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We are asked to find the distance from the point \( (4, 0, 0) \) to the line described by the parametric equations:

\[
x = 4 + t, \quad y = 2t, \quad z = \frac{7\sqrt{5}}{5} + 2t
\]

### Step 1: Vector form of the line

The parametric equations can be written in vector form as:

\[
\mathbf{r}(t) = \langle 4, 0, \frac{7\sqrt{5}}{5} \rangle + t \langle 1, 2, 2 \rangle
\]

Here, \( \langle 4, 0, \frac{7\sqrt{5}}{5} \rangle \) is a point on the line, and \( \langle 1, 2, 2 \rangle \) is the direction vector of the line.

### Step 2: Formula for the distance from a point to a line in 3D

The distance \( d \) from a point \( P_0(x_0, y_0, z_0) \) to a line passing through a point \( P_1(x_1, y_1, z_1) \) with direction vector \( \mathbf{v} = \langle a, b, c \rangle \) is given by:

\[
d = \frac{| \mathbf{PQ} \times \mathbf{v} |}{|\mathbf{v}|}
\]

where \( \mathbf{PQ} \) is the vector from \( P_1 \) to \( P_0 \), and \( \mathbf{v} \) is the direction vector of the line.

### Step 3: Apply the formula

1. **Point on the line**: Let \( P_1 = (4, 0, \frac{7\sqrt{5}}{5}) \), which is a point on the line.
2. **Direction vector**: The direction vector \( \mathbf{v} \) is \( \langle 1, 2, 2 \rangle \).
3. **Point to the line**: The point \( P_0 = (4, 0, 0) \).

Now, calculate the vector \( \mathbf{PQ} \):

\[
\mathbf{PQ} = P_0 - P_1 = (4, 0, 0) - (4, 0, \frac{7\sqrt{5}}{5}) = \langle 0, 0, -\frac{7\sqrt{5}}{5} \rangle
\]

### Step 4: Cross product \( \mathbf{PQ} \times \mathbf{v} \)

\[
\mathbf{PQ} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 0 & -\frac{7\sqrt{5}}{5} \\ 1 & 2 & 2 \end{vmatrix} = \mathbf{i}(0 \cdot 2 - 0 \cdot 2) - \mathbf{j}(0 \cdot 2 - (-\frac{7\sqrt{5}}{5}) \cdot 1) + \mathbf{k}(0 \cdot 2 - 0 \cdot 1)
\]

Simplifying:

\[
\mathbf{PQ} \times \mathbf{v} = \langle 0, \frac{7\sqrt{5}}{5}, 0 \rangle
\]

### Step 5: Magnitude of the cross product

\[
|\mathbf{PQ} \times \mathbf{v}| = \sqrt{0^2 + \left( \frac{7\sqrt{5}}{5} \right)^2 + 0^2} = \frac{7\sqrt{5}}{5}
\]

### Step 6: Magnitude of the direction vector

\[
|\mathbf{v}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3
\]

### Step 7: Distance

Now, the distance \( d \) is:

\[
d = \frac{|\mathbf{PQ} \times \mathbf{v}|}{|\mathbf{v}|} = \frac{\frac{7\sqrt{5}}{5}}{3} = \frac{7\sqrt{5}}{15}
\]

Calculating this:

\[
d \approx 4.526
\]

Thus, the correct answer is \( \boxed{4.526} \).

Would you like more details on any step or further explanation?

Here are 5 questions for further exploration:
1. How do you compute the cross product of two vectors?
2. What is the significance of the direction vector in line equations?
3. Why do we use the magnitude of the cross product for distance?
4. What other methods exist to find the distance from a point to a line in 3D?
5. How would the distance formula change if the point and line were in 2D?

**Tip**: When solving 3D geometry problems, visualizing the vectors and their relationships often helps in understanding the concepts better!

Find the distance from the point (4, 0, 0) to the line: x = 4 + t, y = 2t, z = -\frac{7\sqrt{5}}{5} + 2t.

Discover how to compute the distance from a point to a line in 3D using vector algebra and the cross product method. This tutorial covers essential formulas, including the distance formula in 3D, suitable for high school students.

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