Problem 29
Question
The volume of a tetrahedron is known to be \(\frac{1}{3}(\) area of base )(height). From this, show that the volume of the tetrahedron with edges \(\mathbf{a}, \mathbf{b}\), and \(\mathbf{c}\) is \(\frac{1}{6}|\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})|\).
Step-by-Step Solution
Verified Answer
The volume is \( \frac{1}{6} |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})| \).
1Step 1: Define the Volume Formula for a Tetrahedron
The formula for the volume of a tetrahedron, having base area and height, is given by \[V = \frac{1}{3}\times \text{Area of Base}\times \text{Height}.\] This formula applies to any tetrahedron. The objective is to express this using vector notation involving the edges \(\mathbf{a}, \mathbf{b}, \text{and} \mathbf{c}\).
2Step 2: Visualize Vectors as Edges of a Parallelepiped
Consider vectors \(\mathbf{b}\) and \(\mathbf{c}\) that form a base of the parallelepiped, while \(\mathbf{a}\) is a vector emanating from the common vertex to height. The tetrahedron in question is half a parallelepiped on this base, so its volume can also be derived from half the volume of the corresponding parallelepiped.
3Step 3: Calculate the Volume of the Parallelepiped
The volume of the parallelepiped formed by \(\mathbf{a}, \mathbf{b}, \text{and} \mathbf{c}\) is given by the scalar triple product of the vectors, which is calculated as the absolute value of the dot product of \(\mathbf{a}\) with the cross product of \(\mathbf{b}\) and \(\mathbf{c}\):\[V_{parallelepiped} = \left| \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \right|.\]
4Step 4: Relate Volume of Tetrahedron to Parallelepiped
The tetrahedron is exactly one-sixth of the entire volume of the parallelepiped (since dividing by 3 gives a prism, and dividing by 2 from the prism gives the tetrahedron). Thus, we have:\[V_{tetrahedron} = \frac{1}{6} V_{parallelepiped} = \frac{1}{6} \left| \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \right|.\] This matches the given expression for the volume.
Key Concepts
Tetrahedron Volume FormulaScalar Triple ProductParallelepiped
Tetrahedron Volume Formula
A tetrahedron is a 3-dimensional shape with four triangular faces. To calculate its volume, we employ a simple formula.The volume of a tetrahedron is expressed as \[V = \frac{1}{3} \times \text{Area of Base} \times \text{Height}.\]In this setup, one of the triangular faces acts as the "base," and the perpendicular distance from this base to the opposite vertex is the "height." To connect this geometric concept with vectors, consider three vectors \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) representing the edges of the tetrahedron emanating from a common vertex.
- The idea is to express the volume using these vectors.
- The vectors encapsulate both the shape and orientation of the tetrahedron.
Scalar Triple Product
The scalar triple product is crucial for calculating volumes in vector calculus, especially for shapes like tetrahedrons and parallelepipeds.Given three vectors \(\mathbf{a}, \mathbf{b}, \mathbf{c}\), the scalar triple product is defined as:\[\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}).\]Let's break it down:
- \(\mathbf{b} \times \mathbf{c}\) is the cross product of \(\mathbf{b}\) and \(\mathbf{c}\), which gives a new vector perpendicular to both.
- The dot product of this new vector with \(\mathbf{a}\) yields a scalar value.
- This scalar, in terms of geometry, represents the signed volume of the parallelepiped formed by the three vectors.
Parallelepiped
A parallelepiped is a 3-dimensional geometric figure that can be seen as a skewed box.To visualize it, imagine a box where each face is a parallelogram instead of a square or rectangle. Each set of opposite faces are parallel.
- Consider vectors \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) as the edges meeting at one vertex.
- The parallelepiped formed is the complete figure enveloped by these edges.
Other exercises in this chapter
Problem 29
, find the curvature \(\kappa\), the unit tangent vector \(\mathbf{T}\), the unit normal vector \(\mathbf{N}\), and the binormal vector \(\mathbf{B}\) at \(t=t_
View solution Problem 29
Consider the curve \(\mathbf{r}(t)=2 t \mathbf{i}+t^{2} \mathbf{j}+\left(1-t^{2}\right) \mathbf{k}\) (a) Show that this curve lies on a plane and find the equat
View solution Problem 29
Find each of the given projections if \(\mathbf{u}=3 \mathbf{i}+2 \mathbf{j}+\mathbf{k}, \mathbf{v}=2 \mathbf{i}-\mathbf{k}\), and \(\mathbf{w}=\mathbf{i}+5 \ma
View solution Problem 29
Find the arc length of the given curve. \(x=t^{2}, y=(4 / 3) t^{3 / 2}, z=t ; 0 \leq t \leq 8\)
View solution