Problem 80 - University Physics with Modern Physics

Question

A cube is placed so that one corner is at the origin and three edges are along the \(x\)-, \(y\)-, and \(z\)-axes of a coordinate system (Fig. P1.80). Use vectors to compute (a) the angle between the edge along the z axis (line \(ab\)) and the diagonal from the origin to the opposite corner (line \(ad\)), and (b) the angle between line \(ac\) (the diagonal of a face) and line ad.

Step-by-Step Solution

Answer

(a) \( \theta = cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \); (b) \( \phi = cos^{-1}\left(\frac{\sqrt{6}}{3}\right) \).

Step 1: Define Vectors for Line Segments

Assuming the cube has side length s, then line \( ab \) is represented by \( \vec{ab} \ = \ \langle 0, 0, s \rangle \), line \( ad \) is the diagonal and represented by \( \vec{ad} \ = \ \langle s, s, s \rangle \), and line \( ac \) which is the face diagonal is represented by \( \ \vec{ac} \ = \ \langle s, s, 0 \rangle \).

Step 2: Determine the Angle between \( ab \) and \( ad \)

To find the angle \( \theta \) between vectors, the dot product formula \( \vec{u} \cdot \vec{v} \ = \| \vec{u} \| \| \vec{v} \| cos(\theta) \) is used.The dot product \( \vec{ab} \cdot \vec{ad} \ = 0s + 0s + ss = s^2 \).The magnitudes are \( \|\vec{ab} \| = s \) and \( \|\vec{ad} \| = s\sqrt{3} \).Plug these into the formula to get: \[ cos(\theta) = \frac{s^2}{s \cdot s\sqrt{3}} = \frac{1}{\sqrt{3}}. \]Hence, \( \theta = cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \).

Step 3: Determine the Angle between \( ac \) and \( ad \)

For the angle \( \phi \) between vectors \( \vec{ac} \) and \( \vec{ad} \), use the dot product:\( \vec{ac} \cdot \vec{ad} \ = s^2 + s^2 + 0s = 2s^2 \).The magnitudes are \( \|\vec{ac} \| = s\sqrt{2} \) and \( \|\vec{ad} \| = s\sqrt{3} \).Thus, \[ cos(\phi) = \frac{2s^2}{s\sqrt{2} \cdot s\sqrt{3}} = \frac{2}{\sqrt{6}} \= \frac{\sqrt{6}}{3} \].Therefore, \( \phi = cos^{-1}\left(\frac{\sqrt{6}}{3}\right) \).

Key Concepts

Dot Product

The dot product is a fundamental concept in vector mathematics that helps to understand the interaction between two vectors. It's a mathematical operation that multiplies two equal-length sequences of numbers, usually coordinate vectors.
In simpler terms, the dot product tells us how much of one vector goes in the direction of another. If two vectors are perfectly aligned, the dot product equals the product of their magnitudes. If they are perpendicular, the dot product is zero.

• The formula for the dot product of two vectors \( \vec{u} = \langle u_1, u_2, u_3 \rangle \) and \( \vec{v} = \langle v_1, v_2, v_3 \rangle \) is \( \vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2 + u_3v_3 \).
• This operation provides vital information such as the angle between them, where \( \cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{\|\vec{u}\| \|\vec{v}\|} \).

Understanding the dot product is essential in various applications like physics for calculating work done or even in computer graphics for determining lighting calculations.

Cube Geometry

Cube geometry is a fascinating topic that involves understanding the three-dimensional space. A cube is a special type of rectangular prism where all sides are equal, and each angle between adjacent faces is a right angle.
It forms the basis for problems involving vectors and helps us visualize complex math problems in simpler terms.

• In a cube, every vertex joins three edges, and if one corner is placed at the origin of a coordinate system, the edges often align with the axes.
• The edges can be represented as vectors such as \( \vec{ab} = \langle 0, 0, s \rangle \), indicating alignment along the \(z\)-axis.
• Diagonals of the cube, like the body diagonal \( \vec{ad} = \langle s, s, s \rangle \), provide key insight into vector operations and geometry.

Engaging with cube geometry helps to grasp spatial concepts and solve problems related to structures and symmetry.

Angle Calculation

Calculating the angle between vectors is crucial for understanding the orientation of lines or edges in space. It applies the dot product and the magnitudes of the vectors involved.
The result offers insights into the spatial relationship between different geometric elements.

• To find the angle \( \theta \) between two vectors, use \( \cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{\|\vec{u}\| \|\vec{v}\|} \).
• This equation shows how the dot product relates to the magnitudes of the vectors and the cosine of the angle they form.
• With the information calculated, you can find angles like the one between the edge of a cube and a diagonal, such as \( \theta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \).

Mastery of angle calculations in vector mathematics is useful for tasks ranging from 3D modeling to robotics, enhancing spatial awareness and practical application skills.