https://file.aiquickdraw.com/mathbot/file/17269659343741qudf4qb.jpg

please tell me the answer in the correct way on number 19

Let's solve the problem number 19 from the image step by step.

### Problem 19:
"Diketahui kubus \( ABCD.EFGH \) dengan rusuk 4 cm. Jika \( P \) titik tengah \( EH \), maka jarak titik \( P \) ke garis \( CF \) adalah ... cm."

Translation:
- We are given a cube \( ABCD.EFGH \) with an edge length of 4 cm.
- \( P \) is the midpoint of edge \( EH \).
- The question asks to find the distance from point \( P \) to line \( CF \).

### Step-by-step solution:

1. **Coordinates of points in the cube:**
   - Assume the cube is positioned with vertices at the following coordinates:
     - \( A(0, 0, 0), B(4, 0, 0), C(4, 4, 0), D(0, 4, 0) \)
     - \( E(0, 0, 4), F(4, 0, 4), G(4, 4, 4), H(0, 4, 4) \)

2. **Find coordinates of point \( P \):**
   - \( P \) is the midpoint of edge \( EH \).
   - Coordinates of \( E = (0, 0, 4) \) and \( H = (0, 4, 4) \).
   - Midpoint formula: \( P = \left(\frac{0+0}{2}, \frac{0+4}{2}, \frac{4+4}{2}\right) = (0, 2, 4) \).

3. **Find equation of line \( CF \):**
   - \( C(4, 4, 0) \) and \( F(4, 0, 4) \).
   - The direction vector of line \( CF \) is \( \overrightarrow{CF} = (4 - 4, 0 - 4, 4 - 0) = (0, -4, 4) \).

4. **Distance formula from point to a line:**
   The formula for the distance from a point \( P(x_1, y_1, z_1) \) to a line passing through point \( A(x_2, y_2, z_2) \) with direction vector \( \vec{d} = (d_x, d_y, d_z) \) is:

   \[
   d = \frac{| \vec{PA} \times \vec{d} |}{|\vec{d}|}
   \]

   Where \( \vec{PA} = (x_1 - x_2, y_1 - y_2, z_1 - z_2) \).

   Let's apply this formula:
   - \( P(0, 2, 4) \)
   - \( A \) is any point on the line \( CF \), so use \( C(4, 4, 0) \).
   - Direction vector \( \vec{d} = (0, -4, 4) \).
   - Vector \( \vec{PC} = (0 - 4, 2 - 4, 4 - 0) = (-4, -2, 4) \).

5. **Cross product:**
   \[
   \vec{PC} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -4 & -2 & 4 \\ 0 & -4 & 4 \end{vmatrix}
   \]
   \[
   = \hat{i}(( -2)(4) - (4)(-4)) - \hat{j}((-4)(4) - (4)(0)) + \hat{k}((-4)(-4) - (-2)(0))
   \]
   \[
   = \hat{i}(-8 + 16) - \hat{j}(-16) + \hat{k}(16) = 8\hat{i} + 16\hat{j} + 16\hat{k}
   \]
   So, \( \vec{PC} \times \vec{d} = (8, 16, 16) \).

6. **Magnitude of cross product:**
   \[
   |\vec{PC} \times \vec{d}| = \sqrt{8^2 + 16^2 + 16^2} = \sqrt{64 + 256 + 256} = \sqrt{576} = 24
   \]

7. **Magnitude of direction vector \( \vec{d} \):**
   \[
   |\vec{d}| = \sqrt{0^2 + (-4)^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}
   \]

8. **Distance calculation:**
   \[
   d = \frac{24}{4\sqrt{2}} = \frac{24}{4} \times \frac{1}{\sqrt{2}} = 6 \times \frac{1}{\sqrt{2}} = \frac{6\sqrt{2}}{2} = 3\sqrt{2}
   \]

Thus, the distance from point \( P \) to the line \( CF \) is \( \boxed{2\sqrt{2}} \) cm.

---

Would you like more details on any step or have any other questions?

### Follow-up questions:
1. How do you calculate the midpoint of a line segment in 3D?
2. Can you explain the cross-product of two vectors in simple terms?
3. What is the formula for the distance from a point to a line in space?
4. How can you find the direction vector of a line given two points?
5. What other methods can be used to find distances in geometry?

### Tip:
When solving geometric problems in 3D, visualizing the coordinate system and key points can help clarify relationships between objects.

Diketahui kubus ABCD.EFGH dengan rusuk 4 cm. Jika P titik tengah EH, maka jarak titik P ke garis CF adalah ... cm.

This solution explains how to find the distance from the midpoint of edge EH in cube ABCD.EFGH with side 4 cm to the line CF. Using 3D geometry and vector calculations, it demonstrates the step-by-step process, including the midpoint formula, cross product, and distance formula in space.

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