find the distance from the point (8,0,0) to the line x=8+t, y=2t, z=6 (square root 5) + 2t

To find the distance from a point \((8, 0, 0)\) to the line parameterized as:

\[
x = 8 + t, \quad y = 2t, \quad z = \sqrt{5} + 2t,
\]

we can use the formula for the distance from a point to a line in 3D. The formula is given by:

\[
d = \frac{\|\mathbf{P_0} - \mathbf{L}(t)\| \times \mathbf{v}}{\|\mathbf{v}\|}
\]

where:
- \(\mathbf{P_0} = (8, 0, 0)\) is the point,
- \(\mathbf{L}(t) = (8 + t, 2t, \sqrt{5} + 2t)\) is any point on the line,
- \(\mathbf{v}\) is the direction vector of the line, which we get from the parametric equations as \(\mathbf{v} = (1, 2, 2)\).

Now, let's apply the formula.The distance from the point \((8, 0, 0)\) to the line is approximately \(1.67\) units.

Would you like further details or have any questions?

Here are some related questions to explore:
1. How do you derive the formula for distance from a point to a line in 3D?
2. What happens if the point lies on the line? How does the distance formula change?
3. How do we calculate the distance between two parallel lines in 3D?
4. How can you apply this method to non-linear curves in 3D?
5. What is the geometric interpretation of the cross product used in the distance calculation?

**Tip:** The cross product of two vectors gives a vector perpendicular to the plane they define, which is useful in calculating distances involving lines and planes.

This problem explains how to find the distance from the point (8, 0, 0) to the line parametrized by x=8+t, y=2t, z=6+sqrt(5)+2t. Using key concepts from 3D geometry, vectors, and the cross product, this solution provides a detailed step-by-step explanation. Suitable for advanced high school or early college students.

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