/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 Let \(\mathbf{u}\) be an arbitra... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(\mathbf{u}\) be an arbitrary fixed unit vector and show that any vector \(\mathbf{b}\) satisfies $$b^{2}=(\mathbf{u} \cdot \mathbf{b})^{2}+(\mathbf{u} \times \mathbf{b})^{2}$$ Explain this result in words, with the help of a picture.

Short Answer

Expert verified
This equation shows the magnitude of a vector is the sum of the squared magnitudes of its projections parallel and perpendicular to a unit vector.

Step by step solution

01

Understand the Problem

We need to show that for any vector \( \mathbf{b} \), the equation \( b^{2}=(\mathbf{u} \cdot \mathbf{b})^{2}+(\mathbf{u} \times \mathbf{b})^{2} \) holds true, where \( \mathbf{u} \) is a unit vector.
02

Recall Vector Magnitude and Definitions

Recall that the magnitude of a vector \( \mathbf{b} \) is given by \( b = \| \mathbf{b} \| \) and is equal to \( \sqrt{\mathbf{b} \cdot \mathbf{b}} \). Also, the dot product \( \mathbf{u} \cdot \mathbf{b} \) measures projection along \( \mathbf{u} \), and the magnitude of the cross product \( \| \mathbf{u} \times \mathbf{b} \| \) measures the component orthogonal to \( \mathbf{u} \).
03

Express \( \mathbf{b} \) in Terms of \( \mathbf{u} \)

Any vector \( \mathbf{b} \) can be decomposed into components parallel and perpendicular to \( \mathbf{u} \). Let \( \mathbf{b}_{\parallel} = (\mathbf{u} \cdot \mathbf{b}) \mathbf{u} \) and \( \mathbf{b}_{\perp} = \mathbf{b} - \mathbf{b}_{\parallel} \).
04

Use Pythagorean Theorem in Vector Form

Since \( \mathbf{b} = \mathbf{b}_{\parallel} + \mathbf{b}_{\perp} \), by orthogonality, \( b^{2} = \| \mathbf{b}_{\parallel} \|^{2} + \| \mathbf{b}_{\perp} \|^{2} \). Here, \( \| \mathbf{b}_{\parallel} \| = |\mathbf{u} \cdot \mathbf{b}| \) and \( \| \mathbf{b}_{\perp} \| = \| \mathbf{u} \times \mathbf{b} \| \).
05

Plug into the Original Equation

Plugging these into the equation gives \[ \mathbf{b}^{2} = (\mathbf{u} \cdot \mathbf{b})^{2} + (\mathbf{u} \times \mathbf{b})^{2} \], thus proving the statement.

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Key Concepts

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

Dot Product
The dot product, often referred to as the scalar product, is a way to multiply two vectors to produce a scalar. It is calculated by multiplying the magnitudes of the two vectors and the cosine of the angle between them. The formula is:* \( \mathbf{u} \cdot \mathbf{b} = \|\mathbf{u}\| \times \|\mathbf{b}\| \times \cos(\theta) \)* In simpler terms, the dot product * measures how much one vector extends in the direction of another, * acts like a form of projection measuring lengths.* For unit vectors * like \(\mathbf{u}\),\( \mathbf{u} \cdot \mathbf{b} \) simplifies to just \(\cos(\theta)\) of the components of \(\mathbf{b}\), * in this case also corresponds to \(\mathbf{b}\)'s projection onto \(\mathbf{u}\).* If the vectors are perpendicular, the dot product is zero.* It helps identify the component of \( \mathbf{b} \) along \( \mathbf{u} \) as \((\mathbf{u} \cdot \mathbf{b})\).
Cross Product
The cross product is another way to multiply vectors, but the result is a new vector, not a scalar. It measures how vectors differ from each other in 3D space and considers the area of the parallelogram they span. The cross product formula is as follows:* \( \mathbf{u} \times \mathbf{b} = \|\mathbf{u}\| \times \|\mathbf{b}\| \times \sin(\theta) \mathbf{n} \)* \(\mathbf{n}\) is a unit vector perpendicular to the plane formed by \(\mathbf{u}\) and \(\mathbf{b}\).* This equation results in: * a vector orthogonally encompassing width, * orientation given by right-hand rule.* The magnitude \( \|\mathbf{u} \times \mathbf{b}\| \) corresponds to area information; it also * embodies the component perpendicular to \( \mathbf{u} \).* If vectors are parallel, the cross product yields zero.* Cross product gives crucial insight into vector orientation in space.
Vector Decomposition
Vector decomposition entails breaking down a vector into orthogonal components. This enables us to express any vector as a sum of vectors in distinct directions.* For any vector \( \mathbf{b} \), it can be broken down into two components: * Parallel to a given unit vector \( \mathbf{u} \): \( \mathbf{b}_{\parallel} = (\mathbf{u} \cdot \mathbf{b}) \mathbf{u} \). * Perpendicular to \( \mathbf{u} \): \( \mathbf{b}_{\perp} = \mathbf{b} - \mathbf{b}_{\parallel} \).* The parallel component is the projection on \( \mathbf{u} \), while the perpendicular part * is what remains of \( \mathbf{b} \) after extracting parallel influence.* This method benefits resolving vectors along predetermined axes, * often used for problem solving in physics.
Pythagorean Theorem in Vector Form
The Pythagorean Theorem is a cornerstone in vector calculations, offering a way to sum vector components orthogonally.* Given a vector \( \mathbf{b} \), it acts as * \( \mathbf{b} = \mathbf{b}_{\parallel} + \mathbf{b}_{\perp} \).* Using this for magnitude assessment, * \( b^{2} = \|\mathbf{b}\|^{2} \rightarrow \|\mathbf{b}_{\parallel}\|^{2} + \|\mathbf{b}_{\perp}\|^{2} \) * This aligns with familiar Pythagorean identity in right triangles.* Applied specifically to vectors, it allows splitting terms like: * \( (\mathbf{u} \cdot \mathbf{b})^{2} \) * \( (\mathbf{u} \times \mathbf{b})^{2} \)* This equation tells us how naturally vectors * sum their squared partsencompassing measure of size for both components.

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Most popular questions from this chapter

One of the many uses of the scalar product is to find the angle between two given vectors. Find the angle between the vectors \(\mathbf{b}=(1,2,4)\) and \(\mathbf{c}=(4,2,1)\) by evaluating their scalar product.

A student kicks a frictionless puck with initial speed \(v_{\mathrm{o}},\) so that it slides straight up a plane that is inclined at an angle \(\theta\) above the horizontal. (a) Write down Newton's second law for the puck and solve to give its position as a function of time. (b) How long will the puck take to return to its starting point?

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.

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 \(|\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})|\)

Find expressions for the unit vectors \(\hat{\boldsymbol{\rho}}, \hat{\boldsymbol{\phi}},\) and \(\hat{\mathbf{z}}\) of cylindrical polar coordinates (Problem 1.47) in terms of the Cartesian \(\hat{\mathbf{x}}, \hat{\mathbf{y}}, \hat{\mathbf{z}}\). Differentiate these expressions with respect to time to find \(d \hat{\boldsymbol{\rho}} / d t, d \hat{\boldsymbol{\phi}} / d t,\) and \(d \hat{\mathbf{z}} / d t\)

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