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

The unknown vector \(\mathbf{v}\) satisfies \(\mathbf{b} \cdot \mathbf{v}=\lambda\) and \(\mathbf{b} \times \mathbf{v}=\mathbf{c},\) where \(\lambda, \mathbf{b},\) and \(\mathbf{c}\) are fixed and known. Find \(\mathbf{v}\) in terms of \(\lambda, \mathbf{b},\) and \(\mathbf{c}\)

Short Answer

Expert verified
\( \mathbf{v} = \frac{\lambda}{||\mathbf{b}||^2} \mathbf{b} + \mathbf{x} \), where \( \mathbf{x} = \mathbf{c} \times \frac{\mathbf{b}}{||\mathbf{b}||^2} \).

Step by step solution

01

Understand the Given Conditions

We are given two equations involving the unknown vector \(\mathbf{v}\):1. The dot product equation \( \mathbf{b} \cdot \mathbf{v} = \lambda \).2. The cross product equation \( \mathbf{b} \times \mathbf{v} = \mathbf{c} \).Here, \(\mathbf{b}\) and \(\mathbf{c}\) are known vectors, and \(\lambda\) is a known scalar.
02

Use the Dot Product Equation

The dot product equation \( \mathbf{b} \cdot \mathbf{v} = \lambda \) implies that the component of \( \mathbf{v} \) in the direction of \( \mathbf{b} \) is \( \frac{\lambda}{||\mathbf{b}||} \). Therefore, one component of \( \mathbf{v} \) is parallel to \(\mathbf{b}\).
03

Use the Cross Product Equation

The cross product equation \( \mathbf{b} \times \mathbf{v} = \mathbf{c} \) implies that \( \mathbf{v} \) must have components perpendicular to \( \mathbf{b} \), resulting in the vector \( \mathbf{c} \).
04

Construct the Vector v

Since \( \mathbf{v} \) has a component \( a\mathbf{b} \) parallel to \( \mathbf{b} \) and a component that results in \( \mathbf{c} \) when crossed with \( \mathbf{b} \), we propose \( \mathbf{v} = a\mathbf{b} + \mathbf{x} \) where \( \mathbf{x} \) is perpendicular to \( \mathbf{b} \).
05

Solve for the Scalar a

To ensure the dot product condition, substitute \( \mathbf{v} = a\mathbf{b} + \mathbf{x} \) into the dot product equation:\[ \mathbf{b} \cdot (a\mathbf{b} + \mathbf{x}) = \lambda \]Since \( \mathbf{x} \) is perpendicular to \( \mathbf{b} \), \( \mathbf{b} \cdot \mathbf{x} = 0 \). This simplifies to:\[ a||\mathbf{b}||^2 = \lambda \]Thus, \( a = \frac{\lambda}{||\mathbf{b}||^2} \).
06

Determine Vector x

We need \( \mathbf{b} \times (a\mathbf{b} + \mathbf{x}) = \mathbf{c} \). Since \( \mathbf{b} \times a\mathbf{b} = \mathbf{0} \), it follows that \( \mathbf{b} \times \mathbf{x} = \mathbf{c} \). Here \( \mathbf{x} \) is any vector that satisfies this equation, which can be solved using known methods to find a vector perpendicular to \( \mathbf{b} \).
07

Complete Solution for v

\( \mathbf{v} = a\mathbf{b} + \mathbf{x} \), where \( a = \frac{\lambda}{||\mathbf{b}||^2} \) and \( \mathbf{x} \) satisfies \( \mathbf{b} \times \mathbf{x} = \mathbf{c} \) and is perpendicular to \( \mathbf{b} \). Use \( \mathbf{x} = \mathbf{c} \times \frac{\mathbf{b}}{||\mathbf{b}||^2} \) to satisfy the cross product condition.

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 Vaia!

Key Concepts

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

Dot Product
The dot product is a fundamental operation in vector algebra. It involves two vectors, say \( \mathbf{b} \) and \( \mathbf{v} \), and produces a scalar. The dot product is calculated as:
  • \( \mathbf{b} \cdot \mathbf{v} = ||\mathbf{b}|| ||\mathbf{v}|| \cos(\theta) \)
where \( ||\mathbf{b}|| \) and \( ||\mathbf{v}|| \) are the magnitudes of the vectors, and \( \theta \) is the angle between them.
The dot product tells us how much of one vector goes in the direction of the other. In our exercise, when \( \mathbf{b} \cdot \mathbf{v} = \lambda \), it tells us that the projection of \( \mathbf{v} \) onto \( \mathbf{b} \) is \( \lambda \). This can be visualized as the component of \( \mathbf{v} \) that points in the direction of \( \mathbf{b} \). The condition creates a parallel component in \( \mathbf{v} \), represented as \( a\mathbf{b} \), where \( a = \frac{\lambda}{||\mathbf{b}||^2} \).
This parallel part is just one piece of the full vector \( \mathbf{v} \). The solution process requires us to balance this with the cross product condition to fully define \( \mathbf{v} \).
Cross Product
The cross product is another essential operation in vector algebra, which gives us a vector instead of a scalar, unlike the dot product. When you take the cross product of two vectors, say \( \mathbf{b} \) and \( \mathbf{v} \), it results in:
  • \( \mathbf{b} \times \mathbf{v} \)
This product is a vector perpendicular to both \( \mathbf{b} \) and \( \mathbf{v} \). The direction can be determined using the right-hand rule, and its magnitude is calculated as:
  • \( ||\mathbf{b} \times \mathbf{v}|| = ||\mathbf{b}|| ||\mathbf{v}|| \sin(\theta) \)
where \( \theta \) is the angle between the vectors.
In the exercise we're considering, \( \mathbf{b} \times \mathbf{v} = \mathbf{c} \) implies that the component of \( \mathbf{v} \) that is normal to \( \mathbf{b} \) results in the vector \( \mathbf{c} \).
This leads to the solution part where we need to find a vector \( \mathbf{x} \) such that \( \mathbf{b} \times \mathbf{x} = \mathbf{c} \). This \( \mathbf{x} \) is the orthogonal component needed to make \( \mathbf{v} \) satisfy both the dot and cross product conditions.
Perpendicular Vectors
The concept of perpendicular vectors is an important one in vector algebra. Two vectors are perpendicular when their dot product is zero:
  • \( \mathbf{b} \cdot \mathbf{x} = 0 \)
This means there is no component of \( \mathbf{x} \) in the direction of \( \mathbf{b} \).
In the context of our exercise, finding a vector \( \mathbf{x} \), perpendicular to \( \mathbf{b} \), is crucial to satisfying the cross product condition. If \( \mathbf{x} \) is perpendicular to \( \mathbf{b} \), \( \mathbf{x} \) does not contribute to \( \mathbf{b} \cdot \mathbf{v} \), thus fulfilling the requirement \( \mathbf{b} \times \mathbf{x} = \mathbf{c} \).
This forms an orthogonal component \( \mathbf{x} \) that combines with the parallel part \( a\mathbf{b} \) to give the final vector \( \mathbf{v} \). Finding \( \mathbf{x} \) often involves solving equations that respect these perpendicularity conditions, and various approaches like using cross products help in determining such vectors.

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

By evaluating their dot product, find the values of the scalar \(s\) for which the two vectors \(\mathbf{b}=\hat{\mathbf{x}}+s \hat{\mathbf{y}}\) and \(\mathbf{c}=\hat{\mathbf{x}}-s \hat{\mathbf{y}}\) are orthogonal. (Remember that two vectors are orthogonal if and only if their dot product is zero.) Explain your answers with a sketch.

A particle moves in a circle (center \(O\) and radius \(R\) ) with constant angular velocity \(\omega\) counter-clockwise. The circle lies in the \(x y\) plane and the particle is on the \(x\) axis at time \(t=0 .\) Show that the particle's position is given by $$\mathbf{r}(t)=\hat{\mathbf{x}} R \cos (\omega t)+\hat{\mathbf{y}} R \sin (\omega t)$$ Find the particle's velocity and acceleration. What are the magnitude and direction of the acceleration? Relate your results to well-known properties of uniform circular motion.

In Section 1.5 we proved that Newton's third law implies the conservation of momentum. Prove the converse, that if the law of conservation of momentum applies to every possible group of particles, then the interparticle forces must obey the third law. [Hint: However many particles your system contains, you can focus your attention on just two of them. (Call them 1 and 2.) The law of conservation of momentum says that if there are no external forces on this pair of particles, then their total momentum must be constant. Use this to prove that \(\mathbf{F}_{12}=-\mathbf{F}_{21}\).]

Imagine two concentric cylinders, centered on the vertical \(z\) axis, with radii \(R \pm \epsilon,\) where \(\epsilon\) is very small. A small frictionless puck of thickness \(2 \epsilon\) is inserted between the two cylinders, so that it can be considered a point mass that can move freely at a fixed distance from the vertical axis. If we use cylindrical polar coordinates \((\rho, \phi, z)\) for its position (Problem 1.47 ), then \(\rho\) is fixed at \(\rho=R,\) while \(\phi\) and \(z\) can vary at will. Write down and solve Newton's second law for the general motion of the puck, including the effects of gravity. Describe the puck's motion.

Prove that the two definitions of the scalar product \(\mathbf{r} \cdot\) s as \(r s \cos \theta(1.6)\) and \(\sum r_{i} s_{i}(1.7)\) are equal. One way to do this is to choose your \(x\) axis along the direction of \(\mathbf{r}\). [Strictly speaking you should first make sure that the definition (1.7) is independent of the choice of axes. If you like to worry about such niceties, see Problem 1.16.]

See all solutions

Recommended explanations on Physics 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.

Sign-up for free