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Chapter 7: Problem 30

\(\overline{\mathbf{A}}\) and \(\overline{\mathbf{B}}\) are two vectors in the \(x y\) plane that make angles \(\alpha\) and \(\beta\) with the \(x\) axis respectively. Evaluate the scalar product of \(\overrightarrow{\mathbf{A}}\) and \(\overrightarrow{\mathbf{B}}\) and deduce the following trigonometric identity: \(\cos (\alpha-\beta)=\cos \alpha \cos \beta+\sin \alpha \sin \beta\).

Short Answer

Expert verified
The scalar product leads to the identity: \( \cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \).

Step by step solution

01

Define Vectors A and B

Firstly, identify vectors \( \overrightarrow{\mathbf{A}} \) and \( \overrightarrow{\mathbf{B}} \) in terms of their magnitudes and directions. Assuming magnitudes \( |\overrightarrow{\mathbf{A}}| = A \) and \( |\overrightarrow{\mathbf{B}}| = B \), we define:\[ \overrightarrow{\mathbf{A}} = A \cos \alpha \hat{i} + A \sin \alpha \hat{j} \]\[ \overrightarrow{\mathbf{B}} = B \cos \beta \hat{i} + B \sin \beta \hat{j} \]
02

Compute the Scalar Product

The scalar product (dot product) of \( \overrightarrow{\mathbf{A}} \) and \( \overrightarrow{\mathbf{B}} \) is given by:\[ \overrightarrow{\mathbf{A}} \cdot \overrightarrow{\mathbf{B}} = (A \cos \alpha \hat{i} + A \sin \alpha \hat{j}) \cdot (B \cos \beta \hat{i} + B \sin \beta \hat{j}) \]Expanding the expression, we get:\[ AB \cos \alpha \cos \beta + AB \sin \alpha \sin \beta \]
03

Relate Scalar Product to Cosine of the Angle Between

The scalar product is also given by the formula:\[ \overrightarrow{\mathbf{A}} \cdot \overrightarrow{\mathbf{B}} = AB \cos(\theta) \]where \( \theta \) is the angle between the vectors. According to the problem, \( \theta = \alpha - \beta \), so:\[ \overrightarrow{\mathbf{A}} \cdot \overrightarrow{\mathbf{B}} = AB \cos(\alpha - \beta) \]
04

Equate the Two Expressions for Scalar Product

Equate the two expressions for the scalar product to derive a trigonometric identity:\[ AB \cos(\alpha - \beta) = AB \cos \alpha \cos \beta + AB \sin \alpha \sin \beta \]Dividing through by \( AB \), we obtain:\[ \cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \]
05

Conclude the Derivation of the Identity

The derived expression\[ \cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \]is the required trigonometric identity. It shows how the cosine of a difference of angles can be expressed in terms of the cosine and sine of the individual angles.

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Key Concepts

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

Scalar Product
The scalar product, also known as the dot product, is a fundamental operation in vector algebra. It links the trigonometric and geometric perspectives of vectors.
The scalar product of two vectors is a measure of how much one vector extends in the direction of another vector.
  • Formula: For vectors \( \overrightarrow{\mathbf{A}} = A_x \hat{i} + A_y \hat{j} \) and \( \overrightarrow{\mathbf{B}} = B_x \hat{i} + B_y \hat{j} \), the scalar product is \( \overrightarrow{\mathbf{A}} \cdot \overrightarrow{\mathbf{B}} = A_x B_x + A_y B_y \).

  • Result: A single number, or scalar, that represents the product of the vectors' magnitudes and the cosine of the angle between them.
The scalar product is handy in various applications. It helps in determining the angle between vectors and projecting one vector onto another.
Vectors in the XY-Plane
Vectors in the xy-plane are represented using x and y components. Understanding these vectors is crucial for working in two-dimensional spaces, such as physics and engineering problems.
A vector describes a quantity with both magnitude and direction. Its components along the x and y axes help to specify its position
  • Representation: A typical vector \( \overrightarrow{\mathbf{A}} \) in the xy-plane can be represented as \( A_x \hat{i} + A_y \hat{j} \), where \( \hat{i} \) and \( \hat{j} \) are unit vectors in the x and y directions.

  • Magnitude: Given by \( |\overrightarrow{\mathbf{A}}| = \sqrt{A_x^2 + A_y^2} \).
These components allow vectors to be added, subtracted, or multiplied easily, facilitating problem-solving in two-dimensional motion and force problems.
Dot Product
The dot product is an operation on two vectors that returns a scalar, hence another name for it is "scalar product." It's used to find how much one vector acts in the direction of another.
  • Mathematical Definition: The dot product for two vectors \( \overrightarrow{\mathbf{A}} \) and \( \overrightarrow{\mathbf{B}} \) is defined as \( \overrightarrow{\mathbf{A}} \cdot \overrightarrow{\mathbf{B}} = |\overrightarrow{\mathbf{A}}||\overrightarrow{\mathbf{B}}|\cos(\theta) \), where \( \theta \) is the angle between \( \overrightarrow{\mathbf{A}} \) and \( \overrightarrow{\mathbf{B}} \).

  • Properties: The dot product is commutative, \( \overrightarrow{\mathbf{A}} \cdot \overrightarrow{\mathbf{B}} = \overrightarrow{\mathbf{B}} \cdot \overrightarrow{\mathbf{A}} \), and distributive over vector addition.
The dot product helps in understanding the geometry of vectors, calculating angles, and projecting vectors. It efficiently links vectors' magnitudes and their orientation with respect to each other.
Cosine Rule
The cosine rule is a vital trigonometric identity that relates the cosine of an angle between two vectors to the magnitudes and directions of those vectors.
It emerges naturally when working with the dot product. This relationship helps determine unknown angles and sides in various spatial problems.
  • Identity: The cosine identity we derive using vectors is \( \cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \), demonstrating how a difference in angles can be represented using individual sines and cosines.

  • Application: The cosine rule is used extensively in physics, engineering, and mathematics to solve both theoretical and practical problems involving triangles and forces in two dimensions.
By using the cosine rule, we can efficiently solve problems such as determining the angle between two intersecting lines or analyzing the position and forces in a structure.

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Most popular questions from this chapter

A varying force is given by \(F=A e^{-k x},\) where \(x\) is the position; \(A\) and \(k\) are constants that have units of \(\mathrm{N}\) and \(\mathrm{m}^{-1},\) respectively. What is the work done when \(x\) goes from \(0.10 \mathrm{~m}\) to infinity?

In a certain library the first shelf is 12.0\(\mathrm { cm }\) off the ground, and the remaining 4 shelves are each spaced 33.0\(\mathrm { cm }\) above the previous one. If the average book has a mass of 1.40\(\mathrm { kg }\) with a height of \(22.0 \mathrm { cm } ,\) and an average shelf holds 28 with a height of \(22.0 \mathrm { cm } ,\) and an average shelf holds 28 books (standing vertically), how much work is required to fill all the shelves, assuming the books are all laying flat on the floor to start?

When different masses are suspended from a spring, the spring stretches by different amounts as shown in the Table below. Masses are 卤1.0 gram. \begin{tabular}{lllrrrrrrr} \hline Mass (g) & 0 & 50 & 100 & 150 & 200 & 250 & 300 & 350 & 400 \\ Stretch (cm) & 0 & 5.0 & 9.8 & 14.8 & 19.4 & 24.5 & 29.6 & 34.1 & 39.2 \\ \hline \end{tabular} (a) Graph the applied force (in Newtons) versus the stretch (in meters) of the spring, and determine the best-fit straight line. (b) Determine the spring constant ( \(\mathrm{N} / \mathrm{m}\) ) of the spring from the slope of the best-fit line. \((c)\) If the spring is stretched by \(20.0 \mathrm{~cm},\) estimate the force acting on the spring using the best-fit line.

In a certain library the first shelf is \(12.0 \mathrm{~cm}\) off the ground, and the remaining 4 shelves are each spaced \(33.0 \mathrm{~cm}\) above the previous one. If the average book has a mass of \(1.40 \mathrm{~kg}\) with a height of \(22.0 \mathrm{~cm},\) and an average shelf holds 28 books (standing vertically), how much work is required to fill all the shelves, assuming the books are all laying flat on the floor to start?

(a) What magnitude force is required to give a helicopter of mass \(M\) an acceleration of \(0.10 g\) upward? (b) What work is done by this force as the helicopter moves a distance \(h\) upward?
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