Problem 27
Question
a. Cauchy-Schwarz inequality since \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) show that the inequality \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|\) holds for any vectors \(\mathbf{u}\) and \(\mathbf{v}\). b. Under what circumstances, if any, does \(|\mathbf{u} \cdot \mathbf{v}|\) equal \(|\mathbf{u}||\mathbf{v}| ?\) Give reasons for your answer.
Step-by-Step Solution
Verified Answer
a. The inequality holds due to \(|\cos \theta| \leq 1\). b. Equality occurs when vectors are parallel.
1Step 1: Understand the Dot Product Formula
The dot product of two vectors \(\mathbf{u}\) and \(\mathbf{v}\) is given by \(\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}||\mathbf{v}| \cos \theta\), where \(\theta\) is the angle between the vectors. This formula helps in linking the geometric interpretation of dot product with its algebraic computation.
2Step 2: State the Cauchy-Schwarz Inequality
The Cauchy-Schwarz inequality states that for any vectors \(\mathbf{u}\) and \(\mathbf{v}\), the absolute value of their dot product is less than or equal to the product of their magnitudes: \[|\mathbf{u} \cdot \mathbf{v}| \leq |\mathbf{u}||\mathbf{v}|.\]
3Step 3: Apply Geometry to Understand the Inequality
Since \(|\cos \theta| \leq 1\) for all angles \(\theta\), it follows that \(|\mathbf{u} \cdot \mathbf{v}| = |\mathbf{u}||\mathbf{v}| |\cos \theta| \leq |\mathbf{u}||\mathbf{v}|\). This demonstrates that the inequality holds.
4Step 4: Identify When Equality Holds
Equality \(|\mathbf{u} \cdot \mathbf{v}| = |\mathbf{u}||\mathbf{v}|\) holds when \(|\cos \theta| = 1\). This occurs when \(\theta = 0\degree\) or \(\theta = 180\degree\), meaning \(\mathbf{u}\) and \(\mathbf{v}\) are parallel (same or opposite direction).
Key Concepts
Dot ProductVector MagnitudeGeometric Interpretation
Dot Product
The dot product is an essential operation when dealing with vectors in mathematics and physics. For two vectors, say \( \mathbf{u} = (u_1, u_2) \) and \( \mathbf{v} = (v_1, v_2) \), the dot product is algebraically calculated as \( \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 \). But there's more than just the numbers!
On a geometric level, the dot product also represents the product of the magnitudes of the vectors and the cosine of the angle \( \theta \) between them: \( \mathbf{u} \cdot \mathbf{v} = |\mathbf{u}||\mathbf{v}| \cos \theta \).
This means that the dot product is connected to both the vectors' directions and lengths.
On a geometric level, the dot product also represents the product of the magnitudes of the vectors and the cosine of the angle \( \theta \) between them: \( \mathbf{u} \cdot \mathbf{v} = |\mathbf{u}||\mathbf{v}| \cos \theta \).
This means that the dot product is connected to both the vectors' directions and lengths.
- If the dot product is positive, the angle \( \theta \) is acute.
- If it is zero, the vectors are orthogonal (90 degrees).
- If negative, \( \theta \) is obtuse.
Vector Magnitude
A vector's magnitude, sometimes called its length or norm, is a measure of how 'long' the vector is. It is calculated using the square root of the sum of the vector's components squared. For a vector \( \mathbf{u} = (u_1, u_2) \), the magnitude is \( |\mathbf{u}| = \sqrt{u_1^2 + u_2^2} \).
Magnitude serves as an essential part of understanding vector properties and operations. Specifically, the magnitude is crucial in the context of the Cauchy-Schwarz inequality, where it helps compare the length component across vectors.
When working with the dot product, the magnitude scales the vector product. In simpler terms, if two vectors differ only by length but point in the same direction, their dot product will naturally increase just because their magnitude is larger.
Thus, understanding magnitudes provides a deeper insight into comparing and operating with vectors, especially in considering distances, lengths, and interpretations of vector space.
Magnitude serves as an essential part of understanding vector properties and operations. Specifically, the magnitude is crucial in the context of the Cauchy-Schwarz inequality, where it helps compare the length component across vectors.
When working with the dot product, the magnitude scales the vector product. In simpler terms, if two vectors differ only by length but point in the same direction, their dot product will naturally increase just because their magnitude is larger.
Thus, understanding magnitudes provides a deeper insight into comparing and operating with vectors, especially in considering distances, lengths, and interpretations of vector space.
Geometric Interpretation
The Cauchy-Schwarz inequality is not only a fundamental algebraic tool but also has a tangible geometric interpretation. Busy visualizing vectors in space—think of them as arrows pointing outwards. The inequality tells us about the relationship between these arrows.
Geometrically, the Cauchy-Schwarz inequality \(|\mathbf{u} \cdot \mathbf{v}| \leq |\mathbf{u}||\mathbf{v}|\) ensures that the dot product magnitude cannot exceed the product of the individual vector magnitudes, limited by the maximum value of \( |\cos \theta| \leq 1 \).
In simple terms:
Geometrically, the Cauchy-Schwarz inequality \(|\mathbf{u} \cdot \mathbf{v}| \leq |\mathbf{u}||\mathbf{v}|\) ensures that the dot product magnitude cannot exceed the product of the individual vector magnitudes, limited by the maximum value of \( |\cos \theta| \leq 1 \).
In simple terms:
- If \(\theta = 0\degree\), the vectors are perfectly aligned, thus \(|\cos \theta| = 1\) and the inequality becomes an equality \(|\mathbf{u} \cdot \mathbf{v}| = |\mathbf{u}||\mathbf{v}|\).
- If \(\theta\) is something else, the dot product is smaller due to multiplicative effect of \( \cos(\theta) \) reducing the impactful 'stretch'.
Other exercises in this chapter
Problem 27
Which of the following are always true, and which are not always true? Give reasons for your answers. a. \(|\mathbf{u}|=\sqrt{\mathbf{u} \cdot \mathbf{u}}\) b.
View solution Problem 27
Sketch the surfaces HYPERBOLOIDS $$x^{2}+y^{2}-z^{2}=1$$
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Find the point of intersection of the lines \(x=2 t+1\) \(y=3 t+2, \quad z=4 t+3,\) and \(x=s+2, y=2 s+4, z=\) \(-4 s-1,\) and then find the plane determined by
View solution Problem 27
Express each vector as a product of its length and direction. $$5 \mathbf{k}$$
View solution