Math Problem Statement

2 The diagram shows a cuboid ABCDEFGH with horizontal base ADEH Diagram NOT accurately drawn 20 cm 12 cm 8 cm A B C M H D E G F P AB = 8 cm AD = 12 cm DE = 20 cm M is the midpoint of the base ADEH and P is the midpoint of the edge CF Work out the size of angle BMP Give your answer correct to one decimal place

Solution

To find the size of angle BMP\angle BMP in the cuboid, we will:

  1. Identify key points and coordinates:

    • Place the cuboid in a 3D coordinate system with A(0,0,0)A(0, 0, 0).
    • Define the key points based on the dimensions:
      • A(0,0,0)A(0, 0, 0), B(8,0,0)B(8, 0, 0), C(8,12,0)C(8, 12, 0), D(0,12,0)D(0, 12, 0),
      • E(0,0,20)E(0, 0, 20), F(8,12,20)F(8, 12, 20), G(8,0,20)G(8, 0, 20), H(0,12,20)H(0, 12, 20).
    • MM, the midpoint of the base ADEHADEH: M=(0+8+0+84,0+0+12+124,0+0+0+04)=(4,6,0).M = \left( \frac{0+8+0+8}{4}, \frac{0+0+12+12}{4}, \frac{0+0+0+0}{4} \right) = (4, 6, 0).
    • PP, the midpoint of edge CFCF: P=(8+82,12+122,0+202)=(8,12,10).P = \left( \frac{8+8}{2}, \frac{12+12}{2}, \frac{0+20}{2} \right) = (8, 12, 10).
  2. Calculate vectors BM\overrightarrow{BM} and BP\overrightarrow{BP}:

    • BM=MB=(4,6,0)(8,0,0)=(4,6,0)\overrightarrow{BM} = M - B = (4, 6, 0) - (8, 0, 0) = (-4, 6, 0).
    • BP=PB=(8,12,10)(8,0,0)=(0,12,10)\overrightarrow{BP} = P - B = (8, 12, 10) - (8, 0, 0) = (0, 12, 10).
  3. Find the dot product of BM\overrightarrow{BM} and BP\overrightarrow{BP}:

    • Dot product: BMBP=(4)(0)+(6)(12)+(0)(10)=72.\overrightarrow{BM} \cdot \overrightarrow{BP} = (-4)(0) + (6)(12) + (0)(10) = 72.
  4. Find the magnitudes of BM\overrightarrow{BM} and BP\overrightarrow{BP}:

    • Magnitude of BM\overrightarrow{BM}: BM=(4)2+62+02=16+36=52=213.|\overrightarrow{BM}| = \sqrt{(-4)^2 + 6^2 + 0^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}.
    • Magnitude of BP\overrightarrow{BP}: BP=(0)2+122+102=0+144+100=244=261.|\overrightarrow{BP}| = \sqrt{(0)^2 + 12^2 + 10^2} = \sqrt{0 + 144 + 100} = \sqrt{244} = 2\sqrt{61}.
  5. Calculate cosθ\cos \theta:

    • Using the formula for the angle between two vectors: cosθ=BMBPBMBP.\cos \theta = \frac{\overrightarrow{BM} \cdot \overrightarrow{BP}}{|\overrightarrow{BM}| \cdot |\overrightarrow{BP}|}.
    • Substitute the values: cosθ=72(213)(261)=724793=18793.\cos \theta = \frac{72}{(2\sqrt{13})(2\sqrt{61})} = \frac{72}{4\sqrt{793}} = \frac{18}{\sqrt{793}}.
  6. Find θ\theta:

    • Use θ=cos1(18793)\theta = \cos^{-1} \left( \frac{18}{\sqrt{793}} \right): θcos1(0.6385).\theta \approx \cos^{-1}(0.6385). θ50.1.\theta \approx 50.1^\circ.

Final Answer:

The size of angle BMP\angle BMP is 50.1°, correct to one decimal place.


Would you like further clarification on any step, or assistance with another problem? 😊

Relative Questions:

  1. What is the formula for the dot product of two vectors in 3D space?
  2. How do you calculate the midpoint of a line segment in a 3D coordinate system?
  3. What is the significance of the angle between two vectors in geometry?
  4. How do you simplify square roots involving large numbers in magnitude calculations?
  5. Why is it important to verify units and scales in a problem involving 3D geometry?

Tip:

When working with vectors in 3D, always double-check your coordinates and calculations for midpoint, as errors propagate quickly!

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Math Problem Analysis

Mathematical Concepts

3D Geometry
Vectors
Dot Product
Midpoint
Angle between Vectors

Formulas

Dot product of vectors: \vec{A} \cdot \vec{B} = |A| |B| \cos(\theta)
Magnitude of a vector: |\vec{A}| = \sqrt{x^2 + y^2 + z^2}
Midpoint formula in 3D: M = (x_1 + x_2)/2, (y_1 + y_2)/2, (z_1 + z_2)/2

Theorems

Dot Product Theorem
Cosine Rule for Angle Between Vectors

Suitable Grade Level

Grades 10-12