametric equations?
5. How would you solve problems involving three-dimensional lines in vector form?

Tip: When checking if lines are parallel, remember that direction vectors should be proportional (i.e., one vector is a scalar multiple of the other).