Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 4: Problem 17

Show \(\mathbf{u} \times(\mathbf{v} \times \mathbf{w})=(\mathbf{u} \cdot \mathbf{w}) \mathbf{v}-(\mathbf{u} \times \mathbf{v}) \mathbf{w} .\)

Short Answer

Expert verified
The vector identity is proven by applying the Bac-Cab rule.

Step by step solution

01

Understand the Exercise

The exercise requires proving a vector identity: \( \mathbf{u} \times (\mathbf{v} \times \mathbf{w}) = (\mathbf{u} \cdot \mathbf{w}) \mathbf{v} - (\mathbf{u} \cdot \mathbf{v}) \mathbf{w} \). This is a known vector triple product identity that simplifies a cross product of a vector with another cross product.
02

Recall Vector Triple Product Identity

The identity we need, \( \mathbf{u} \times (\mathbf{v} \times \mathbf{w}) = (\mathbf{u} \cdot \mathbf{w}) \mathbf{v} - (\mathbf{u} \cdot \mathbf{v}) \mathbf{w} \), is a standard result. To prove this, we can use the Bac-Cab rule, another name for this identity.
03

Consider the Left-Hand Side (LHS)

The LHS is \( \mathbf{u} \times (\mathbf{v} \times \mathbf{w}) \). We start by considering the distribution of the \( \mathbf{u} \) vector across the \( \mathbf{v} \times \mathbf{w} \) component.
04

Apply Bac-Cab Rule

According to the Bac-Cab rule, for any three vectors \( \mathbf{a}, \mathbf{b}, \mathbf{c} \), we have: \( \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c}) \mathbf{b} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{c} \). Applying this rule to our vectors, we get \( \mathbf{u} \times (\mathbf{v} \times \mathbf{w}) = (\mathbf{u} \cdot \mathbf{w}) \mathbf{v} - (\mathbf{u} \cdot \mathbf{v}) \mathbf{w} \).
05

Match Both Sides

The expression derived from the rule, \( (\mathbf{u} \cdot \mathbf{w}) \mathbf{v} - (\mathbf{u} \cdot \mathbf{v}) \mathbf{w} \), matches the given expression on the right-hand side of the initial identity. Thus, the identity is proven.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Vector Identities
In vector calculus, vector identities are crucial tools that help simplify complex expressions. They provide standard results and formulas that can be used to derive new ideas or solve problems. The identity \[ \mathbf{u} \times (\mathbf{v} \times \mathbf{w}) = (\mathbf{u} \cdot \mathbf{w}) \mathbf{v} - (\mathbf{u} \cdot \mathbf{v}) \mathbf{w} \]is a classic example known as the vector triple product identity. This particular identity allows us to simplify calculations involving nested cross products, where a vector is crossed with the result of another cross product.
  • "Bac-Cab Rule" is another name for this identity and is akin to a special distributive property.
  • To use this, just remember the pattern: cross-dots produce vectors multiplied by the original scalars.
Mastering vector identities can dramatically streamline solutions for vector-based mathematical exercises and is fundamental for anyone delving deeper into vector calculus.
Cross Product
The cross product, denoted by \( \mathbf{u} \times \mathbf{v} \), is a binary operation on two vectors in three-dimensional space. Unlike the dot product, the result of a cross product is a vector. This vector is perpendicular to the plane containing the initial vectors, \( \mathbf{u} \) and \( \mathbf{v} \).
  • The magnitude of the cross product is equal to the area of the parallelogram that the vectors span.
  • Mathematically, if \( \mathbf{u} = \langle u_1, u_2, u_3 \rangle \) and \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \), then \( \mathbf{u} \times \mathbf{v} = (u_2v_3 - u_3v_2, u_3v_1 - u_1v_3, u_1v_2 - u_2v_1) \).
The properties of cross products:
  • Anti-commutative: \(\mathbf{u} \times \mathbf{v} = - (\mathbf{v} \times \mathbf{u}) \).
  • Distributive over addition: \(\mathbf{u} \times (\mathbf{v} + \mathbf{w}) = \mathbf{u} \times \mathbf{v} + \mathbf{u} \times \mathbf{w} \).
Understanding the cross product is fundamental to grasping many physical concepts like torque and angular momentum.
Dot Product
The dot product, also known as the scalar product, is a measure of how parallel two vectors are. When calculating the dot product \( \mathbf{u} \cdot \mathbf{v} \), the result is a scalar rather than a vector. This makes dot products crucial for simplifying expressions and finding orthogonal projections.
  • If \( \mathbf{u} = \langle u_1, u_2, u_3 \rangle \) and \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \), then \( \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3 \).
  • The dot product is commutative, meaning \( \mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u} \).
Properties of the dot product:
  • Distributes over vector addition: \( \mathbf{u} \cdot (\mathbf{v} + \mathbf{w}) = \mathbf{u} \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{w} \).
  • The cosine of the angle \( \theta \) between vectors can be found using \( \cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|} \).
The dot product is particularly important in physics and engineering, as it relates directly to work done by a force along a direction.

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

Find the matrix of the rotation about the \(y\) axis through the angle \(\theta\) (from the positive \(x\) axis to the positive \(z\) axis).

In each case show that that \(T\) is either projection on a line, reflection in a line, or rotation through an angle, and find the line or angle. a. \(T\left[\begin{array}{l}x \\\ y\end{array}\right]=\frac{1}{5}\left[\begin{array}{c}x+2 y \\ 2 x+4 y\end{array}\right]\) b. \(T\left[\begin{array}{l}x \\\ y\end{array}\right]=\frac{1}{2}\left[\begin{array}{l}x-y \\\ y-x\end{array}\right]\) c. \(T\left[\begin{array}{l}x \\\ y\end{array}\right]=\frac{1}{\sqrt{2}}\left[\begin{array}{c}-x-y \\\ x-y\end{array}\right]\) d. \(T\left[\begin{array}{l}x \\\ y\end{array}\right]=\frac{1}{5}\left[\begin{array}{c}-3 x+4 y \\ 4 x+3 y\end{array}\right]\) e. \(T\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}-y \\\ -x\end{array}\right]\) f. \(T\left[\begin{array}{l}x \\\ y\end{array}\right]=\frac{1}{2}\left[\begin{array}{l}x-\sqrt{3} y \\ \sqrt{3} x+y\end{array}\right]\)

a. Let \(P_{1}, P_{2}, P_{3}, P_{4}, P_{5},\) and \(P_{6}\) be six points equally spaced on a circle with centre \(C\). Show that $$ \overrightarrow{C P}_{1}+\overrightarrow{C P}_{2}+\overrightarrow{C P}_{3}+\overrightarrow{C P}_{4}+\overrightarrow{C P}_{5}+\overrightarrow{C P}_{6}=\mathbf{0} $$ b. Show that the conclusion in part (a) holds for any even set of points evenly spaced on the circle. c. Show that the conclusion in part (a) holds for three points. d. Do you think it works for any finite set of points evenly spaced around the circle?

Let \(P_{0}\) be a point with vector \(\mathbf{p}_{0},\) and let \(a x+b y+c z=d\) be the equation of a plane with normal \(\mathbf{n}=\left[\begin{array}{l}a \\\ b \\ c\end{array}\right]\) a. Show that the point on the plane closest to \(P_{0}\) has vector \(\mathbf{p}\) given by $$ \mathbf{p}=\mathbf{p}_{0}+\frac{d-\left(\mathbf{p}_{0} \cdot \mathbf{n}\right)}{\|\mathbf{n}\|^{2}} \mathbf{n} $$ \(\left[\right.\) Hint \(: \mathbf{p}=\mathbf{p}_{0}+t \mathbf{n}\) for some \(t,\) and \(\left.\mathbf{p} \cdot \mathbf{n}=\mathbf{d} .\right]\) b. Show that the shortest distance from \(P_{0}\) to the plane is \(\frac{\left|d-\left(\mathbf{p}_{0} \cdot \mathbf{n}\right)\right|}{\|\mathbf{n}\|}\). c. Let \(P_{0}^{\prime}\) denote the reflection of \(P_{0}\) in the planethat is, the point on the opposite side of the plane such that the line through \(P_{0}\) and \(P_{0}^{\prime}\) is perpendicular to the plane. Show that \(\mathbf{p}_{0}+2 \frac{d-\left(\mathbf{p}_{0} \cdot \mathbf{n}\right)}{\|\mathbf{n}\|^{2}} \mathbf{n}\) is the vector of \(P_{0}^{\prime}\)

Show that the volume of the parallelepiped determined by \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{u} \times \mathbf{v}\) is \(\|\mathbf{u} \times \mathbf{v}\|^{2}\).
See all solutions

Recommended explanations on Economics Textbooks

Taxation Economics

Read Explanation
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.

Sign-up for free