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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

Find the angle between a body diagonal of a cube and any one of its face diagonals. [Hint: Choose a cube with side 1 and with one corner at \(O\) and the opposite corner at the point (1,1,1) . Write down the vector that represents a body diagonal and another that represents a face diagonal, and then find the angle between them as in Problem 1.4.]

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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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.

The hallmark of an inertial reference frame is that any object which is subject to zero net force will travel in a straight line at constant speed. To illustrate this, consider the following: I am standing on a level floor at the origin of an inertial frame \(\mathcal{S}\) and kick a frictionless puck due north across the floor. (a) Write down the \(x\) and \(y\) coordinates of the puck as functions of time as seen from my inertial frame. (Use \(x\) and \(y\) axes pointing east and north respectively.) Now consider two more observers, the first at rest in a frame \(\mathcal{S}^{\prime}\) that travels with constant velocity \(v\) due east relative to \(\mathcal{S},\) the second at rest in a frame \(\mathcal{S}^{\prime \prime}\) that travels with constant acceleration due east relative to \(\mathcal{S}\). (All three frames coincide at the moment when I kick the puck, and \(\mathcal{S}^{\prime \prime}\) is at rest relative to \(\mathcal{S}\) at that same moment.) (b) Find the coordinates \(x^{\prime}, y^{\prime}\) of the puck and describe the puck's path as seen from \(\mathcal{S}^{\prime} .\) (c) Do the same for \(\mathcal{S}^{\prime \prime}\) Which of the frames is inertial?

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