Chapter 10: Problem 12
Find (a) \(\mathbf{u} \cdot \mathbf{v}\) and (b) the angle between \(\mathbf{u}\) and \(\mathbf{v}\) to the nearest degree. $$ \mathbf{u}=\mathbf{i}+\mathbf{j}, \quad \mathbf{v}=\mathbf{i}-\mathbf{j} $$
Short Answer
Expert verified
(a) 0, (b) 90 degrees.
Step by step solution
01
Understand Vectors
Given vectors \( \mathbf{u} = \mathbf{i} + \mathbf{j} \) and \( \mathbf{v} = \mathbf{i} - \mathbf{j} \). These are unit vectors along the x and y axes.
02
Calculate Dot Product
Use the formula for the dot product of two vectors: \( \mathbf{u} \cdot \mathbf{v} = (a_1b_1 + a_2b_2) \). Substitute \( \mathbf{u} = (1, 1) \) and \( \mathbf{v} = (1, -1) \):\[ (1 \cdot 1) + (1 \cdot (-1)) = 1 - 1 = 0 \]
03
Determine Magnitudes
Find the magnitudes of \( \mathbf{u} \) and \( \mathbf{v} \). They both have magnitudes:\[ \sqrt{1^2 + 1^2} = \sqrt{2} \]
04
Calculate Angle Between Vectors
Use the formula for the cosine of the angle \( \theta \) between two vectors:\[ \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|} \]Since \( \mathbf{u} \cdot \mathbf{v} = 0 \) and both magnitudes are \( \sqrt{2} \):\[ \cos \theta = \frac{0}{2} = 0 \]Thus, \( \theta = 90^\circ \), computed using the inverse cosine.
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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, also known as the scalar product, is a fundamental operation in vector algebra. It's a way to combine two vectors to produce a single scalar value. Here's how it works:
- The formula for the dot product of two vectors \( \mathbf{a} = (a_1, a_2) \) and \( \mathbf{b} = (b_1, b_2) \) is: \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \).
- The result is a scalar, which is why it’s sometimes called the scalar product.
Angle Between Vectors
Determining the angle between two vectors is a critical skill in vector algebra, providing insights into their directional relationship.### Calculating the AngleTo find the angle \( \theta \) between two vectors \( \mathbf{a} \) and \( \mathbf{b} \), you use the cosine formula:\[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} \]- Here, \( \mathbf{a} \cdot \mathbf{b} \) is the dot product.- \( \|\mathbf{a}\| \) and \( \|\mathbf{b}\| \) are the magnitudes of the vectors.In the given example:- The dot product \( \mathbf{u} \cdot \mathbf{v} = 0 \).- Both magnitudes \( \|\mathbf{u}\| \) and \( \|\mathbf{v}\| \) are \( \sqrt{2} \).Plug these into the formula:\[ \cos \theta = \frac{0}{2} = 0 \]The inverse cosine of 0 is \( 90^\circ \), indicating the vectors are perpendicular. An angle of \( 90^\circ \) confirms that the vectors meet at a right angle, a valuable insight for determining vector directionality and relationships.
Magnitude of Vectors
The magnitude of a vector is equivalent to its length or size in space. Calculating it is essential for a variety of applications, from physics to engineering.### How to Compute MagnitudeThe magnitude \( \|\mathbf{a}\| \) of a vector \( \mathbf{a} = (a_1, a_2) \) is given by:\[ \|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2} \]For example, for the vector \( \mathbf{u} = (1, 1) \):\[ \|\mathbf{u}\| = \sqrt{1^2 + 1^2} = \sqrt{2} \]The same process applies to vector \( \mathbf{v} = (1, -1) \), resulting in the same magnitude.### Importance of Magnitude- **Normalization**: Magnitude is used to scale vectors to unit vectors.- **Stability**: Helps in maintaining stability and consistency within vector operations.In many real-world scenarios, understanding and computing vector magnitudes helps predict the behavior of forces and objects in space, making it a foundational concept in vector algebra.