https://file.aiquickdraw.com/mathbot/file/17263040558088oc0vk5f.jpg

We are tasked with finding the distance between the point \( (8, 0, 0) \) and the line given by the parametric equations:

\[
x = 8 + t, \quad y = 2t, \quad z = 21\sqrt{5} + 2t.
\]

To find the distance between a point and a line in 3D space, we use the formula:

\[
d = \frac{| \mathbf{PQ} \cdot (\mathbf{v} \times \mathbf{PQ})|}{|\mathbf{v}|},
\]
where:
- \( \mathbf{P} = (8, 0, 0) \) is the point,
- \( \mathbf{Q} \) is a point on the line,
- \( \mathbf{v} \) is the direction vector of the line.

### Step 1: Direction vector of the line
From the parametric equations:
- The direction vector \( \mathbf{v} \) of the line is \( \langle 1, 2, 2 \rangle \), because these are the coefficients of \( t \) in each parametric equation.

### Step 2: Point on the line
For simplicity, we can choose \( t = 0 \), which gives us a point \( \mathbf{Q} \) on the line:
\[
\mathbf{Q} = (8, 0, 21\sqrt{5}).
\]

### Step 3: Vector from \( P \) to \( Q \)
The vector \( \mathbf{PQ} \) is:
\[
\mathbf{PQ} = \mathbf{Q} - \mathbf{P} = (8, 0, 21\sqrt{5}) - (8, 0, 0) = \langle 0, 0, 21\sqrt{5} \rangle.
\]

### Step 4: Cross product \( \mathbf{v} \times \mathbf{PQ} \)
Now, compute the cross product:
\[
\mathbf{v} \times \mathbf{PQ} = \langle 1, 2, 2 \rangle \times \langle 0, 0, 21\sqrt{5} \rangle.
\]

The determinant of the matrix will give us the cross product:
\[
\mathbf{v} \times \mathbf{PQ} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 2 \\ 0 & 0 & 21\sqrt{5} \end{vmatrix}
= \mathbf{i} \left( 2 \cdot 21\sqrt{5} - 2 \cdot 0 \right) - \mathbf{j} \left( 1 \cdot 21\sqrt{5} - 2 \cdot 0 \right) + \mathbf{k} \left( 1 \cdot 0 - 2 \cdot 0 \right).
\]

This simplifies to:
\[
\mathbf{v} \times \mathbf{PQ} = \langle 42\sqrt{5}, -21\sqrt{5}, 0 \rangle.
\]

### Step 5: Magnitude of \( \mathbf{v} \times \mathbf{PQ} \)
Now, find the magnitude of the cross product:
\[
|\mathbf{v} \times \mathbf{PQ}| = \sqrt{(42\sqrt{5})^2 + (-21\sqrt{5})^2 + 0^2}
= \sqrt{42^2 \cdot 5 + (-21)^2 \cdot 5}
= \sqrt{1764 \cdot 5 + 441 \cdot 5}
= \sqrt{2205 \cdot 5} = \sqrt{11025} = 105.
\]

### Step 6: Magnitude of \( \mathbf{v} \)
The magnitude of \( \mathbf{v} = \langle 1, 2, 2 \rangle \) is:
\[
|\mathbf{v}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3.
\]

### Step 7: Final distance calculation
Finally, the distance from the point \( P \) to the line is:
\[
d = \frac{|\mathbf{v} \times \mathbf{PQ}|}{|\mathbf{v}|} = \frac{105}{3} = 35.
\]

Thus, the distance from the point \( (8, 0, 0) \) to the line is \( \boxed{35} \).

Do you need further details or have any questions?

Here are 5 related questions you might find useful:
1. How do we find the parametric equation of a line given two points in 3D?
2. What is the geometric interpretation of the cross product in 3D?
3. Can we find the distance between two parallel lines in space?
4. How is the distance between two skew lines in 3D space calculated?
5. What is the role of the direction vector in determining the properties of a line?

**Tip**: The magnitude of the cross product between two vectors gives the area of the parallelogram formed by those vectors.

Find the distance from the point (8, 0, 0) to the line: x = 8 + t, y = 2t, z = 21√5 + 2t.

This problem demonstrates how to calculate the distance from a point (8, 0, 0) to a line given by the parametric equations: x = 8 + t, y = 2t, z = 21√5 + 2t. Using vector cross products and magnitudes, this step-by-step solution is ideal for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation