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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.]

Short Answer

Expert verified
The two definitions are equivalent: both equal \(rs_x\) when \(\mathbf{r}\) is along the \(x\)-axis.

Step by step solution

01

Define the Scalar Product via Angle and Magnitude

The scalar product of two vectors \(\mathbf{r}\) and \(\mathbf{s}\) is given by the formula \(\mathbf{r} \cdot \mathbf{s} = rs \cos \theta\), where \(r\) and \(s\) are the magnitudes of vectors \(\mathbf{r}\) and \(\mathbf{s}\), and \(\theta\) is the angle between them.
02

Express Vectors in Component Form

Assume the vector \(\mathbf{r}\) is aligned with the \(x\)-axis. Then, \(\mathbf{r} = (r, 0, 0)\). The vector \(\mathbf{s}\) in component form can be \(\mathbf{s} = (s_x, s_y, s_z)\).
03

Compute the Scalar Product Using Components

According to the component form definition, the scalar product \(\mathbf{r} \cdot \mathbf{s}\) is given by the sum of products of the corresponding components: \(\mathbf{r} \cdot \mathbf{s} = r_x s_x + r_y s_y + r_z s_z\). Given \(\mathbf{r} = (r, 0, 0)\), it follows that \(\mathbf{r} \cdot \mathbf{s} = rs_x\).
04

Relate to the Magnitude and Angle Definition

Since \(\mathbf{r} = (r, 0, 0)\), the angle \(\theta\) between \(\mathbf{r}\) and \(\mathbf{s}\) is such that \(\cos \theta = \frac{s_x}{s}\), where \(s = \sqrt{s_x^2 + s_y^2 + s_z^2}\) is the magnitude of \(\mathbf{s}\). Therefore, \(r s \cos \theta = rs \frac{s_x}{s} = rs_x\).
05

Conclusion

Both definitions of the scalar product yield \(rs_x\) in the special case where \(\mathbf{r}\) is along the \(x\)-axis. This proves that \(\mathbf{r} \cdot \mathbf{s} = \sum r_i s_i\) and \(\mathbf{r} \cdot \mathbf{s} = rs \cos \theta\) are equivalent definitions.

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Key Concepts

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

Vector Components
Understanding vector components is crucial when dealing with vectors in any mathematical context, especially in physics. Every vector can be broken down into its components, which are projections of the vector onto the coordinate axes. This breakdown helps in simplifying many problems and calculations.

A vector \( extbf{r}\) in three dimensional space can be expressed as \((r_x, r_y, r_z)\), where each component represents the vector's extent in the direction of the respective axis. To get these components:
  • Use the formulas \(r_x = r \cos \alpha, r_y = r \cos \beta, r_z = r \cos \gamma\)\
  • Here, \( heta\) terms \(\alpha, \beta, \gamma\) are the angles the vector makes with the axis.
This representation helps in performing vector operations such as addition or scalar multiplication by handling each component separately.
Angle Between Vectors
The angle between two vectors is a measure of their orientation with respect to each other. It plays a significant role in computing the scalar product and resolving components.

The angle \(\theta\) between two vectors \( extbf{r}\) and \( extbf{s}\) can be found using the dot product formula: \( extbf{r} \cdot extbf{s} = rs \cos \theta\). This equation effectively relates the magnitudes of the vectors, the angle between them, and their scalar product.

When the vectors are aligned in the same direction \((\theta = 0)\), the cosine is 1, making the dot product maximized. If the vectors are perpendicular \((\theta = 90^\circ)\), the cosine is 0, leading the dot product to equal zero. Understanding these angles helps in applications ranging from physics to engineering.
Dot Product Definition
The dot product, also known as the scalar product, is a fundamental vector operation that combines two vectors to produce a scalar. It quantifies the extent to which two vectors point in the same direction.

There are two main ways to define the dot product:
  • Using Magnitude and Angle: The dot product is defined as \(\textbf{r} \cdot \textbf{s} = rs \cos \theta\), where \(r\) and \(s\) are magnitudes, and \(\theta\) is the angle between the vectors.
  • Using Component Form: In terms of components, \(\textbf{r} \cdot \textbf{s} = r_xs_x + r_ys_y + r_zs_z\), where each part represents the product of respective components.
Both methods provide the same result and are used based on the situation's convenience. Understanding these definitions encourages flexibility in problem-solving and applying mathematical techniques.
Coordinate Axes Alignment
Aligning a vector along a coordinate axis can simplify vector calculations, especially when trying to prove equivalences like those in the scalar product definitions. It is a helpful strategy in vector mathematics.

Typically, when a vector \(\textbf{r}\) is aligned with the \(x\)-axis, its components become \(\textbf{r} = (r, 0, 0)\). This special case simplifies calculations such as the scalar product, since many terms disappear due to the presence of zeros.

This alignment is not just a mathematical trick, but practical for proofs and clear visualization of vector operations. By setting one vector along an axis, one can easily relate it to other vectors and proceed with simplified algebra. This technique is especially advantageous when tackling complex three-dimensional vector problems.

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Most popular questions from this chapter

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?

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

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.

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.

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.

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