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This problem involves finding the components of the force \( F \) along the coordinate axes \( x \), \( y \), and \( z \) in a cubic structure where each edge is parallel to the respective axis.

### Problem Breakdown:

1. **Given**:
   - A cube with side length \( L \).
   - A force \( F \) acting along the diagonal from point \( D_1 \) to point \( B \).
   - The coordinate system is aligned such that the edges of the cube are parallel to the \( x \), \( y \), and \( z \) axes.

2. **Objective**:
   - Determine the components of the force \( F \) along the \( x \), \( y \), and \( z \) axes.

### Step-by-Step Solution:

1. **Position Vectors**:
   - Let the origin \( O \) be at point \( O \) where all axes intersect.
   - The coordinates of points \( D_1 \) and \( B \) are:
     - \( D_1 \) is at \( (0, 0, L) \).
     - \( B \) is at \( (L, L, 0) \).

2. **Direction Vector of Force \( F \)**:
   - The force \( F \) is along the line \( D_1B \).
   - The direction vector \( \vec{D_1B} \) is found by subtracting the position vectors of \( D_1 \) from \( B \):
     \[
     \vec{D_1B} = \vec{B} - \vec{D_1} = (L, L, 0) - (0, 0, L) = (L, L, -L)
     \]

3. **Magnitude of the Direction Vector**:
   - The magnitude of \( \vec{D_1B} \) is:
     \[
     |\vec{D_1B}| = \sqrt{L^2 + L^2 + (-L)^2} = \sqrt{3L^2} = L\sqrt{3}
     \]

4. **Unit Vector Along the Direction of \( F \)**:
   - The unit vector in the direction of \( F \) is:
     \[
     \hat{u} = \frac{1}{L\sqrt{3}} (L, L, -L) = \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right)
     \]

5. **Components of Force \( F \)**:
   - The components of the force \( F \) along the \( x \), \( y \), and \( z \) axes are obtained by multiplying the magnitude \( F \) by the unit vector components:
     \[
     F_x = F \cdot \frac{1}{\sqrt{3}}, \quad F_y = F \cdot \frac{1}{\sqrt{3}}, \quad F_z = F \cdot \left(-\frac{1}{\sqrt{3}}\right)
     \]
   - Therefore, the components of \( F \) are:
     \[
     \boxed{F_x = \frac{F}{\sqrt{3}}, \quad F_y = \frac{F}{\sqrt{3}}, \quad F_z = -\frac{F}{\sqrt{3}}}
     \]

This gives you the components of the force \( F \) along the \( x \), \( y \), and \( z \) axes.

Would you like further details or explanations?

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### Relative Questions:
1. What is the significance of the unit vector in calculating force components?
2. How would the components change if the force were applied along a different diagonal of the cube?
3. What happens to the force components if the cube's side length \( L \) changes?
4. How can this method be applied to other geometric shapes, like tetrahedrons?
5. What are the practical applications of breaking down forces into components?

### Tip:
Always ensure the direction vector is normalized (converted to a unit vector) before finding the components of the force to avoid errors in magnitude.

在图示正立方体中，已知：力F，边长L，正方体各边分别平行于x, y, z轴。试求力F对各坐标轴之矩。

Learn how to decompose a force F along the x, y, and z axes in a cube where each edge is parallel to the coordinate axes. This problem involves vector decomposition, force components, and unit vectors. Suitable for university-level students studying physics or engineering.

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