https://file.aiquickdraw.com/mathbot/file/1727400455172846tksqq.jpg

The problem in the image asks to find the distance from point \( S \) to the line \( CD \) in a cube with a side length of 8 cm.

### Step-by-step Solution:

1. **Identify the position of point \( S \)**:  
   In the diagram, point \( S \) is the midpoint of the top edge \( HG \) of the cube. Since the cube has a side length of 8 cm, the position of \( S \) relative to the other vertices can be determined.

2. **Coordinates of key points**:  
   Let's set the origin at point \( A \) with the following coordinates:
   - \( A(0, 0, 0) \)
   - \( B(8, 0, 0) \)
   - \( D(0, 8, 0) \)
   - \( C(8, 8, 0) \)
   - \( H(0, 8, 8) \)
   - \( G(8, 8, 8) \)
   - Point \( S \) is the midpoint of \( HG \), so its coordinates are:
     \[
     S\left( \frac{0 + 8}{2}, \frac{8 + 8}{2}, \frac{8 + 8}{2} \right) = S(4, 8, 8)
     \]

3. **Distance from point \( S(4, 8, 8) \) to line \( CD \)**:  
   The line \( CD \) lies on the plane \( z = 0 \) and is a vertical line between \( C(8, 8, 0) \) and \( D(0, 8, 0) \). The direction vector of this line is:
   \[
   \overrightarrow{CD} = D(0, 8, 0) - C(8, 8, 0) = (-8, 0, 0)
   \]

   To find the perpendicular distance from point \( S(4, 8, 8) \) to the line \( CD \), we use the formula for the distance between a point and a line in 3D space:
   \[
   d = \frac{| \overrightarrow{CS} \times \overrightarrow{CD} |}{|\overrightarrow{CD}|}
   \]
   where \( \overrightarrow{CS} = S(4, 8, 8) - C(8, 8, 0) = (-4, 0, 8) \).

4. **Calculating the cross product** \( \overrightarrow{CS} \times \overrightarrow{CD} \):
   \[
   \overrightarrow{CS} \times \overrightarrow{CD} = 
   \begin{vmatrix}
   \hat{i} & \hat{j} & \hat{k} \\
   -4 & 0 & 8 \\
   -8 & 0 & 0
   \end{vmatrix}
   = \hat{i}(0 - 0) - \hat{j}(-32 - 0) + \hat{k}(0 - 0) = 32\hat{j}
   \]

   So, \( \overrightarrow{CS} \times \overrightarrow{CD} = (0, 32, 0) \).

5. **Magnitude of the cross product**:
   \[
   | \overrightarrow{CS} \times \overrightarrow{CD} | = \sqrt{0^2 + 32^2 + 0^2} = 32
   \]

6. **Magnitude of the direction vector \( \overrightarrow{CD} \)**:
   \[
   |\overrightarrow{CD}| = \sqrt{(-8)^2 + 0^2 + 0^2} = 8
   \]

7. **Final distance**:
   \[
   d = \frac{32}{8} = 4 \text{ cm}
   \]

Thus, the distance from point \( S \) to line \( CD \) is **4 cm**.

---

Would you like more details on any of the steps or have additional questions? 

Here are five related questions for further practice:
1. How would you find the distance from point \( S \) to line \( AB \)?
2. What is the distance from point \( H \) to line \( CD \)?
3. How can you calculate the distance between two skew lines in 3D?
4. What is the length of the diagonal \( AC \) of the cube?
5. How do you find the angle between two lines in 3D?

**Tip**: When dealing with cubes, try to set up a coordinate system to simplify the calculation of distances and vectors.

Perhatikanlah gambar kubus ABCDEFGH dengan panjang rusuk 8 cm. Tentukan jarak antara titik S ke garis CD!

Learn how to find the distance from point S to line CD in a cube with a side length of 8 cm using vector calculations and distance formulas in 3D geometry. This problem involves cross products and vector magnitudes, suitable 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