Problem 9
Question
Find the length of \(\mathbf{x}=[0,1,5]^{\prime}\).
Step-by-Step Solution
Verified Answer
The length of the vector \(\mathbf{x} = [0, 1, 5]^\prime\) is \(\sqrt{26}\).
1Step 1: Understanding the Length of a Vector
The length of a vector, also known as the vector's magnitude or norm, is a measure of the size of the vector. For a vector \(\mathbf{x} = [x_1, x_2, x_3]^\prime\), the length is calculated using the formula \(\|\mathbf{x}\| = \sqrt{x_1^2 + x_2^2 + x_3^2}\). In this exercise, \(\mathbf{x} = [0, 1, 5]^\prime\).
2Step 2: Substituting Vector Components into the Formula
Using the formula for the length of a vector, substitute the components of \(\mathbf{x}\) into the formula: \(\|\mathbf{x}\| = \sqrt{0^2 + 1^2 + 5^2}\).
3Step 3: Calculate the Squares of Each Component
Calculate the square of each component: \(0^2 = 0\), \(1^2 = 1\), and \(5^2 = 25\).
4Step 4: Add the Squared Components
Add the squared components together: \(0 + 1 + 25 = 26\).
5Step 5: Take the Square Root of the Sum
Find the square root of the sum: \(\|\mathbf{x}\| = \sqrt{26}\).
6Step 6: Finalize the Answer
The length of the vector \(\mathbf{x}\) is \(\sqrt{26}\).
Key Concepts
Vector NormVector LengthEuclidean DistanceVector Components
Vector Norm
The vector norm, often referred to simply as the norm, is an essential concept in vector mathematics. It gives us a sense of how long or large a vector is. When you find the norm of a vector, you're essentially finding its length in the geometric space.
The most common vector norm is the Euclidean norm, which is calculated as:
The most common vector norm is the Euclidean norm, which is calculated as:
- Given a vector \(\mathbf{x} = [x_1, x_2, x_3 ]^\prime \), the formula is \(\|\mathbf{x}\| = \sqrt{x_1^2 + x_2^2 + x_3^2} \).
Vector Length
The vector length is another term often interchangeable with vector magnitude or vector norm. It's a measure of the straight-line distance from the initial point of the vector to its terminal point. This idea is intuitive as it reflects how long the vector "stretches" across its space.
To calculate the vector length, you essentially use the same Euclidean formula applied in the norm:
To calculate the vector length, you essentially use the same Euclidean formula applied in the norm:
- \(\|\mathbf{x}\| = \sqrt{x_1^2 + x_2^2 + x_3^2} \)
Euclidean Distance
The concept of Euclidean distance stems from Euclidean geometry, which is the mathematical study of shapes, space, and their relationships. The Euclidean distance between two points in space is the length of a straight line connecting them. When considering a vector, it describes the distance from the vector's endpoint to the origin point.
For a vector originating from the origin, the length of the vector itself is also its Euclidean distance. The formula you use for this is identical to the vector norm and length:
For a vector originating from the origin, the length of the vector itself is also its Euclidean distance. The formula you use for this is identical to the vector norm and length:
- \(\|\mathbf{x}\| = \sqrt{x_1^2 + x_2^2 + x_3^2} \)
Vector Components
Vector components are the building blocks of a vector. Each component points in a different direction of the vector's space, commonly corresponding to the x, y, and z dimensions. In our original exercise, the vector \([0, 1, 5]^\prime\) has components 0, 1, and 5, representing its movement along those respective axes.
Understanding each component helps reveal how the vector behaves in a Cartesian coordinate system:
Understanding each component helps reveal how the vector behaves in a Cartesian coordinate system:
- The vector \([0, 1, 5]^\prime\) means no movement along x, a unit movement along y, and five units along z.
Other exercises in this chapter
Problem 9
$$ \begin{array}{r} A=\left[\begin{array}{rrr} 1 & 3 & -1 \\ 2 & 4 & 1 \\ 0 & -2 & 2 \end{array}\right], \quad B=\left[\begin{array}{rrr} 5 & -1 & 4 \\ 2 & 0 &
View solution Problem 9
In Problems 9-16, reduce the system of linear equations to upper triangular form and solve. $$ \begin{array}{r} 2 x-y=3 \\ x-3 y=7 \end{array} $$
View solution Problem 9
In Problems 9-12, vectors are given in their polar coordinate representation (length \(r\), and angle \(\alpha\) measured counterclockwise from the positive \(x
View solution Problem 10
$$ \begin{array}{r} A=\left[\begin{array}{rrr} 1 & 3 & -1 \\ 2 & 4 & 1 \\ 0 & -2 & 2 \end{array}\right], \quad B=\left[\begin{array}{rrr} 5 & -1 & 4 \\ 2 & 0 &
View solution