Vectors \(\overrightarrow{\mathbf{a}}\) and \(\overrightarrow{\mathbf{b}}\) lie in the \(x y\) plane. The angle between \(\overrightarrow{\mathbf{a}}\) and \(\overrightarrow{\mathbf{b}}\) is \(\phi\), which is less than \(90^{\circ}\). Let \(\overrightarrow{\mathbf{c}}=\overrightarrow{\mathbf{a}} \times(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{a}})\). Find the magnitude of \(\overrightarrow{\mathbf{c}}\) and the angle between \(\overrightarrow{\mathbf{b}}\) and \(\overrightarrow{\mathbf{c}}\).

Short Answer

Expert verified
The magnitude of \(\overrightarrow{\mathbf{c}}\) is \(\|\overrightarrow{\mathbf{a}}\| \|\overrightarrow{\mathbf{b}}\| (1 - \cos{\phi})\), and the angle between \(\overrightarrow{\mathbf{b}}\) and \(\overrightarrow{\mathbf{c}}\) is \(\cos^{-1}(1 - \cos{\phi})\)

Step by step solution

01

Define the given

Given: vectors \(\overrightarrow{\mathbf{a}}\) and \(\overrightarrow{\mathbf{b}}\) with angle \(\phi\) between them. The vector \(\overrightarrow{\mathbf{c}}\) is defined as \(\overrightarrow{\mathbf{c}}=\overrightarrow{\mathbf{a}} \times(\overrightarrow{\mathbf{b}} \times \(\overrightarrow{\mathbf{a}}\)\).
02

Simplify the expression for vector \(\overrightarrow{\mathbf{c}}\)

Using the properties of the cross product and the BAC-CAB rule, \(\overrightarrow{\mathbf{c}}=\overrightarrow{\mathbf{a}} \times(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{a}})= (\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{a}}) \overrightarrow{\mathbf{b}} - (\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}) \overrightarrow{\mathbf{a}}\).
03

Find the magnitude of \(\overrightarrow{\mathbf{c}}\)

The magnitude of \(\overrightarrow{\mathbf{c}}\) is \(\|\overrightarrow{\mathbf{c}}\| = \|\overrightarrow{\mathbf{a}}\|^2 \|\overrightarrow{\mathbf{b}}\| - \|\overrightarrow{\mathbf{a}}\| \|\overrightarrow{\mathbf{b}}\| \cos{\phi} = \|\overrightarrow{\mathbf{a}}\| \|\overrightarrow{\mathbf{b}}\| (1 - \cos{\phi})\).
04

Find the angle between \(\overrightarrow{\mathbf{b}}\) and \(\overrightarrow{\mathbf{c}}\)

The angle \(\theta\) between \(\overrightarrow{\mathbf{b}}\) and \(\overrightarrow{\mathbf{c}}\) is given by \(\cos{\theta} = \frac{\(\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}}\)}{\|\overrightarrow{\mathbf{b}}\| \|\overrightarrow{\mathbf{c}}\|\}, which simplifies to \(\cos{\theta} = 1 - \cos{\phi}\). Thus, \(\theta = \cos^{-1}(1 - \cos{\phi})\)

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Key Concepts

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

Vector Mathematics
Vector mathematics is an essential branch of mathematics concerned with quantities having both magnitude and direction. These quantities are represented as arrows in space, where the length of the arrow indicates the magnitude, and the arrowhead points in the direction of the vector. Vectors are commonly used in physics, engineering, and computer science to describe forces, velocities, and positions.

Two crucial operations involving vectors are addition and multiplication. Vector addition combines two vectors to produce a third one, which represents the cumulative effect. Multiplication, on the other hand, includes the dot product, yielding a scalar, and the cross product, which gives another vector perpendicular to the plane formed by the original two vectors.
Magnitude of a Vector
The magnitude of a vector is a measure of its length or size, which is always a non-negative number. It is calculated by taking the square root of the sum of the squares of its components along each axis in a Cartesian coordinate system. For a vector \(\overrightarrow{\mathbf{v}}} = (x, y, z)\), the magnitude, denoted by \(\|overrightarrow{\mathbf{v}}\|\), is calculated as \(\sqrt{x^2 + y^2 + z^2}\). In the context of the given exercise, it is crucial to understand how to compute the magnitude of a vector to find the size of \(\overrightarrow{\mathbf{c}}\) after applying the cross product.
Angle Between Vectors
The angle between two vectors is measured in the plane where the two vectors originate from the same point. It is a measure of how much one vector needs to be rotated around a common origin to align with another vector. The angle can be found using the dot product formula, \(\overrightarrow{\mathbf{a}} \cdot overrightarrow{\mathbf{b}} = \|overrightarrow{\mathbf{a}}\|\|overrightarrow{\mathbf{b}}\| \cos{\phi}\), where \(\phi\) is the angle between vectors \(\overrightarrow{\mathbf{a}}\) and \(\overrightarrow{\mathbf{b}}\). Knowing the angle between vectors is essential for solving problems involving vector cross product and determining the orientation of vectors relative to one another.
BAC-CAB Rule
The BAC-CAB rule is a mnemonic for remembering the result of the vector triple product \(\overrightarrow{\mathbf{a}}\times (\overrightarrow{\mathbf{b}}\times \overrightarrow{\mathbf{c}})\). The rule states that this operation is equal to \(\overrightarrow{\mathbf{b}}(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{c}}) - \overrightarrow{\mathbf{c}}(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}})\), which gives another vector. It is named after the cyclic permutation of the letters A, B, and C. In this rule, 'BAC' represents that \(\overrightarrow{\mathbf{b}}\) is multiplied by the dot product of \(\overrightarrow{\mathbf{a}}\) and \(\overrightarrow{\mathbf{c}}\), while 'CAB' means that \(\overrightarrow{\mathbf{c}}\) is multiplied by the dot product of \(\overrightarrow{\mathbf{a}}\) and \(\overrightarrow{\mathbf{b}}\). This rule is pivotal in simplifying expressions involving multiple vector products, as demonstrated in the solution of our exercise.

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Most popular questions from this chapter

Vector \(\overrightarrow{\mathbf{a}}\) has magnitude \(3.20\) units and lies in the \(y z\) plane \(63.0^{\circ}\) from the \(+y\) axis with a positive \(z\) component. Vector \(\overrightarrow{\mathbf{b}}\) has magnitude \(1.40\) units and lies in the \(x z\) plane \(48.0^{\circ}\) from the \(+x\) axis with a positive \(z\) component. Find \(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}\).

Figure \(9-54\) shows the massive shield door at a neutron test facility at Lawrence Livermore Laboratory; this is the world's heaviest hinged door. The door has a mass of \(44,000 \mathrm{~kg}\), a rotational inertia about its hinge line of \(8.7 \times 10^{4} \mathrm{~kg} \cdot \mathrm{m}^{2}\), and a width of \(2.4 \mathrm{~m}\). What steady force, applied at its outer edge at right angles to the door, can move it from rest through an angle of \(90^{\circ}\) in \(30 \mathrm{~s}\) ?

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