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Chapter 11: Problem 29

Consider the parallelepiped with adjacent edges $$ \begin{array}{l}{\mathbf{u}=3 \mathbf{i}+2 \mathbf{j}+\mathbf{k}} \\\ {\mathbf{v}=\mathbf{i}+\mathbf{j}+2 \mathbf{k}} \\ {\mathbf{w}=\mathbf{i}+3 \mathbf{j}+3 \mathbf{k}}\end{array} $$ (a) Find the volume. (b) Find the area of the face determined by \(\mathbf{u}\) and \(\mathbf{w}\). (c) Find the angle between \(\mathbf{u}\) and the plane containing the face determined by \(\mathbf{v}\) and \(\mathbf{w} .\)

Short Answer

Expert verified
(a) Volume: 7 units鲁. (b) Area: \(\sqrt{122}\) units虏. (c) Angle: \(\cos^{-1}\left(\frac{7}{\sqrt{140}}\right)\).

Step by step solution

01

Calculate the Volume of the Parallelepiped

The volume of a parallelepiped with vectors \( \mathbf{u} \), \( \mathbf{v} \), and \( \mathbf{w} \) as adjacent edges is given by the scalar triple product \( \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) \). 1. First, compute the cross product \( \mathbf{v} \times \mathbf{w} \). Set up the determinant:\[\mathbf{v} \times \mathbf{w} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 1 & 2 \ 1 & 3 & 3 \end{vmatrix}\]Solving this, we have:\[\mathbf{v} \times \mathbf{w} = (1\cdot3 - 2\cdot3)\mathbf{i} - (1\cdot3 - 2\cdot1)\mathbf{j} + (1\cdot3 - 1\cdot3)\mathbf{k} = -3\mathbf{i} + \mathbf{j} + 0\mathbf{k}\]2. Now, take the dot product of \( \mathbf{u} \) and \( \mathbf{v} \times \mathbf{w} \):\[\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = (3\mathbf{i} + 2\mathbf{j} + \mathbf{k}) \cdot (-3\mathbf{i} + \mathbf{j}) = 3(-3) + 2(1) = -9 + 2 = -7\]3. The volume of the parallelepiped is 7 cubic units (taking the absolute value).
02

Calculate the Area of the Face Determined by \(\mathbf{u}\) and \(\mathbf{w}\)

The area of a parallelogram formed by two vectors \( \mathbf{u} \) and \( \mathbf{w} \) can be found using the magnitude of their cross product:Set up the determinant to find \( \mathbf{u} \times \mathbf{w} \):\[\mathbf{u} \times \mathbf{w} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 3 & 2 & 1 \ 1 & 3 & 3 \end{vmatrix}\]Solving this, we have:\[\mathbf{u} \times \mathbf{w} = ((2\cdot3 - 1\cdot3)\mathbf{i} - (3\cdot3 - 1\cdot1)\mathbf{j} + (3\cdot3 - 2\cdot1)\mathbf{k}) = 3\mathbf{i} - 8\mathbf{j} + 7\mathbf{k}\]The magnitude of this vector is:\[|\mathbf{u} \times \mathbf{w}| = \sqrt{3^2 + (-8)^2 + 7^2} = \sqrt{9 + 64 + 49} = \sqrt{122}\]The area of the face is \( \sqrt{122} \) square units.
03

Calculate the Angle Between \(\mathbf{u}\) and the Plane Containing \(\mathbf{v}\) and \(\mathbf{w}\)

The angle between a vector \( \mathbf{u} \) and a plane defined by vectors \( \mathbf{v} \) and \( \mathbf{w} \) can be found using the normal vector of the plane, which is \( \mathbf{n} = \mathbf{v} \times \mathbf{w} \).Using the cross product from Step 1, \( \mathbf{n} = -3\mathbf{i} + \mathbf{j} \), let's calculate the angle:\[\cos(\theta) = \frac{|\mathbf{u} \cdot \mathbf{n}|}{|\mathbf{u}| \cdot |\mathbf{n}|}\]where \( \mathbf{u} = 3\mathbf{i} + 2\mathbf{j} + \mathbf{k} \):\( |\mathbf{u}| = \sqrt{3^2 + 2^2 + 1^2} = \sqrt{14} \)\( |\mathbf{n}| = \sqrt{(-3)^2 + 1^2 + 0^2 } = \sqrt{10} \)Compute \( \mathbf{u} \cdot \mathbf{n} = (3\mathbf{i} + 2\mathbf{j} + \mathbf{k}) \cdot (-3\mathbf{i} + \mathbf{j}) = -9 + 2 = -7 \)Thus, \( \cos(\theta) = \frac{|-7|}{\sqrt{14} \times \sqrt{10}} = \frac{7}{\sqrt{140}} \)\(\theta = \cos^{-1}\left(\frac{7}{\sqrt{140}}\right)\) and solving gives the angle.

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Key Concepts

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

Scalar Triple Product
The scalar triple product of vectors is a useful tool in vector calculus. It can determine the volume of a parallelepiped with three vectors as its adjacent edges. The scalar triple product is represented by the formula \( \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) \). Here, \( \mathbf{u} \), \( \mathbf{v} \), and \( \mathbf{w} \) are vectors representing the edges of the parallelepiped. This operation produces a scalar quantity, which can be interpreted as the volume of the parallelepiped.
  • The structure of this formula includes both a cross product and a dot product, combining two essential vector operations.
  • In the context of vector spaces, the magnitude of the scalar triple product is equal to the absolute volume of the parallelepiped described.
The beauty of this approach lies in its simplicity and directness, allowing for an efficient calculation of the volume. It's important to remember that, although sometimes negative, the volume is always taken as a positive value by noting its absolute magnitude.
Cross Product
The cross product of two vectors gives a vector that is perpendicular to the plane containing the initial vectors. In three-dimensional space, this is particularly useful for calculations like those needed for determining the volume of a parallelepiped or the area of a parallelogram.
  • The cross product \( \mathbf{v} \times \mathbf{w} \) represents a vector that is orthogonal to both \( \mathbf{v} \) and \( \mathbf{w} \).
  • The magnitude of this resultant vector represents the area of the parallelogram spanned by the initial two vectors.
To compute a cross product, you can use determinants. These determinants involve solving a matrix set up with the unit vectors \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) in the first row, and the components of the two vectors in the subsequent rows. The cross product is fundamental in various applications such as physics and engineering, where it often determines torque or rotational dynamics.
Vector Dot Product
The vector dot product, or simply the dot product, results in a scalar quantity. It measures the extent to which two vectors point in the same direction. In the problem of finding the angle between a vector and a plane, the dot product is essential.
  • Given two vectors \( \mathbf{u} \) and \( \mathbf{v} \), the dot product is calculated as \( \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3 \), where \( u_i \) and \( v_i \) are the components of the vectors.
  • It is also defined as the product of the magnitudes of the vectors and the cosine of the angle between them: \( \mathbf{u} \cdot \mathbf{v} = |\mathbf{u}| |\mathbf{v}| \cos(\theta) \).
In the context of this exercise, the dot product allows us to calculate the angle between \( \mathbf{u} \) and the plane by breaking down vectors and helping find their relative orientation. It greatly facilitates understanding vector projections and interactions in multidimensional spaces.
Magnitude of a Vector
The magnitude of a vector is akin to its length or size. When performing calculations involving vectors, knowing their magnitude is essential for both theoretical understanding and practical application.
  • The magnitude of a vector \( \mathbf{a} \) with components \( (a_1, a_2, a_3) \) is given by \( |\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \).
  • This formula is derived from the Pythagorean theorem, adapted into three dimensions.
In this problem, the magnitude helps in determining not only the length of individual vectors but also in calculating areas, volumes, and angles. For instance, when finding the angle between a vector and a plane, the magnitudes of the involved vectors are crucial to apply trigonometric concepts accurately. Understanding vector magnitudes allows for clear visualization of vector properties and spatial relationships.

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Most popular questions from this chapter

An equation is given in cylindrical coordinates. Express the equation in rectangular coordinates and sketch the graph. $$ r=2 \sec \theta $$

A vector w is a linear combination of the vectors \(\mathbf{v}_{1}, \mathbf{v}_{2},\) and \(\mathbf{v}_{3}\) if \(\mathbf{w}\) can be expressed as \(\mathbf{w}=c_{1} \mathbf{v}_{1}+c_{2} \mathbf{v}_{2}+c_{3} \mathbf{v}_{3},\) where \(c_{1}, c_{2},\) and \(c_{3}\) are scalars. (a) Find scalars \(c_{1}, c_{2},\) and \(c_{3}\) to express \(\langle- 1,1,5\rangle\) as a linear combination of \(\mathbf{v}_{1}=\langle 1,0,1\rangle, \mathbf{v}_{2}=\langle 3,2,0\rangle,\) and \(\mathbf{v}_{3}=\langle 0,1,1\rangle\) (b) Show that the vector \(2 \mathbf{i}+\mathbf{j}-\mathbf{k}\) cannot be expressed as a linear combination of \(\mathbf{v}_{1}=\mathbf{i}-\mathbf{j}, \mathbf{v}_{2}=3 \mathbf{i}+\mathbf{k},\) and \(\mathbf{v}_{3}=4 \mathbf{i}-\mathbf{j}+\mathbf{k} .\)k} .$

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