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The problem involves a cube \( ABCD.EFGH \) with a side length of 6 cm, and we are asked to find the distance from point \( C \) to the plane \( AFH \).

### Steps to Solve:

1. **Identify Points and Plane**:
   - The cube has vertices labeled as \( A, B, C, D, E, F, G, H \).
   - The plane \( AFH \) passes through points \( A \), \( F \), and \( H \).

2. **Determine Coordinates**:
   Let's assign coordinates to the vertices assuming the origin at \( A \):
   - \( A(0, 0, 0) \)
   - \( B(6, 0, 0) \)
   - \( C(6, 6, 0) \)
   - \( D(0, 6, 0) \)
   - \( E(0, 0, 6) \)
   - \( F(6, 0, 6) \)
   - \( G(6, 6, 6) \)
   - \( H(0, 6, 6) \)

3. **Find the Equation of Plane \( AFH \)**:
   The plane passes through \( A(0, 0, 0) \), \( F(6, 0, 6) \), and \( H(0, 6, 6) \). To find the equation of the plane:
   - First, find two vectors in the plane:
     - \( \overrightarrow{AF} = (6, 0, 6) \)
     - \( \overrightarrow{AH} = (0, 6, 6) \)
   - The normal vector \( \vec{n} \) to the plane is the cross product of \( \overrightarrow{AF} \) and \( \overrightarrow{AH} \):
     \[
     \vec{n} = \overrightarrow{AF} \times \overrightarrow{AH} = \left| \begin{array}{ccc}
     \hat{i} & \hat{j} & \hat{k} \\
     6 & 0 & 6 \\
     0 & 6 & 6 \end{array} \right|
     = (-36, -36, 36)
     \]
     Simplified, \( \vec{n} = (-1, -1, 1) \).

   - The equation of the plane is:
     \[
     -x - y + z = 0
     \]

4. **Calculate Distance from \( C \) to Plane \( AFH \)**:
   The coordinates of point \( C \) are \( (6, 6, 0) \). The distance from a point \( (x_1, y_1, z_1) \) to a plane \( Ax + By + Cz + D = 0 \) is given by:
   \[
   d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}
   \]
   Substituting \( A = -1 \), \( B = -1 \), \( C = 1 \), \( D = 0 \), and \( (x_1, y_1, z_1) = (6, 6, 0) \):
   \[
   d = \frac{|(-1)(6) + (-1)(6) + (1)(0) + 0|}{\sqrt{(-1)^2 + (-1)^2 + (1)^2}} = \frac{| -6 - 6 |}{\sqrt{1 + 1 + 1}} = \frac{12}{\sqrt{3}} = 4\sqrt{3} \text{ cm}
   \]

Thus, the distance from point \( C \) to the plane \( AFH \) is \( 4\sqrt{3} \) cm.

Would you like further clarification or more details on any step?

Here are five related questions to expand on this:
1. How do you find the equation of a plane given three points?
2. What is the geometric meaning of the normal vector to a plane?
3. How do you compute the cross product of two vectors?
4. What are the properties of distances from points to planes?
5. How does the Pythagorean theorem apply to 3D geometry?

**Tip**: When working with 3D geometry, always try to visualize the problem by assigning coordinates to key points and using vector operations to simplify calculations.

Diketahui kubus ABCD.EFGH dengan rusuk 6 cm. Jarak titik C ke bidang AFH adalah ....

Find the distance from point C to the plane AFH in a cube ABCD.EFGH with 6 cm edge length. This problem involves vector operations and 3D geometry, suitable for students in Grades 10-12.

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