Math Problem Statement

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

Solution

The problem involves a cube ABCD.EFGHABCD.EFGH with a side length of 6 cm, and we are asked to find the distance from point CC to the plane AFHAFH.

Steps to Solve:

  1. Identify Points and Plane:

    • The cube has vertices labeled as A,B,C,D,E,F,G,HA, B, C, D, E, F, G, H.
    • The plane AFHAFH passes through points AA, FF, and HH.
  2. Determine Coordinates: Let's assign coordinates to the vertices assuming the origin at AA:

    • A(0,0,0)A(0, 0, 0)
    • B(6,0,0)B(6, 0, 0)
    • C(6,6,0)C(6, 6, 0)
    • D(0,6,0)D(0, 6, 0)
    • E(0,0,6)E(0, 0, 6)
    • F(6,0,6)F(6, 0, 6)
    • G(6,6,6)G(6, 6, 6)
    • H(0,6,6)H(0, 6, 6)
  3. Find the Equation of Plane AFHAFH: The plane passes through A(0,0,0)A(0, 0, 0), F(6,0,6)F(6, 0, 6), and H(0,6,6)H(0, 6, 6). To find the equation of the plane:

    • First, find two vectors in the plane:
      • AF=(6,0,6)\overrightarrow{AF} = (6, 0, 6)
      • AH=(0,6,6)\overrightarrow{AH} = (0, 6, 6)
    • The normal vector n\vec{n} to the plane is the cross product of AF\overrightarrow{AF} and AH\overrightarrow{AH}: \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: xy+z=0-x - y + z = 0
  4. Calculate Distance from CC to Plane AFHAFH: The coordinates of point CC are (6,6,0)(6, 6, 0). The distance from a point (x1,y1,z1)(x_1, y_1, z_1) to a plane Ax+By+Cz+D=0Ax + By + Cz + D = 0 is given by: d=Ax1+By1+Cz1+DA2+B2+C2d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} Substituting A=1A = -1, B=1B = -1, C=1C = 1, D=0D = 0, and (x1,y1,z1)=(6,6,0)(x_1, y_1, z_1) = (6, 6, 0): d=(1)(6)+(1)(6)+(1)(0)+0(1)2+(1)2+(1)2=661+1+1=123=43 cmd = \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 CC to the plane AFHAFH is 434\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.

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Math Problem Analysis

Mathematical Concepts

3D Geometry
Planes
Distance from Point to Plane

Formulas

Cross product of two vectors: \vec{AF} \times \vec{AH}
Equation of a plane: Ax + By + Cz + D = 0
Distance from point to plane: d = |Ax_1 + By_1 + Cz_1 + D| / sqrt(A^2 + B^2 + C^2)

Theorems

Distance from point to plane in 3D geometry

Suitable Grade Level

Grade 10-12