find the arc length parametrization of the line x = 7+7t, y = 3-4t, z = 4+4t that has the same direction as the given line and has reference point (7,3,4). use an arc length "s" as parameter x = ? y = ? z = ? use the parametric equations obtained on the first question to find the point on the line that is 11 units from the reference point in the direction of increasing parameter. (? , ? , ?)

To find the arc length parameterization of the line given by:

\[ x = 7 + 7t, \quad y = 3 - 4t, \quad z = 4 + 4t \]

we start by finding the direction vector of the line. The direction vector \(\mathbf{d}\) can be extracted from the coefficients of \(t\):

\[ \mathbf{d} = \langle 7, -4, 4 \rangle \]

Next, we calculate the magnitude of this direction vector, \(|\mathbf{d}|\):

\[ |\mathbf{d}| = \sqrt{7^2 + (-4)^2 + 4^2} = \sqrt{49 + 16 + 16} = \sqrt{81} = 9 \]

The arc length \(s\) is related to the parameter \(t\) by:

\[ s = |\mathbf{d}| \cdot t = 9t \]
\[ t = \frac{s}{9} \]

Now, substituting \(t = \frac{s}{9}\) into the original parametric equations gives us the arc length parameterization:

\[ x(s) = 7 + 7\left(\frac{s}{9}\right) = 7 + \frac{7s}{9} \]
\[ y(s) = 3 - 4\left(\frac{s}{9}\right) = 3 - \frac{4s}{9} \]
\[ z(s) = 4 + 4\left(\frac{s}{9}\right) = 4 + \frac{4s}{9} \]

Therefore, the arc length parameterization is:

\[ x(s) = 7 + \frac{7s}{9} \]
\[ y(s) = 3 - \frac{4s}{9} \]
\[ z(s) = 4 + \frac{4s}{9} \]

Now, to find the point on the line that is 11 units from the reference point \((7, 3, 4)\) in the direction of increasing parameter \(s\), we set \(s = 11\):

\[ x(11) = 7 + \frac{7 \cdot 11}{9} = 7 + \frac{77}{9} = 7 + 8.555... = 15.555... \]
\[ y(11) = 3 - \frac{4 \cdot 11}{9} = 3 - \frac{44}{9} = 3 - 4.888... = -1.888... \]
\[ z(11) = 4 + \frac{4 \cdot 11}{9} = 4 + \frac{44}{9} = 4 + 4.888... = 8.888... \]

Therefore, the point on the line that is 11 units from the reference point in the direction of increasing parameter is approximately:

\[ \left(15.56, -1.89, 8.89\right) \]

Would you like more details or have any questions? Here are some related questions you might want to ask:

1. How do you find the direction vector of a line?
2. What is the significance of arc length parametrization in vector calculus?
3. How do you calculate the magnitude of a vector?
4. How do you convert between different parameterizations of a line?
5. How can you verify that a point lies on a given line?

**Tip:** Always check your calculations carefully when dealing with vectors and their magnitudes to ensure accuracy in your results.

Explore how to find the arc length parametrization of a line and calculate distances in vector calculus. This problem involves parametric equations, vector direction, arc length parameterization, and magnitude of vectors. Suitable for advanced undergraduate students.

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