Problem 7
Question
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.]
Step-by-Step Solution
Verified Answer
The two definitions are equivalent: both equal \(rs_x\) when \(\mathbf{r}\) is along the \(x\)-axis.
1Step 1: 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.
2Step 2: 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 \(\mathbf{s} = (s_x, s_y, s_z)\).
3Step 3: 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 \(\mathbf{r} \cdot \mathbf{s} = rs_x\).
4Step 4: 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, \(r s \cos \theta = rs \frac{s_x}{s} = rs_x\).
5Step 5: 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.
Key Concepts
Vector ComponentsAngle Between VectorsDot Product DefinitionCoordinate Axes Alignment
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:
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.
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.
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:
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.
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.
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.
Other exercises in this chapter
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