Problem 63
Question
If \(u\) is a nonzero vector, for what values of \(k\) does the equation \(\|k \mathbf{u}\|=k\|\mathbf{u}\|\) hold? Explain.
Step-by-Step Solution
Verified Answer
The equation holds for all \(k \geq 0\).
1Step 1: Understand the norm of a scalar multiple of a vector
The norm of a scalar multiple of a vector is given by the absolute value of the scalar times the norm of the vector. In this case, the norm of \(ku\) is \(\|ku\| = |k|\|\mathbf{u}\|\).
2Step 2: Compare with the given equation
We are asked for which \(k\) the equation \(\|ku\| = k\|\mathbf{u}\|\) holds. By comparison with the formula from Step 1, we can see that the equation holds if \(|k| = k\).
3Step 3: Determine the possible values for k
The absolute value function, \(|k|\), outputs only nonnegative values. The equation \(|k| = k\) holds for \(k \geq 0\), since for these values, \(k\) equals its own absolute value.
Key Concepts
Scalar Multiplication of VectorsAbsolute ValueNorm of a VectorPrecalculus
Scalar Multiplication of Vectors
When we talk about scalar multiplication in the context of vectors, we refer to scaling a vector by a certain amount, called a scalar. This action changes the magnitude of the vector but not its direction (unless the scalar is negative, which flips the direction).
Mathematically, when you multiply a vector \textbf{u} by a scalar value 'k', you get a new vector, expressed as \(k\textbf{u}\). Each component of the original vector is multiplied by 'k'. For example, if we have a vector \( \textbf{u}= [u_1, u_2, ... , u_n]\), multiplying it by scalar 'k' yields the vector \(k\textbf{u} = [ku_1, ku_2, ... , ku_n]\). This operation can stretch or compress the length of the vector, and if 'k' is negative, it also inverts the vector's orientation.
Mathematically, when you multiply a vector \textbf{u} by a scalar value 'k', you get a new vector, expressed as \(k\textbf{u}\). Each component of the original vector is multiplied by 'k'. For example, if we have a vector \( \textbf{u}= [u_1, u_2, ... , u_n]\), multiplying it by scalar 'k' yields the vector \(k\textbf{u} = [ku_1, ku_2, ... , ku_n]\). This operation can stretch or compress the length of the vector, and if 'k' is negative, it also inverts the vector's orientation.
Absolute Value
The absolute value function is essential in mathematics, especially when explaining the magnitude of numbers regardless of their sign. It can be thought of as the distance of a number from zero on the number line.
For any real number 'x', the absolute value, denoted by \( |x| \), is defined as follows:
Thus, the absolute value of a number is always nonnegative. This concept becomes particularly important when we're dealing with vectors and wanting to express their magnitude without regards for direction.
For any real number 'x', the absolute value, denoted by \( |x| \), is defined as follows:
- If \( x \geq 0 \) , then \( |x| = x \).
- If \( x < 0 \), then \( |x| = -x \), which is a positive value.
Thus, the absolute value of a number is always nonnegative. This concept becomes particularly important when we're dealing with vectors and wanting to express their magnitude without regards for direction.
Norm of a Vector
In the world of vectors, the term 'norm' refers to a measure of a vector's length or magnitude. For a given vector \( \textbf{u} \), the norm is denoted as \( \|\textbf{u}\| \). It's equivalent to the distance from the origin to the point defined by the vector in a coordinate space. To calculate the norm of a vector with components \( \textbf{u}= [u_1, u_2, ... , u_n] \), you would use the formula \( \|\textbf{u}\| = \sqrt{u_1^2 + u_2^2 + ... + u_n^2} \).
The norm of a vector is always a nonnegative number since it is derived from the square root of the sum of squared vector components, which cannot be negative. It is critical to note that scalar multiplication affects the norm, according to the relationship \( \|k\textbf{u}\| = |k|\|\textbf{u}\| \), which reflects the influence of the scalar's absolute value on the vector's magnitude.
The norm of a vector is always a nonnegative number since it is derived from the square root of the sum of squared vector components, which cannot be negative. It is critical to note that scalar multiplication affects the norm, according to the relationship \( \|k\textbf{u}\| = |k|\|\textbf{u}\| \), which reflects the influence of the scalar's absolute value on the vector's magnitude.
Precalculus
Precalculus serves as the groundwork for calculus, providing the necessary concepts and tools used extensively in higher mathematics. It covers various topics like functions, algebra, trigonometry, and yes, vectors. Understanding how to manipulate vectors, including operations like scalar multiplication, and concepts like absolute value and norms, are fundamental in precalculus.
Students learn about the behaviors of these concepts in two-dimensional and three-dimensional spaces and how to apply them in more advanced mathematical fields. In particular, the ability to understand the relationship between scalars and vectors, and how scaling affects vector norms, is crucial when delving into the depths of calculus where these ideas are extended to cover limits, derivatives, and integrals - the core concepts of change and motion in mathematics.
Students learn about the behaviors of these concepts in two-dimensional and three-dimensional spaces and how to apply them in more advanced mathematical fields. In particular, the ability to understand the relationship between scalars and vectors, and how scaling affects vector norms, is crucial when delving into the depths of calculus where these ideas are extended to cover limits, derivatives, and integrals - the core concepts of change and motion in mathematics.
Other exercises in this chapter
Problem 62
Use a graphing utility to graph \(r_{1}=1+\cos \theta\) and \(r_{2}=1+\cos \left(\theta-\frac{\pi}{2}\right) .\) Explain the relationship between the two graphs
View solution Problem 62
Convert each of the given polar equations to rectangular form. $$r \cos \theta-3 r \sin \theta=5$$
View solution Problem 63
Show that the formula Area \((A B C)=\frac{1}{2} a b \sin C\) holds if \(A B C\) is a right triangle.
View solution Problem 63
Convert each of the given polar equations to rectangular form. $$r=2 \cos \theta$$
View solution