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Write each of the given vectors in terms of the unit vectors \(\mathbf{i}\) and \(\mathbf{j}\). $$\mathbf{u}=\langle-4,6\rangle$$

Short Answer

Expert verified
\(\mathbf{u} = -4\mathbf{i} + 6\mathbf{j}\)

Step by step solution

01

Understand the structure of vector representation

For a two dimensional vector represented by \(\mathbf{u} = \langle x, y \rangle\), it can be mapped to unit vectors as \(\mathbf{u} = x\mathbf{i} + y\mathbf{j}\). Here \(\mathbf{i}\) and \(\mathbf{j}\) are unit vectors along x and y axis respectively.
02

Apply vector components to the formula

Substitute the given vector components into the formula. Here, x = -4 and y = 6 from vector \(\mathbf{u}\). Therefore, the expression in terms of \(i\) and \(j\) unit vectors will be \(\mathbf{u}= -4\mathbf{i} + 6\mathbf{j}\).

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Key Concepts

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

Unit Vectors in Vector Representation
Unit vectors are essential in breaking down and simplifying vector equations. They are vectors with a magnitude of exactly one and indicate direction in a space. Typically, in two-dimensional vector systems, these unit vectors are denoted as \(\mathbf{i}\) and \(\mathbf{j}\). The vector \(\mathbf{i}\) represents a unit vector along the horizontal axis (x-axis), while \(\mathbf{j}\) represents a unit vector along the vertical axis (y-axis).

These vectors serve as basic building blocks in vector mathematics, enabling the representation of any two-dimensional vector as a combination of these unit vectors. For example, a vector \(\mathbf{u} = \langle x, y \rangle\) can be decomposed into \(x\mathbf{i} + y\mathbf{j}\). This simplifies operations like addition and scalar multiplication, leading to more intuitive and visual manipulations.
Understanding unit vectors allows us to comprehend how much a vector is stretched and in which direction, making them powerful tools in physics and engineering, as well as in simple graph calculations.
Understanding Two-Dimensional Vectors
Two-dimensional vectors, often depicted as \(\mathbf{v} = \langle x, y \rangle\), reside in a plane defined by both the x-axis and y-axis. Essentially, these vectors describe movement or pointing in both horizontal and vertical directions at once.

They are a fundamental part of vector analysis because they allow the representation of any point or line segment within this flat space. When analyzing vectors in two dimensions, it's crucial to visualize them on what you may refer to as the Cartesian plane, where each vector is defined by its component parts.
  • The horizontal component (x-value) tells you how far to move left or right.
  • The vertical component (y-value) tells you how far to move up or down.
In practical applications, two-dimensional vectors are used in topics ranging from navigation to art, effectively any field requiring an analysis of points and lines in a flat space. Recognizing the components allows you to manipulate vectors, calculate their length (magnitude), and even project them onto other lines. Understanding two-dimensional vectors is fundamental to gaining insights into spatial relationships in many scientific and engineering fields.
Vector Components and Their Role
Vector components play an integral role in determining the direction and magnitude of a vector. Put simply, every vector breaks down into components that represent its directional parts. In the two-dimensional space, this means a vector \(\mathbf{u} = \langle x, y \rangle\) can be split into two key parts: the x-component and the y-component.

The x-component corresponds to how much the vector moves in the horizontal direction, and similarly, the y-component corresponds to the vertical movement. This way, any vector can be represented as a combination of its x and y components using the formula \(x\mathbf{i} + y\mathbf{j}\).
  • These components help to easily identify the effect of the vector on each axis. For instance, in our given example, with \(x = -4\) and \(y = 6\), the vector \(\mathbf{u}\) affects the x-direction by -4 units, moving left, and the y-direction by 6 units, moving upward.
  • Identifying vector components is crucial for performing vector addition, finding resultant vectors, and other operations like rotations and scalings, making these calculations straightforward and methodical.
Utilizing these components can dramatically simplify many practical problems in physics, mechanics, and engineering, bridging the gap between abstract math and real-world applications.

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Most popular questions from this chapter

Find \(\mathbf{u}-\mathbf{v}, \mathbf{u}+2 \mathbf{v},\) and \(-3 \mathbf{u}+\mathbf{v}\). $$\mathbf{u}=6 \mathbf{i}-2 \mathbf{j}, \mathbf{v}=-5 \mathbf{i}+3 \mathbf{j}$$

This set of exercises will draw on the ideas presented in this section and your general math background. Determine the set of positive values of \(a\) for which there is exactly one triangle \(A B C\) with \(A=60^{\circ}\) and \(b=10,\) where \(a\) and \(b\) are the sides opposite angles \(A\) and \(B\), respectively. Then find the set of positive values of \(a\) for which exactly two such triangles \(A B C\) exist, and the set of positive values of \(a\) for which no such triangle exists.

Find \(\mathbf{u}-\mathbf{v}, \mathbf{u}+2 \mathbf{v},\) and \(-3 \mathbf{u}+\mathbf{v}\). $$\mathbf{u}=-1.1 \mathbf{i}+4 \mathbf{j}, \mathbf{v}=4 \mathbf{i}+2.4 \mathbf{j}$$

Round your answers to two decimal places. Lucas pulls a 40 -pound box along a level surface from left to right by attaching a piece of rope to the box and pulling on it with a force \(\mathbf{F}_{1}\) of 20 pounds in the direction \(25^{\circ}\) above the horizontal. A friction force \(\mathbf{F}_{2}\) of 5 pounds is acting on the box as it is being pulled. (A friction force acts in the direction opposite to the direction of motion.) (a) Find the \(x\) and \(y\) components of \(\mathbf{F}_{1}\) (b) Find the \(x\) and \(y\) components of \(\mathbf{F}_{2}\) (c) Use your answers to parts (a) and (b) to express the vector sum \(\mathbf{F}_{1}+\mathbf{F}_{2}\) in terms of its \(x\) and \(y\) components. (d) Give the magnitude and direction of each of the other forces acting on the box. (e) Find the magnitude and direction of the net force acting on the box.

Multiply or divide as indicated, and leave the answer in trigonometric form. $$\frac{6\left(\cos \frac{\pi}{12}+i \sin \frac{\pi}{12}\right)}{3\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)}$$

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