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A parallelepiped (a six-faced solid with opposite faces parallel) has one corner at the origin \(O\) and the three edges that emanate from \(O\) defined by vectors \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) Show that the volume of the parallelepiped is \(|\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})|\)

Short Answer

Expert verified
The volume is found using vector operations: \(|\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})|\).

Step by step solution

01

Understanding the Geometric Definition of Volume

The volume of a parallelepiped is defined as the product of the base area and the height. In the context of vectors, the base formed by vectors \( \mathbf{b} \) and \( \mathbf{c} \) defines a parallelogram, and the height is the perpendicular distance from the tip of vector \( \mathbf{a} \) to the plane containing the base.
02

Calculating the Base Area Using the Cross Product

The magnitude of the cross product \( \mathbf{b} \times \mathbf{c} \) is equal to the area of the parallelogram formed by the vectors \( \mathbf{b} \) and \( \mathbf{c} \). This is given by \(|\mathbf{b} \times \mathbf{c}|\).
03

Determining the Height Using Dot Product

The height of the parallelepiped is the projection of vector \( \mathbf{a} \) onto the normal vector to the base. This normal vector is \( \mathbf{b} \times \mathbf{c} \). The projection, or height, is given by \( |\mathbf{a} \cdot \frac{\mathbf{b} \times \mathbf{c}}{|\mathbf{b} \times \mathbf{c}|}|\).
04

Deriving the Volume Formula

The volume of the parallelepiped is thus \(|\mathbf{a} \cdot \mathbf{n}| \times \text{Base Area}\), where \( \mathbf{n} = \mathbf{b} \times \mathbf{c} \). Plugging in the values, we find the volume to be \( |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})|\) since the height is scalar and already reflects the full magnitude through a dot product.

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Key Concepts

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

Volume Calculation
Calculating the volume of a three-dimensional shape like a parallelepiped can be understood in terms of geometry and vectors. A parallelepiped is a solid where opposite faces are parallel and can be visualized using vectors. In vector calculus, the volume of a parallelepiped with vectors \( \mathbf{a}, \mathbf{b}, \mathbf{c} \) is given by \( |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})| \). This formula is derived from considering the base area and the height.
The base is a parallelogram formed by two vectors, and its area is calculated using the cross product. The height is determined by projecting one vector onto the normal of the base surface. This combination of the base area and height gives us the formula for volume.
Cross Product
The cross product is a vector operation that helps in finding a vector perpendicular to two given vectors. For vectors \( \mathbf{b} \) and \( \mathbf{c} \), the cross product is represented as \( \mathbf{b} \times \mathbf{c} \). This operation results in a new vector that is orthogonal to both \( \mathbf{b} \) and \( \mathbf{c} \).
The magnitude of this cross product vector is equal to the area of the parallelogram formed by the original vectors. Using the right-hand rule, you can find the direction of the resulting vector, which acts as the normal to the plane of the parallelogram.
  • This concept is crucial in determining the base area of a parallelepiped.
  • The formula for its magnitude is \( |\mathbf{b} \times \mathbf{c}| = |\mathbf{b}||\mathbf{c}| \sin(\theta) \), where \( \theta \) is the angle between \( \mathbf{b} \) and \( \mathbf{c} \).
Dot Product
The dot product is another important vector operation that measures how much one vector extends in the direction of another. For vectors \( \mathbf{a} \) and \( \mathbf{n} \) (where \( \mathbf{n} \) is the cross product \( \mathbf{b} \times \mathbf{c} \)), the dot product is represented as \( \mathbf{a} \cdot \mathbf{n} \).
This value helps in calculating the height of the parallelepiped. It projects \( \mathbf{a} \) onto the normal vector \( \mathbf{n} \) and thus, allows us to calculate how tall the parallelepiped stands over its base.
  • The formula for the dot product is \( \mathbf{a} \cdot \mathbf{n} = |\mathbf{a}||\mathbf{n}| \cos(\phi) \), where \( \phi \) is the angle between \( \mathbf{a} \) and \( \mathbf{n} \).
  • It's essential for relating the geometric height to the vector context.
Parallelepiped
A parallelepiped is a geometric figure consisting of six faces (all parallelograms), where opposite faces are parallel. It can be visualized using vectors \( \mathbf{a}, \mathbf{b}, \mathbf{c} \), which form the edges emanating from one vertex.
The understanding of this shape in vector calculus revolves around how these vectors combine to form the space inside the shape. With one vertex at the origin, these vectors define the edges of the shape.
  • The volume formula \( |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})| \) comes from the need to calculate a three-dimensional scalene volume based on the orientation and length of these vectors.
  • Visualizing a parallelepiped helps in understanding why cross and dot products are used to analyze its volume.

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Most popular questions from this chapter

Prove that if \(\mathbf{v}(t)\) is any vector that depends on time (for example the velocity of a moving particle) but which has constant magnitude, then \(\dot{\mathbf{v}}(t)\) is orthogonal to \(\mathbf{v}(t) .\) Prove the converse that if \(\dot{\mathbf{v}}(t)\) is orthogonal to \(\mathbf{v}(t),\) then \(|\mathbf{v}(t)|\) is constant. [Hint: Consider the derivative of \(\mathbf{v}^{2}\).] This is a very handy result. It explains why, in two- dimensional polars, \(d \hat{\mathbf{r}} / d t\) has to be in the direction of \(\hat{\boldsymbol{\phi}}\) and vice versa. It also shows that the speed of a charged particle in a magnetic field is constant, since the acceleration is perpendicular to the velocity.

[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”and how to plot the solution that it provides. This takes a bit of time, but is something that is very well worth learning.

Verify by direct substitution that the function \(\phi(t)=A \sin (\omega t)+B \cos (\omega t)\) of (1.56) is a solution of the second-order differential equation \((1.55), \ddot{\phi}=-\omega^{2} \phi .\) (since this solution involves two arbitrary constants - the coefficients of the sine and cosine functions \(-\) it is in fact the general solution.)

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 \(r^{2}=x^{2}+y^{2}+z^{2}\)

The position of a moving particle is given as a function of time \(t\) to be $$\mathbf{r}(t)=\hat{\mathbf{x}} b \cos (\omega t)+\hat{\mathbf{y}} c \sin (\omega t)+\hat{\mathbf{z}} v_{\mathrm{o}} t$$ where \(b, c, v_{\mathrm{o}}\) and \(\omega\) are constants. Describe the particle's orbit.

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