Math Problem Statement

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

Solution

The problem in the image asks to find the distance from point SS to the line CDCD in a cube with a side length of 8 cm.

Step-by-step Solution:

  1. Identify the position of point SS:
    In the diagram, point SS is the midpoint of the top edge HGHG of the cube. Since the cube has a side length of 8 cm, the position of SS relative to the other vertices can be determined.

  2. Coordinates of key points:
    Let's set the origin at point AA with the following coordinates:

    • A(0,0,0)A(0, 0, 0)
    • B(8,0,0)B(8, 0, 0)
    • D(0,8,0)D(0, 8, 0)
    • C(8,8,0)C(8, 8, 0)
    • H(0,8,8)H(0, 8, 8)
    • G(8,8,8)G(8, 8, 8)
    • Point SS is the midpoint of HGHG, so its coordinates are: S(0+82,8+82,8+82)=S(4,8,8)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)S(4, 8, 8) to line CDCD:
    The line CDCD lies on the plane z=0z = 0 and is a vertical line between C(8,8,0)C(8, 8, 0) and D(0,8,0)D(0, 8, 0). The direction vector of this line is: CD=D(0,8,0)C(8,8,0)=(8,0,0)\overrightarrow{CD} = D(0, 8, 0) - C(8, 8, 0) = (-8, 0, 0)

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

  4. Calculating the cross product CS×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: CS×CD=02+322+02=32| \overrightarrow{CS} \times \overrightarrow{CD} | = \sqrt{0^2 + 32^2 + 0^2} = 32

  6. Magnitude of the direction vector CD\overrightarrow{CD}: CD=(8)2+02+02=8|\overrightarrow{CD}| = \sqrt{(-8)^2 + 0^2 + 0^2} = 8

  7. Final distance: d=328=4 cmd = \frac{32}{8} = 4 \text{ cm}

Thus, the distance from point SS to line CDCD 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 SS to line ABAB?
  2. What is the distance from point HH to line CDCD?
  3. How can you calculate the distance between two skew lines in 3D?
  4. What is the length of the diagonal ACAC 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.

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Math Problem Analysis

Mathematical Concepts

Geometry
Distance in 3D Space
Vector Calculations

Formulas

Distance between a point and a line in 3D space
Cross product of vectors
Magnitude of a vector

Theorems

Distance Formula in 3D Geometry
Properties of Vectors and Cross Product

Suitable Grade Level

Grades 10-12