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Chapter 9: Problem 10

Find the angle between the diagonal of a cube of side length 8 and the diagonal of one of its faces. The angle should be measured in radians.

Short Answer

Expert verified
The angle between the cube's diagonal and the face diagonal, measured in radians, is \(\theta = \arccos\left(\frac{2}{\sqrt{6}}\right)\).

Step by step solution

01

Find the length of the cube's diagonal.

To find the length of the cube's diagonal (d), use the Pythagorean theorem in three dimensions. The cube has side length l = 8. \[d = \sqrt{l^2 + l^2 + l^2} \\ d = \sqrt{3l^2}\] Substitute the value of l: \[d = \sqrt{3(8)^2} \\ d = \sqrt{3(64)} \\ d = 8\sqrt{3}\] The length of the cube's diagonal is \(8\sqrt{3}\).
02

Find the length of the face diagonal.

Similarly, we can find the face diagonal's length (f) using the Pythagorean theorem in two dimensions. We already have the side length (l = 8). \[f = \sqrt{l^2 + l^2} \\ f = \sqrt{2l^2}\] Substitute the value of l: \[f = \sqrt{2(8)^2} \\ f = \sqrt{2(64)} \\ f = 8\sqrt{2}\] The length of the face diagonal is \(8\sqrt{2}\).
03

Calculate the dot product between the two diagonals.

The dot product between the cube's diagonal vector (A) and the face diagonal vector (B) is given by the formula: \[A \cdot B = |A||B| \cos{\theta}\] where \(\theta\) is the angle between the vectors A and B. The cube's diagonal vector A can be written as (8, 8, 8) and the face diagonal vector B as (8, 8, 0). Calculate the dot product: \[A \cdot B = (8)(8) + (8)(8) + (8)(0) = 128\]
04

Find the angle between the two diagonals using the dot product formula.

Using the dot product formula and plugging in the lengths and dot product: \[128 = (8\sqrt{3})(8\sqrt{2})\cos{\theta}\] Divide both sides by the magnitudes (in this case, \(8\sqrt{3}\) and \(8\sqrt{2}\)): \[\cos{\theta} = \frac{128}{(8\sqrt{3})(8\sqrt{2})}\] \[cos{\theta} = \frac{128}{64\sqrt{6}}\] \[cos{\theta} = \frac{2}{\sqrt{6}}\] Now, find the angle \(\theta\): \[\theta = \arccos\left(\frac{2}{\sqrt{6}}\right)\] The angle between the cube's diagonal and the face diagonal, measured in radians, is: \[\theta = \arccos\left(\frac{2}{\sqrt{6}}\right)\]

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Key Concepts

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

Cube Diagonal
Understanding the concept of a cube diagonal is much simpler than it may initially seem. A cube is a three-dimensional shape with equal side lengths. Imagine a line stretching from one corner of the cube to the opposite corner that does not lie on the same face. This line is the diagonal of the cube.
To find the length of this diagonal, we apply the three-dimensional version of the Pythagorean theorem. The formula is:
  • First, consider one side of the cube as a right-angle triangle with three edges forming it: length, width, and height (all equal to the side of the cube, let's call this \( l \)).
  • The diagonal is the hypotenuse of this right triangle.
  • According to the Pythagorean theorem in 3D: \[ d = \sqrt{l^2 + l^2 + l^2} = \sqrt{3l^2} = l\sqrt{3} \]
So for a cube with side length 8, the length of the diagonal becomes \( 8 \sqrt{3} \).
Understanding this helps in multiple geometry problems where you need to calculate distances within 3D shapes.
Pythagorean Theorem
The Pythagorean theorem is an essential principle of geometry used for determining the length of sides in right triangles. The simplest formule is expressed as \( a^2 + b^2 = c^2 \), where \( c \) is the hypotenuse. However, let's explore how it applies in more dimensions like in a cube.
  • In two dimensions, this theorem helps us determine the diagonal of a face of a cube (essentially a square).
  • For the face of a cube, if each side is of length \( l \), the diagonal \( f \) can be computed as: \[ f = \sqrt{l^2 + l^2} = \sqrt{2l^2} = l\sqrt{2} \]
  • In three dimensions, as seen with the cube diagonal, all three dimensions are considered, helping to find the body diagonal of the cube.
The Pythagorean theorem truly extends its versatility from simple 2D applications to more complex 3D contexts.
Dot Product
The dot product is a valuable operation in vector mathematics that allows us to find an angle between two vectors. It also helps to determine if two vectors are perpendicular or parallel. When working with diagonals of a cube, the dot product comes into play significantly.
  • For two vectors \( \mathbf{A} \) and \( \mathbf{B} \), the dot product is defined as: \[ \mathbf{A} \cdot \mathbf{B} = |A||B|\cos{\theta} \] where \( \theta \) is the angle between them.
  • To compute the dot product of the cube's diagonal \( (8,8,8) \) with the face diagonal \( (8,8,0) \), arithmetic simplifies to: \[ 128 = 8 \cdot 8 + 8 \cdot 8 + 8\cdot 0 \]
Finding the dot product and understanding its formula is crucial to solve problems related to angles between vectors in any geometric setting.
Vector Angles
When analyzing angles between vectors, particularly in three-dimensional space, understanding vector angles becomes essential. This applies to a cube where you examine the angle between diagonals. Calculating vector angles through the dot product is not only insightful but straightforward.
  • The formula \( A \cdot B = |A||B|\cos{\theta} \) is used to find the angle \( \theta \) between two vectors.
  • By isolating \( \cos{\theta} \), determine: \[ \cos{\theta} = \frac{A \cdot B}{|A||B|} \] which helps in evaluating \( \theta \).
  • Subsequently, utilize the inverse cosine function, \( \arccos \), to find the angle in radians.
Using these calculations ensures a deeper comprehension of spatial relationships and how angles are computed in various geometric problems.

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Most popular questions from this chapter

A force (like gravity) has both a magnitude and a direction. If two forces \(\mathbf{u}\) and \(\mathbf{v}\) are applied to an object at the same point, the resultant force on the object is the vector sum of the two forces. When a force is applied by a rope or a cable, we call that force tension. Vectors can be used to determine tension. As an example, suppose a painting weighing 50 pounds is to be hung from wires attached to the frame as illustrated in Figure \(9.2 .10 .\) We need to know how much tension will be on the wires to know what kind of wire to use to hang the picture. Assume the wires are attached to the frame at point \(O\). Let \(\mathbf{u}\) be the vector emanating from point \(O\) to the left and \(\mathbf{v}\) the vector emanating from point \(O\) to the right. Assume \(\mathbf{u}\) makes a \(60^{\circ}\) angle with the horizontal at point \(O\) and \(\mathbf{v}\) makes a \(45^{\circ}\) angle with the horizontal at point \(O\). Our goal is to determine the vectors \(\mathbf{u}\) and \(\mathbf{v}\) in order to calculate their magnitudes. a. Treat point \(O\) as the origin. Use trigonometry to find the components \(u_{1}\) and \(u_{2}\) so that \(\mathbf{u}=u_{1} \mathbf{i}+u_{2} \mathbf{j} .\) Since we don't know the magnitude of \(\mathbf{u},\) your components will be in terms of \(|\mathbf{u}|\) and the cosine and sine of some angle. Then find the components \(v_{1}\) and \(v_{2}\) so that \(\mathbf{v}=v_{1} \mathbf{i}+v_{2} \mathbf{j}\). Again, your components will be in terms of \(|\mathbf{v}|\) and the cosine and sine of some angle. b. The total force holding the picture up is given by \(\mathbf{u}+\mathbf{v}\). The force acting to pull the picture down is given by the weight of the picture. Find the force vector \(\mathbf{w}\) acting to pull the picture down. c. The picture will hang in equilibrium when the force acting to hold it up is equal in magnitude and opposite in direction to the force acting to pull it down. Equate these forces to find the components of the vectors \(\mathbf{u}\) and \(\mathbf{v}\).

(a) Describe the set of points whose distance from the x-axis equals the distance from the yz-plane. \(\odot\) A cylinder opening along the \(x\) -axis \(\odot\) A cone opening along the x-axis \(\odot\) A cone opening along the z-axis \(\odot\) A cylinder opening along the z-axis \(\odot\) A cone opening along the y-axis \(\odot\) A cylinder opening along the y-axis (b) Find the equation for the set of points whose distance from the x-axis equals the distance from the yz-plane. \(\odot x^{2}+y^{2}=r^{2}\) \(\odot x^{2}+z^{2}=r^{2}\) \(\odot y^{2}=x^{2}+z^{2}\) (\odot) \(y^{2}+z^{2}=r^{2}\) \(\odot z^{2}=x^{2}+y^{2}\) \(\odot x^{2}=y^{2}+z^{2}\)

Consider the function \(h\) defined by \(h(x, y)=8-\sqrt{4-x^{2}-y^{2}}\). a. What is the domain of \(h ?\) (Hint: describe a set of ordered pairs in the plane by explaining their relationship relative to a key circle.) b. The range of a function is the set of all outputs the function generates. Given that the range of the square root function \(g(t)=\sqrt{t}\) is the set of all nonnegative real numbers, what do you think is the range of \(h ?\) Why? c. Choose 4 different values from the range of \(h\) and plot the corresponding level curves in the plane. What is the shape of a typical level curve? d. Choose 5 different values of \(x\) (including at least one negative value and zero), and sketch the corresponding traces of the function \(h\). e. Choose 5 different values of \(y\) (including at least one negative value and zero), and sketch the corresponding traces of the function \(h\). f. Sketch an overall picture of the surface generated by \(h\) and write at least one sentence to describe how the surface appears visually. Does the surface remind you of a familiar physical structure in nature?

Use the geometric definition of the cross product and the properties of the cross product to make the following calculations. (a) \(((\vec{i}+\vec{j}) \times \vec{i}) \times \vec{j}=\)_________ (b) \((\vec{j}+\vec{k}) \times(\vec{j} \times \vec{k})=\)_________ (c) \(5 \vec{i} \times(\vec{i}+\vec{j})=\)_________ (d) \((\vec{k}+\vec{j}) \times(\vec{k}-\vec{j})=\)_________

By setting one variable constant, find a plane that intersects the graph of \(z=6 x^{2}-3 y^{2}+3\) in a: (a) Parabola opening upward: the plane ________=_________ (Give your answer by specifying the variable in the first answer blank and a value for it in the second.) (b) Parabola opening downward: the plane ________=_______ (Give your answer by specifying the variable in the first answer blank and a value for it in the second.) (c) Pair of intersecting straight lines: the plane _________=_________ (Give your answer by specifying the variable in the first answer blank and a value for it in the second.)
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