Problem 57 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the smallest angle between the main diagonals of a rectangular box 4 feet by 6 feet by 10 feet.

Step-by-Step Solution

Answer

The smallest angle between the main diagonals is approximately 66.42 degrees.

Step 1: Identify Vertices of the Box

Consider the rectangular box with vertices represented as points: - Point A at (0, 0, 0), - Point B at (4, 0, 0), - Point C at (4, 6, 0), - Point D at (0, 6, 0), - Point E at (0, 0, 10), - Point F at (4, 0, 10), - Point G at (4, 6, 10), - Point H at (0, 6, 10). The box has two main diagonals, AG and CF, spanning from opposite corners.

Step 2: Calculate Vector of Diagonal AG

The vector from A to G is formed by subtracting the coordinates of point A from point G. Thus, the vector AG is:\[ \vec{AG} = (4 - 0, 6 - 0, 10 - 0) = (4, 6, 10) \]

Step 3: Calculate Vector of Diagonal CF

The vector from C to F is formed by subtracting the coordinates of point C from point F. Thus, the vector CF is:\[ \vec{CF} = (4 - 4, 0 - 6, 10 - 0) = (0, -6, 10) \]

Step 4: Apply the Dot Product Formula

The dot product of vectors \( \vec{AG} \) and \( \vec{CF} \) is calculated by:\[ \vec{AG} \cdot \vec{CF} = (4)(0) + (6)(-6) + (10)(10) = -36 + 100 = 64 \]

Step 5: Compute Magnitudes of the Vectors

Calculate the magnitudes of the vectors \( \vec{AG} \) and \( \vec{CF} \):\[ |\vec{AG}| = \sqrt{4^2 + 6^2 + 10^2} = \sqrt{16 + 36 + 100} = \sqrt{152} \]\[ |\vec{CF}| = \sqrt{0^2 + (-6)^2 + 10^2} = \sqrt{0 + 36 + 100} = \sqrt{136} \]

Step 6: Use the Dot Product to Find the Cosine of the Angle

Cosine of the angle \( \theta \) between \( \vec{AG} \) and \( \vec{CF} \) is given by:\[ \cos \theta = \frac{\vec{AG} \cdot \vec{CF}}{|\vec{AG}| |\vec{CF}|} = \frac{64}{\sqrt{152} \cdot \sqrt{136}} \]

Step 7: Calculate the Angle \( \theta \)

Using the cosine value from Step 6, calculate the angle \( \theta \) by taking the inverse cosine:\[ \theta = \cos^{-1}\left(\frac{64}{\sqrt{152} \cdot \sqrt{136}}\right) \]

Step 8: Approximate \( \theta \)

Calculate approximate values of the magnitudes and use a calculator to approximate:\[ \theta \approx \cos^{-1}\left(\frac{64}{\sqrt{152} \cdot \sqrt{136}}\right) \approx 66.42^\circ \]

Key Concepts

Rectangular Box

A rectangular box, also known as a rectangular prism, is a three-dimensional shape with all angles being right angles, and faces are rectangles. In this exercise, the box has dimensions 4 feet by 6 feet by 10 feet. The shape is defined by eight vertices, which are the corner points where the edges meet. These vertices are critical in identifying the strategic locations to calculate vectors for the diagonals.

• Each pair of opposite faces are congruent rectangles.
• The vertices for this box are: (0,0,0), (4,0,0), (4,6,0), (0,6,0), (0,0,10), (4,0,10), (4,6,10), and (0,6,10).
• Main diagonals are the longest paths from one corner of the box to the opposite corner, crossing through the interior of the box.

Understanding these properties of a rectangular box helps in examining its geometry and applying mathematical formulas to solve problems.

Main Diagonals

The main diagonals of a rectangular box are significant lines, stretching from one vertex to its opposite vertex. They traverse through the interior, providing insight into the spatial dimensions and symmetry of the shape.
For our specific problem:

• The main diagonal AG runs from vertex A (0,0,0) to vertex G (4,6,10).
• The other main diagonal CF runs from vertex C (4,6,0) to vertex F (4,0,10).

The investigation of these diagonals allows us to ascertain more than just their length. They are the key to exploring angles formed and other spatial relationships within the box.

Dot Product

The dot product, or scalar product, is a crucial operation in vector mathematics, used to find the angle between two vectors. It is obtained by multiplying corresponding components of two vectors and summing these products.
In our problem:

• We calculate the vector AG as (4,6,10) and CF as (0,-6,10).
• The dot product is calculated by: \[ \vec{AG} \cdot \vec{CF} = (4)(0) + (6)(-6) + (10)(10) = -36 + 100 = 64 \]

This operation provides an essential value in determining the cosine of the angle between these two diagonals, an important step in angle calculation.

Angle Calculation

Calculating the angle between two vectors involves using the dot product result and the magnitudes (lengths) of the vectors. The cosine of the angle θ between them is given by the formula:
\[ \cos \theta = \frac{\vec{AG} \cdot \vec{CF}}{|\vec{AG}| |\vec{CF}|} \]Determining this angle helps in several geometry applications, such as understanding the orientation of shapes.
Steps involved in the calculation:

• Find magnitudes of vectors AG and CF:\[ |\vec{AG}| = \sqrt{4^2 + 6^2 + 10^2} = \sqrt{152} \]\[ |\vec{CF}| = \sqrt{0^2 + (-6)^2 + 10^2} = \sqrt{136} \]
• Plug these into the cosine formula:\[ \cos \theta = \frac{64}{\sqrt{152} \cdot \sqrt{136}} \]
• Finally, calculate the angle by taking the inverse cosine.\[ \theta = \cos^{-1}\left(\frac{64}{\sqrt{152} \cdot \sqrt{136}}\right) \approx 66.42^\circ \]

This thorough approach equips students to solve similar problems involving angles between vectors.