/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 194 (a) Let \(A B C D\) be a regular... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(a) Let \(A B C D\) be a regular tetrahedron with edges of length 2. Calculate the (exact) angle between the two faces \(A B C\) and \(D B C\). (b) We know that in \(2 \mathrm{D}\) five equilateral triangles fit together at a point leaving just enough of an angle to allow a sixth triangle to fit. How many identical regular tetrahedra can one fit together, without overlaps around an edge, so that they all share the edge \(\underline{B C}\) (say)?

Short Answer

Expert verified
(a) The angle is \(70.53^\circ\). (b) Three tetrahedra fit around an edge.

Step by step solution

01

Understand Regular Tetrahedron

A regular tetrahedron is a polyhedron composed of four equilateral triangular faces, six equal edges, and four vertices. In this problem, you are interested in two faces: \(ABC\) and \(DBC\).
02

Geometry of the Tetrahedron

For a regular tetrahedron, the length of each edge is 2 as given. We need vectors that form the edges of faces \(ABC\) and \(DBC\). Assume vertex \(A\) is at \((1, 1, 1)\), \(B\) at \((1, 0, 0)\), \(C\) at \((0, 1, 0)\), and \(D\) at \((0, 0, 1)\).
03

Calculate Normal Vectors to Faces

The face \(ABC\) can be determined by vectors \( \mathbf{AB} = (0, -1, -1) \) and \( \mathbf{AC} = (-1, 0, -1) \). Using the cross product, the normal to \(ABC\) is \(\mathbf{N}_{ABC} = (1, 1, 1)\). Similarly, for face \(DBC\), the vectors are \( \mathbf{DB} = (1, 0, -1) \) and \( \mathbf{DC} = (0, 1, -1) \) with normal \(\mathbf{N}_{DBC} = (1, 1, 1) \).
04

Determine Angle Between Planes

The angle \(\theta\) between the normals \(\mathbf{N}_{ABC}\) and \(\mathbf{N}_{DBC}\) is calculated using the dot product: \( \cos \theta = \frac{\mathbf{N}_{ABC} \cdot \mathbf{N}_{DBC}}{\|\mathbf{N}_{ABC}\| \|\mathbf{N}_{DBC}\|} \). This results in \( \cos \theta = \frac{3}{\sqrt{3} \times \sqrt{3}} = \frac{1}{\sqrt{3}} \), so \( \theta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) = 70.53^\circ \).
05

Understand 2D Fit of Equilateral Triangles

In 2D, five equilateral triangles around a common point leave an angle of \(60^\circ\). A sixth can fit because \(30^\circ\) is an equilateral triangle angle.
06

3D Fit of Regular Tetrahedra Around Edge

For regular tetrahedra, more than three tetrahedra cannot fit perfectly around an edge as each face makes an angle of \(70.53^\circ\), totaling \(211.59^\circ\) for three and leaving slightly more than the sum of the dihedral angle of one more, which is a fit for three around an edge, "almost" closing with exactly zero gap.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Regular Tetrahedron
A regular tetrahedron is a type of polyhedron that consists of four equilateral triangular faces. This means that all sides of each triangle are equal, resulting in a very symmetric and balanced 3D shape. In a regular tetrahedron, there are six equal edges and four vertices. Each vertex is where three edges meet. This particular property gives the tetrahedron a unique place among other polyhedral shapes because of its uniformity and simplicity.

In this specific problem, we are dealing with a regular tetrahedron named \(ABCD\) with all edges measuring 2 units. The focus is on calculating the angle between the two triangular faces \(ABC\) and \(DBC\). Understanding the symmetry and equal measurement of the edges is crucial, as it allows for uniform calculations and consistent application in geometric problems. Regular tetrahedrons are a staple in geometry due to their predictability and ease of calculations when determining angles and other properties.
Dihedral Angle Calculation
Dihedral angles occur between two intersecting planes, which in this case are the faces of a regular tetrahedron. In simple terms, it's the 'twist' or the 'tilt' from one face to another adjacent face.

For the tetrahedron \(ABCD\), we are calculating the angle between the faces \(ABC\) and \(DBC\). This can be done by finding the normal vectors of each face and then working with these vectors to find the angle. Normal vectors are perpendicular to the surface they originate from and give a clear sense of the direction of a face.

To calculate the dihedral angle \(\theta\), the vectors forming the faces must be known. In this problem, assume \(A\) is at \((1, 1, 1)\), \(B\) at \((1, 0, 0)\), \(C\) at \((0, 1, 0)\), and \(D\) at \((0, 0, 1)\). The normal vectors can be calculated using the cross product technique, which will be detailed in the next section. Once the normals are known, the dot product helps in finding the cosine of the angle by applying the formula: \[ \cos \theta = \frac{\mathbf{N}_{ABC} \cdot \mathbf{N}_{DBC}}{\|\mathbf{N}_{ABC}\| \|\mathbf{N}_{DBC}\|} \] resulting in \(\theta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right)\), which gives approximately \(70.53^\circ\). Understanding this calculation is vital as it demonstrates the application of vector operations in evaluating geometric properties.
Vector Cross Product
The vector cross product is a mathematical operation on two vectors that results in another vector, which is orthogonal to both original vectors. This is incredibly useful in geometry, especially for finding normal vectors of a plane.

In the problem, we need to find normal vectors for the faces \(ABC\) and \(DBC\). For face \(ABC\), vectors \( \mathbf{AB} = (0, -1, -1) \) and \( \mathbf{AC} = (-1, 0, -1) \) are used. The vector cross product is calculated by taking the determinant of a 3x3 matrix that includes the unit vectors \(i, j, k\) and the components of the vectors. The resulting normal vector for \(ABC\) is \(\mathbf{N}_{ABC} = (1, 1, 1)\).

Similarly, to find the normal for face \(DBC\), use \( \mathbf{DB} = (1, 0, -1) \) and \( \mathbf{DC} = (0, 1, -1) \), resulting in another normal vector \(\mathbf{N}_{DBC} = (1, 1, 1)\). This cross product technique is instrumental in calculations that involve plane angles, as it provides the necessary tools to handle and interpret 3-dimensional data efficiently.
3D Geometric Configurations
Three-dimensional geometric configurations involve understanding how shapes like tetrahedrons fit together in space. It’s the art of navigating complex shapes and their interactions in a 3D environment.

In 3D configurations concerning regular tetrahedra, there’s an interesting spatial limitation. When you try to fit multiple tetrahedra around a common edge, you encounter constraints dictated by their dihedral angles. In a 3D space, using the given angle of approximately \(70.53^\circ\) between faces, we gather that only three identical tetrahedra can align precisely around a shared edge, like \(BC\).

This is because the total angle around the edge accumulates to \(211.59^\circ\) when three tetrahedra are combined, with each contributing its dihedral angle. Such an arrangement closes with almost zero gap, effectively fitting like puzzle pieces. However, adding a fourth would surpass the circle's \(360^\circ\), causing overlap. Understanding these spatial relationships is vital for mastering the fit and symmetry of 3D geometric shapes, which can apply in numerous fields, from architecture to molecular chemistry.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(A B C D\) be a parallelogram. (i) Prove that \(\triangle A B C \equiv \triangle C D A,\) so that each triangle has area exactly half of area \((A B C D)\) (ii) Conclude that opposite sides of \(A B C D\) are equal in pairs and that opposite angles are equal in pairs. (iii) Let \(A C\) and \(B D\) meet at \(X\). Prove that \(X\) is the midpoint of both \(\underline{A C}\) and \(\underline{B D} .\)

Prove that the only regular polyhedron that tiles \(3 \mathrm{D}\) (without gaps or overlaps) is the cube.

(a) Let \(A B C D E F\) be a regular octahedron with vertices \(B, C, D, E\) adjacent to \(A\) forming a square \(B C D E,\) and with edges of length \(2 .\) Calculate the (exact) angle between the two faces \(A B C\) and \(F B C\). (b) How many identical regular octahedra can one fit together around an edge, without overlaps, so that they all share the edge \(B C\) (say)?

(a) A regular tetrahedron \(A B C D\) has edges of length \(2,\) and sits with its base \(B C D\) on the table. Find the height of \(A\) above the base. (b) A regular octahedron \(A B C D E F\) has four triangles meeting at each vertex. (i) Let the four triangles which meet at \(A\) be \(A B C, A C D, A D E, A E B\). Prove that \(B C D E\) must be a square. (ii) Suppose that all the triangles have edges of length \(2,\) and that the octahedron sits with one face \(B C F\) on the table \(-\) next to the regular tetrahedron from part (a). Which of these two solids is the taller? \(\triangle\)

A vertex figure is to be formed by fitting regular \(p\) -gons together, edge- to-edge, for a fixed \(p .\) If there are \(q\) of these \(p\) -gons at a vertex, we denote the vertex figure by \(p^{q} .\) If the angles at each vertex add to less than \(360^{\circ}\), prove that the only possible vertex figures are \(3^{3}, 3^{4}, 3^{5}\), \(4^{3}, 5^{3}\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.