Chapter 11: Problem 16
Is it possible to have \(\|\mathbf{u}-\mathbf{v}\|=\|\mathbf{u}+\mathbf{v}\|\) if \(\mathbf{u}\) and \(\mathbf{v}\) are nonzero vectors? Justify your conclusion geometrically.
Short Answer
Expert verified
Yes, if \(\mathbf{u}\) and \(\mathbf{v}\) are perpendicular.
Step by step solution
01
Understand the Meaning of the Equation
The equation \(\|\mathbf{u}-\mathbf{v}\|=\|\mathbf{u}+\mathbf{v}\|\) means that the magnitude (or length) of the vector \(\mathbf{u}-\mathbf{v}\) is equal to the magnitude of the vector \(\mathbf{u}+\mathbf{v}\). This suggests a geometric condition between the vectors \(\mathbf{u}\) and \(\mathbf{v}\).
02
Visualize Geometrically
Geometrically, \(\mathbf{u}-\mathbf{v}\) and \(\mathbf{u}+\mathbf{v}\) would be vectors forming the diagonals of a parallelogram. The equation indicates that these diagonals have the same length. Such a condition is satisfied if and only if the parallelogram is a rectangle.
03
Recognize Perpendicularity
For a parallelogram to be a rectangle, the vectors \(\mathbf{u}\) and \(\mathbf{v}\) must be perpendicular to each other. This implies that their dot product is zero, i.e., \(\mathbf{u} \cdot \mathbf{v} = 0\).
04
Confirm with Dot Product
Given \(\mathbf{u} \cdot \mathbf{v} = 0\), the computation \((\mathbf{u}+\mathbf{v}) \cdot (\mathbf{u}+\mathbf{v}) = \|\mathbf{u}\|^2 + \|\mathbf{v}\|^2\) results in the same value as \((\mathbf{u}-\mathbf{v}) \cdot (\mathbf{u}-\mathbf{v}) = \|\mathbf{u}\|^2 + \|\mathbf{v}\|^2\). Thus, the magnitudes are equal, confirming the condition geometrically.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Magnitude
Understanding vector magnitude is crucial in vector geometry. The magnitude of a vector, often referred to as its length or norm, represents how long the vector is. For any vector \(\mathbf{a} = (a_1, a_2, ..., a_n)\) in an n-dimensional space, its magnitude is computed using the formula \(\|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2 + ... + a_n^2}\).
This formula is essentially an extension of the Pythagorean theorem to multiple dimensions. By understanding this concept, you can determine the distance a vector covers from the origin to its endpoint.
Magnitude is always a non-negative value as it represents a length, which cannot be negative.
This formula is essentially an extension of the Pythagorean theorem to multiple dimensions. By understanding this concept, you can determine the distance a vector covers from the origin to its endpoint.
Magnitude is always a non-negative value as it represents a length, which cannot be negative.
- In problems like the one given, comparing the magnitudes can reveal important geometric relationships.
- When two vector expressions, like \(\mathbf{u} - \mathbf{v}\) and \(\mathbf{u} + \mathbf{v}\) have equal magnitudes, their geometric implications become evident, leading to further exploration of concepts like perpendicular vectors.
Perpendicular Vectors
Two vectors are considered perpendicular, or orthogonal, if they meet at a right angle. In vector mathematics, this relationship is revealed through the dot product.
If vectors \(\mathbf{u}\) and \(\mathbf{v}\) are perpendicular, their dot product equals zero: \(\mathbf{u} \cdot \mathbf{v} = 0\). This condition arises due to the angle between them being 90 degrees.
In a geometric context, if the vectors \(\mathbf{u}\) and \(\mathbf{v}\) form the diagonals of a parallelogram and the magnitude condition is met as per the original problem, the parallelogram must be a rectangle. In a rectangle, diagonals are equal, highlighting that \(\mathbf{u}\) and \(\mathbf{v}\) must be perpendicular.
If vectors \(\mathbf{u}\) and \(\mathbf{v}\) are perpendicular, their dot product equals zero: \(\mathbf{u} \cdot \mathbf{v} = 0\). This condition arises due to the angle between them being 90 degrees.
In a geometric context, if the vectors \(\mathbf{u}\) and \(\mathbf{v}\) form the diagonals of a parallelogram and the magnitude condition is met as per the original problem, the parallelogram must be a rectangle. In a rectangle, diagonals are equal, highlighting that \(\mathbf{u}\) and \(\mathbf{v}\) must be perpendicular.
- Understanding perpendicular vectors helps in visualizing and solving problems related to geometric structures.
- This concept is critical when ensuring balance and symmetry in spatial computations.
Dot Product
The dot product, or scalar product, is a mathematical operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. This number provides insights into the geometric relationship between the vectors.
The dot product of two vectors \(\mathbf{a} = (a_1, a_2,..., a_n)\) and \(\mathbf{b} = (b_1, b_2,..., b_n)\) is given by: \(\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + ... + a_nb_n\).
It can also be described as \(\|\mathbf{a}\| \|\mathbf{b}\| \cos \theta\), where \(\theta\) is the angle between the two vectors. When the vectors are perpendicular, \(\cos \theta = 0\) which results in a zero dot product.
The dot product of two vectors \(\mathbf{a} = (a_1, a_2,..., a_n)\) and \(\mathbf{b} = (b_1, b_2,..., b_n)\) is given by: \(\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + ... + a_nb_n\).
It can also be described as \(\|\mathbf{a}\| \|\mathbf{b}\| \cos \theta\), where \(\theta\) is the angle between the two vectors. When the vectors are perpendicular, \(\cos \theta = 0\) which results in a zero dot product.
- This operation is key for determining if two vectors are perpendicular, as used in the solution of the given exercise.
- In physics and engineering, the dot product helps in calculating work done by force or determining projections of one vector onto another.