Math Problem Statement

Find the angle between the line QT and the plane PQRS of a rectangular-based pyramid PQRST, where the pyramid has a height of 24 cm, the base PQRS is a rectangle with PQ = 22 cm and PS = 25 cm.

Solution

To solve this problem, we need to find the angle between the line QTQT and the plane PQRSPQRS of the rectangular-based pyramid PQRSTPQRST.

Step-by-Step Solution:

  1. Understand the Geometry of the Pyramid:

    • PQRSPQRS is the base of the pyramid, and it is a rectangle.
    • The height of the pyramid TT is 24 cm above the center of the rectangle PQRSPQRS.
    • The dimensions of the rectangle are PQ=22PQ = 22 cm and PS=25PS = 25 cm.
  2. Find the Position of QQ:

    • QQ is the midpoint of PRPR. To find QQ, we first need the coordinates of PP and RR.
    • Assume PP is at the origin (0,0)(0, 0), QQ will be at (222,0)=(11,0)\left(\frac{22}{2}, 0\right) = (11, 0).
  3. Coordinates of Points:

    • P(0,0,0)P(0, 0, 0)
    • Q(11,0,0)Q(11, 0, 0)
    • R(22,0,0)R(22, 0, 0)
    • S(0,25,0)S(0, 25, 0)
    • The point TT, being directly above the center of the base rectangle, has coordinates T(11,12.5,24)T(11, 12.5, 24).
  4. Find the Line QTQT:

    • QTQT connects the point Q(11,0,0)Q(11, 0, 0) to T(11,12.5,24)T(11, 12.5, 24).
    • The direction vector of QTQT is TQ=(1111,12.50,240)=(0,12.5,24)T - Q = (11 - 11, 12.5 - 0, 24 - 0) = (0, 12.5, 24).
  5. Normal to Plane PQRSPQRS:

    • The plane PQRSPQRS is horizontal, so its normal vector is along the zz-axis, which is (0,0,1)(0, 0, 1).
  6. Find the Cosine of the Angle:

    • The angle θ\theta between line QTQT and the plane PQRSPQRS can be found using the dot product: cosθ=Vector QTNormal to PQRSQTNormal\cos \theta = \frac{\text{Vector } QT \cdot \text{Normal to } PQRS}{\|QT\| \cdot \|Normal\|}
    • The dot product (0,12.5,24)(0,0,1)=24(0, 12.5, 24) \cdot (0, 0, 1) = 24.
    • Magnitude of QTQT, QT=02+12.52+242=0+156.25+576=732.25\|QT\| = \sqrt{0^2 + 12.5^2 + 24^2} = \sqrt{0 + 156.25 + 576} = \sqrt{732.25}.
    • Magnitude of the normal vector Normal=1\|Normal\| = 1.

    cosθ=24732.25\cos \theta = \frac{24}{\sqrt{732.25}}

  7. Calculate the Angle: cosθ=24732.25\cos \theta = \frac{24}{\sqrt{732.25}}

    Let's calculate the angle θ\theta in degrees:

θ=cos1(24732.25)\theta = \cos^{-1}\left(\frac{24}{\sqrt{732.25}}\right)

Let's compute this.The angle between the line QTQT and the plane PQRSPQRS is approximately 27.5 degrees when rounded to one decimal place.

Would you like more details on this calculation or have any questions?

Here are some related questions you might find interesting:

  1. How do you calculate the dot product between two vectors?
  2. What is the significance of the normal vector to a plane?
  3. How can you find the height of a pyramid given the dimensions of its base and its slant height?
  4. What are some other methods to find the angle between a line and a plane?
  5. How can you determine if a line is perpendicular or parallel to a plane?

Tip: Always double-check your geometric interpretations, especially when working with 3D objects, to ensure the correct coordinates and vectors are used.

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Math Problem Analysis

Mathematical Concepts

Geometry
Trigonometry
Vectors

Formulas

Cosine of angle between a line and a plane: cos(θ) = (Vector · Normal) / (|Vector| · |Normal|)
Magnitude of a vector: |v| = sqrt(x^2 + y^2 + z^2)

Theorems

Dot Product
Vector Magnitude

Suitable Grade Level

Grades 10-12