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Chapter 5: Problem 2

Find the length of the vector. \(\mathbf{v}=(0,1)\)

Short Answer

Expert verified
The length of the vector \(\mathbf{v}=(0,1)\) is 1.

Step by step solution

01

Identify the components of the vector

The vector \(\mathbf{v}\) is defined as (0,1). As such, the x-component (a) is 0 and the y-component (b) is 1.
02

Substitute the vector components into the length formula

Using the formula \(\|\mathbf{v}\|=\sqrt{a^2+b^2}\), substitute a = 0 and b = 1 to get \(\|\mathbf{v}\|=\sqrt{0^2+1^2}\).
03

Calculate the length of the vector

Perform the calculations to get \(\|\mathbf{v}\|=\sqrt{0+1} = \sqrt{1} = 1.\

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Components
When dealing with vectors, it's crucial to understand vector components. A vector in two-dimensional space, like \(\mathbf{v} = (0,1)\), is often broken down into two parts, known as components. These are the x-component and the y-component. These components tell us how far the vector stretches along the x-axis and y-axis respectively.
For \(\mathbf{v} = (0, 1)\), the vector has:
  • An x-component of 0: This indicates there is no movement along the x-axis.
  • A y-component of 1: This shows the vector moves 1 unit along the y-axis.
By identifying these components, you lay the groundwork for calculating the vector's length.
Magnitude Formula
The magnitude or length of a vector is like finding out how long a line segment is. You use the magnitude formula, which is \[\|\mathbf{v}\| = \sqrt{a^2 + b^2}\]to calculate this distance.
Here, \(a\) and \(b\) are the x and y components of the vector. For our example, \(\mathbf{v} = (0, 1)\), substituting the components gives:
  • \(\|\mathbf{v}\| = \sqrt{0^2 + 1^2} = \sqrt{1} = 1\)
This formula essentially uses the Pythagorean theorem, as it calculates the hypotenuse of a right-angled triangle formed by the vector components.
Coordinate Geometry
Coordinate geometry is a branch of mathematics that studies geometric figures through a coordinate system. It鈥檚 fundamental when working with vectors, as it allows you to visualize and calculate distances.
In coordinate geometry, each point in space can be described using a pair (or triplet) of numbers. For vectors defined as \(\mathbf{v} = (x, y)\), these pair of numbers are crucial:
  • The x-component shows horizontal distance from the origin.
  • The y-component shows vertical distance from the origin.
By applying these principles, finding vector lengths or magnitudes becomes more intuitive and tied to the geometric interpretation of space.

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Most popular questions from this chapter

Show that the volume of a parallelepiped having \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) as adjacent sides is the triple scalar product \(|\mathbf{u} \cdot(\mathbf{v} \times \mathbf{w})|\).

Prove that the angle \(\theta\) between \(\mathbf{u}\) and \(\mathbf{v}\) is given by \(\|\mathbf{u} \times \mathbf{v}\|=\|\mathbf{u}\|\|\mathbf{v}\| \sin \theta\).

Use a graphing utility with vector capabilities to find \(\mathbf{u} \times \mathbf{v}\) and then show that it is orthogonal to both \(\mathbf{u}\) and \(\mathbf{v}\). $$\mathbf{u}=(0,1,-1), \quad \mathbf{v}=(1,2,0)$$

Find bases for the four fundamental subspaces of the matrix \(A\). $$A=\left[\begin{array}{rrr} 1 & 0 & -1 \\ 0 & -1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right]$$

Find the least squares solution of the system \(A \mathbf{x}=\mathbf{b}\). $$A=\left[\begin{array}{rrr} 1 & -1 & 1 \\ 1 & 1 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{array}\right] \quad \mathbf{b}=\left[\begin{array}{l} 2 \\ 1 \\ 0 \\ 2 \end{array}\right]$$
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