Math Problem Statement

Find the angle between the line ST and the plane PQRS in a rectangular-based pyramid with a height of 20 cm and a square base side length of 22 cm.

Solution

We are given a rectangular-based pyramid PQRSTPQRST, where the base PQRSPQRS is a square with side lengths of 22 cm, and the height of the pyramid from the apex TT to the center of the base is 20 cm. We are tasked with finding the angle between the line STST and the plane PQRSPQRS.

Approach:

  1. Coordinates Setup:

    • Place the center of the square base OO at the origin (0,0,0)(0, 0, 0).
    • Let the vertices P,Q,R,SP, Q, R, S of the square base be at:
      • P(11,11,0)P(-11, -11, 0)
      • Q(11,11,0)Q(11, -11, 0)
      • R(11,11,0)R(11, 11, 0)
      • S(11,11,0)S(-11, 11, 0)
    • The apex TT is directly above OO with coordinates (0,0,20)(0, 0, 20).
  2. Vector for Line STST: The vector ST\overrightarrow{ST} can be calculated using the coordinates of SS and TT. ST=(0(11),011,200)=(11,11,20)\overrightarrow{ST} = (0 - (-11), 0 - 11, 20 - 0) = (11, -11, 20)

  3. Normal to the Plane PQRSPQRS: Since PQRSPQRS is in the xyxy-plane, the normal vector to the plane is simply in the zz-direction, which is: n=(0,0,1)\mathbf{n} = (0, 0, 1)

  4. Angle Between Line and Plane: The angle θ\theta between the line and the plane can be found using the dot product formula between the line's direction vector v=ST\mathbf{v} = \overrightarrow{ST} and the normal vector n\mathbf{n}: cosθ=vnvn\cos \theta = \frac{|\mathbf{v} \cdot \mathbf{n}|}{|\mathbf{v}| |\mathbf{n}|} where:

    • v=(11,11,20)\mathbf{v} = (11, -11, 20)
    • n=(0,0,1)\mathbf{n} = (0, 0, 1)

    The dot product is: vn=(11)(0)+(11)(0)+(20)(1)=20\mathbf{v} \cdot \mathbf{n} = (11)(0) + (-11)(0) + (20)(1) = 20

    The magnitude of v\mathbf{v} is: v=112+(11)2+202=121+121+400=64225.33|\mathbf{v}| = \sqrt{11^2 + (-11)^2 + 20^2} = \sqrt{121 + 121 + 400} = \sqrt{642} \approx 25.33

    The magnitude of n\mathbf{n} is: n=02+02+12=1|\mathbf{n}| = \sqrt{0^2 + 0^2 + 1^2} = 1

    So, we have: cosθ=2025.33×12025.330.7895\cos \theta = \frac{|20|}{25.33 \times 1} \approx \frac{20}{25.33} \approx 0.7895

    Therefore, the angle θ\theta is: θ=cos1(0.7895)37.8\theta = \cos^{-1}(0.7895) \approx 37.8^\circ

Final Answer:

The angle between the line STST and the plane PQRSPQRS is approximately 37.837.8^\circ.

Would you like more details on any of the steps?

Relative Questions:

  1. How do you find the direction vector of a line given two points?
  2. What is the geometric interpretation of the dot product in vector math?
  3. How do you find the normal vector of a plane in 3D space?
  4. What is the physical meaning of an angle between a line and a plane?
  5. Can you calculate angles between two lines in space using similar methods?

Tip:

When dealing with geometric figures in 3D, assigning coordinates to points and using vector operations like dot products and cross products can simplify finding angles and distances.

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Math Problem Analysis

Mathematical Concepts

3D Geometry
Vector Mathematics
Dot Product
Angle Between Line and Plane

Formulas

Dot Product: a • b = |a| |b| cos(θ)
Magnitude of Vector: |v| = √(x^2 + y^2 + z^2)
Cosine of Angle Between Line and Plane: cos(θ) = |v • n| / (|v| |n|)

Theorems

Dot Product Angle Theorem

Suitable Grade Level

Grades 10-12