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Chapter 1: Problem 5

Find the angle between a body diagonal of a cube and any one of its face diagonals. [Hint: Choose a cube with side 1 and with one corner at \(O\) and the opposite corner at the point (1,1,1) . Write down the vector that represents a body diagonal and another that represents a face diagonal, and then find the angle between them as in Problem 1.4.]

Short Answer

Expert verified
The angle between a body diagonal and a face diagonal of the cube is \(\cos^{-1}\left(\frac{\sqrt{6}}{3}\right)\).

Step by step solution

01

Define Vectors

To solve the problem, we begin by defining vectors for the body diagonal and face diagonal. Consider a cube with a side length of 1. Place the cube with one corner at the origin, makes its coordinates \(0,0,0\). The opposite corner will be at \(1,1,1\). The body diagonal of the cube is a vector from \(0,0,0\) to \(1,1,1\), so this vector can be denoted as \vec{d_b} = (1,1,1)\. Next, consider a face of the cube. For example, the face along the x-y plane would extend from \(0,0,0\) to \(1,1,0\) and from \(0,0,0\) to \(0,1,0\). The face diagonal in this plane is \vec{d_f} = (1,1,0)\.
02

Calculate the Dot Product

The angle between two vectors can be found using the dot product formula \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos{\theta}\). First, calculate the dot product of \vec{d_b} = (1,1,1)\ and \vec{d_f} = (1,1,0)\. The dot product is given by \( \vec{d_b} \cdot \vec{d_f} = 1 \times 1 + 1 \times 1 + 1 \times 0 = 2 \).
03

Calculate Magnitudes

Find the magnitudes of both \vec{d_b}\ and \vec{d_f}\. The magnitude of \vec{d_b}\ is \( |\vec{d_b}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \). The magnitude of \vec{d_f}\ is \( |\vec{d_f}| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2} \).
04

Find the Cosine of the Angle

Using the formula \( \cos{\theta} = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \; |\vec{b}|} \), calculate \cos{\theta}\. Substitute the previously found values: \( \cos{\theta} = \frac{2}{\sqrt{3} \times \sqrt{2}} = \frac{2}{\sqrt{6}} \). Simplify \( \frac{2}{\sqrt{6}} \) to get \( \frac{\sqrt{6}}{3} \).
05

Calculate the Angle

Finally, find the angle \(\theta\) by computing \(\theta = \cos^{-1}\left(\frac{\sqrt{6}}{3}\right)\). This will give us the angle we are looking for in either radians or degrees, depending on the function used (calculator setting).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dot Product
The dot product is a fundamental operation in vector calculus, used to calculate the angle between two vectors. It's an algebraic operation that takes two equal-length sequences of numbers, usually coordinate vectors, and returns a single number.
Understanding the dot product is essential in vector analysis because it can show how much of one vector goes in the direction of another. It is defined as:
  • \( \vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3 \)
  • In a physical sense, this result is a scalar representing the projection of one vector onto another.
  • For example, in this exercise, the dot product of vectors \( \vec{d_b} = (1,1,1) \) and \( \vec{d_f} = (1,1,0) \) is calculated as:
  • \( \vec{d_b} \cdot \vec{d_f} = 1\times1 + 1\times1 + 1\times0 = 2 \)
This scalar reflects the overlap in direction between the body diagonal and face diagonal of the cube. The greater this value, the smaller the angle between the vectors. Conversely, a dot product of zero indicates perpendicularity.
Cube Geometry
Understanding cube geometry is crucial for visualizing the problem at hand. A cube is a three-dimensional shape with six square faces, all of the same size, with edges of equal length. Here, the cube is positioned at the origin of a coordinate system, providing a straightforward way to define positions as vectors.
When we talk about the body diagonal of a cube, we refer to the line connecting opposite corners, such as from \((0,0,0)\) to \((1,1,1)\). This line cuts through the interior of the cube and is longer than any face diagonal.
  • The face diagonal extends across one of the cube's square faces. It can be visualized as stretching from one corner of the square to the opposite corner.
  • In this case, for the face of the cube on the x-y plane, the diagonal runs from \((0,0,0)\) to \((1,1,0)\).
Each type of diagonal provides key insights into the cube's geometry, and understanding their properties is critical for solving spatial problems like finding angles between vectors.
Magnitude of Vectors
Magnitude (or length) of a vector is another vital concept in vector calculus. It signifies the "size" or "length" of the vector and is calculated using the Pythagorean theorem in three dimensions.
For any vector \( \vec{v} = (v_1, v_2, v_3) \), its magnitude \( |\vec{v}| \) is expressed as follows:
  • \( |\vec{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2} \)
Finding the magnitude is essential when working with the dot product, especially when determining angles:
  • For the body diagonal \( \vec{d_b} = (1, 1, 1) \), the magnitude is \( \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \).
  • For the face diagonal \( \vec{d_f} = (1, 1, 0) \), the magnitude is \( \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2} \).
These magnitudes are necessary to solve the cosine of the angle formula using the dot product, showing how much of one vector aligns with another鈥檚 direction.

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Most popular questions from this chapter

[Computer] The differential equation (1.51) for the skateboard of Example 1.2 cannot be solved in terms of elementary functions, but is easily solved numerically. (a) If you have access to software, such as Mathematica, Maple, or Matlab, that can solve differential equations numerically, solve the differential equation for the case that the board is released from \(\phi_{\mathrm{o}}=20\) degrees, using the values \(R=5 \mathrm{m}\) and \(g=9.8 \mathrm{m} / \mathrm{s}^{2} .\) Make a plot of \(\phi\) against time for two or three periods. ( \(\mathbf{b}\) ) On the same picture, plot the approximate solution (1.57) with the same \(\phi_{\mathrm{o}}=20^{\circ}\) Comment on your two graphs. Note: If you haven't used the numerical solver before, you will need to learn the necessary syntax. For example, in Mathematica you will need to learn the syntax for "NDSolve鈥漚nd how to plot the solution that it provides. This takes a bit of time, but is something that is very well worth learning.

By applying Pythagoras's theorem (the usual two-dimensional version) twice over, prove that the length \(r\) of a three-dimensional vector \(\mathbf{r}=(x, y, z)\) satisfies \(r^{2}=x^{2}+y^{2}+z^{2}\)

Prove that the two definitions of the scalar product \(\mathbf{r} \cdot\) s as \(r s \cos \theta(1.6)\) and \(\sum r_{i} s_{i}(1.7)\) are equal. One way to do this is to choose your \(x\) axis along the direction of \(\mathbf{r}\). [Strictly speaking you should first make sure that the definition (1.7) is independent of the choice of axes. If you like to worry about such niceties, see Problem 1.16.]

Imagine two concentric cylinders, centered on the vertical \(z\) axis, with radii \(R \pm \epsilon,\) where \(\epsilon\) is very small. A small frictionless puck of thickness \(2 \epsilon\) is inserted between the two cylinders, so that it can be considered a point mass that can move freely at a fixed distance from the vertical axis. If we use cylindrical polar coordinates \((\rho, \phi, z)\) for its position (Problem 1.47 ), then \(\rho\) is fixed at \(\rho=R,\) while \(\phi\) and \(z\) can vary at will. Write down and solve Newton's second law for the general motion of the puck, including the effects of gravity. Describe the puck's motion.

A particle moves in a circle (center \(O\) and radius \(R\) ) with constant angular velocity \(\omega\) counter-clockwise. The circle lies in the \(x y\) plane and the particle is on the \(x\) axis at time \(t=0 .\) Show that the particle's position is given by $$\mathbf{r}(t)=\hat{\mathbf{x}} R \cos (\omega t)+\hat{\mathbf{y}} R \sin (\omega t)$$ Find the particle's velocity and acceleration. What are the magnitude and direction of the acceleration? Relate your results to well-known properties of uniform circular motion.
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