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What is the sum of the following four vectors in (a) unitvector notation, and as (b) a magnitude and (c) an angle? $$\begin{array}{ll}\vec{A}=(2.00 \mathrm{~m}) \hat{\mathrm{i}}+(3.00 \mathrm{~m}) \hat{\mathrm{j}} & \vec{B}: 4.00 \mathrm{~m}, \text { at }+65.0^{\circ} \\ \vec{C}=(-4.00 \mathrm{~m}) \hat{\mathrm{i}}+(-6.00 \mathrm{~m}) \hat{\mathrm{j}} & \vec{D}: 5.00 \mathrm{~m}, \text { at }-235^{\circ}\end{array}$$

Short Answer

Expert verified
In unit vector notation: \(-3.1776\hat{i} - 3.4708\hat{j}\); magnitude: 4.705 m; angle: 227.77°.

Step by step solution

01

Convert Vector B to Unit Vector Notation

Vector \( \vec{B} \) has a magnitude of 4.00 m and an angle of +65.0 degrees. To convert into unit vector notation:\( \vec{B} = (4.00 \cos(65.0^{\circ}))\hat{i} + (4.00 \sin(65.0^{\circ}))\hat{j} \).Calculating these components:\( \vec{B} = (4.00 \times 0.4226)\hat{i} + (4.00 \times 0.9063)\hat{j} = 1.6904\hat{i} + 3.6252\hat{j} \).
02

Convert Vector D to Unit Vector Notation

Vector \( \vec{D} \) has a magnitude of 5.00 m and an angle of -235 degrees. To convert it into unit vector notation:\( \vec{D} = (5.00 \cos(-235^{\circ}))\hat{i} + (5.00 \sin(-235^{\circ}))\hat{j} \).Calculating the components:\( \vec{D} = (5.00 \times -0.5736)\hat{i} + (5.00 \times -0.8192)\hat{j} = -2.8680\hat{i} + -4.0960\hat{j} \).
03

Add All Vectors in Unit Vector Notation

Now, add all vectors, \( \vec{A}, \vec{B}, \vec{C}, \) and \( \vec{D} \) in unit vector notation:\[ \vec{A} + \vec{B} + \vec{C} + \vec{D} = (2.00\hat{i} + 3.00\hat{j}) + (1.6904\hat{i} + 3.6252\hat{j}) + (-4.00\hat{i} - 6.00\hat{j}) + (-2.8680\hat{i} - 4.0960\hat{j}) \].Simplify by combining like terms:\[ \vec{R} = (2.00 + 1.6904 - 4.00 - 2.8680)\hat{i} + (3.00 + 3.6252 - 6.00 - 4.0960)\hat{j} \]\[ \vec{R} = (-3.1776)\hat{i} + (-3.4708)\hat{j} \].
04

Calculate the Magnitude of the Resultant Vector

The magnitude \( R \) of the resultant vector \( \vec{R} \) is given by the formula:\[ R = \sqrt{(R_x)^2 + (R_y)^2} \]\[ R = \sqrt{(-3.1776)^2 + (-3.4708)^2} \]\[ R = \sqrt{10.0984 + 12.0475} = \sqrt{22.1459} = 4.705 \text{ m} \].
05

Determine the Angle of the Resultant Vector

The angle \( \theta \) of the resultant vector from the positive x-axis is:\[ \theta = \tan^{-1}\left(\frac{R_y}{R_x}\right) \]Using \( R_x = -3.1776 \) and \( R_y = -3.4708 \):\[ \theta = \tan^{-1}\left(\frac{-3.4708}{-3.1776}\right) = \tan^{-1}(1.0923) \]\[ \theta = 47.77^{\circ} \].Since both \( R_x \) and \( R_y \) are negative, the vector is in the third quadrant, so adjust the angle:\[ \theta = 180^{\circ} + 47.77^{\circ} = 227.77^{\circ} \].

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Key Concepts

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

Unit Vector Notation
Unit vector notation is a convenient way to express vectors, especially in two or three dimensions. It involves breaking down a vector into its horizontal and vertical components along the x and y axes, respectively. These components are represented using the unit vectors \( \hat{i} \) and \( \hat{j} \).
  • The \( \hat{i} \) component corresponds to the x-axis.
  • The \( \hat{j} \) component corresponds to the y-axis.
To convert a vector expressed with a certain magnitude and angle into unit vector notation, you calculate the horizontal and vertical components:
  1. For the x-component: Multiply the magnitude by the cosine of the angle.
  2. For the y-component: Multiply the magnitude by the sine of the angle.
For example, a vector \( \vec{B} \) with a magnitude of 4.00 meters at an angle of 65 degrees would have:
  • an \( \hat{i} \) component of \( 4.00 \cos(65^{\circ}) \)
  • and a \( \hat{j} \) component of \( 4.00 \sin(65^{\circ}) \)
Hence, in unit vector notation, such a vector can be expressed as: \( \vec{B} = 1.6904 \hat{i} + 3.6252 \hat{j} \). This method allows for easy addition and subtraction of vectors, acting like simple algebraic quantities.
Magnitude of a Vector
The magnitude of a vector is a measure of its length, providing a scalar quantity that describes how large or small a vector is. Magnitude is always a positive value and is represented by the symbol \( |\vec{V}| \) or simply \( V \).
In a two-dimensional space (2D), you can calculate the magnitude of a vector \( \vec{R} = a\hat{i} + b\hat{j} \) using the Pythagorean theorem:\[R = \sqrt{a^2 + b^2}\] This formula is derived from considering the vector's components as part of a right triangle, where the magnitude is the hypotenuse. For example, if our resultant vector \( \vec{R} \) has components \( -3.1776 \) along \( \hat{i} \) and \( -3.4708 \) along \( \hat{j} \), then:\[R = \sqrt{(-3.1776)^2 + (-3.4708)^2} = 4.705 \text{ m}\] The magnitude provides us with essential information regarding the vector's overall size, independent of its direction, useful in various applications from physics to engineering.
Angle of a Vector
The angle of a vector is crucial as it provides direction in relation to a known reference line, typically the positive x-axis. Understanding the angle helps in graphically representing the vector's orientation.
To find the angle \( \theta \) of a vector with components \( R_x \) and \( R_y \), use the tangent function:\[\theta = \tan^{-1}\left(\frac{R_y}{R_x}\right)\] This inverse tangent function calculates the angle opposite one of the vector's components.
For the vector \( \vec{R} \) with \( R_x = -3.1776 \) and \( R_y = -3.4708 \), the angle is:\[\theta = \tan^{-1}\left(\frac{-3.4708}{-3.1776}\right) = 47.77^{\circ}\] Since both components are negative, the vector lies in the third quadrant. This requires an adjustment to find the correct angle:\[\theta = 180^{\circ} + 47.77^{\circ} = 227.77^{\circ}\] Recognizing the vector's quadrant ensures accurate direction, which is vital in applications like navigation and vector field analysis.
Quadrant Identification
Quadrant identification is an essential skill when working with vectors, as it helps determine the vector's direction based on its components. The coordinate plane is divided into four quadrants:
  • **First Quadrant**: Both \( x \) and \( y \) components are positive.
  • **Second Quadrant**: The \( x \) component is negative, and \( y \) is positive.
  • **Third Quadrant**: Both \( x \) and \( y \) components are negative.
  • **Fourth Quadrant**: \( x \) is positive, and \( y \) is negative.
Identifying the correct quadrant helps in defining the vector's angle more precisely. For instance, if a vector has negative \( x \) and \( y \) components, it means the vector points toward the third quadrant.
This quadrant consideration is essential when calculating vectors' angles, as it influences the direction measure from the positive \( x \)-axis. Knowing the quadrant ensures that any angle calculations are adjusted accordingly, making the vector representation full and correct, which is crucial in fields like cartography and robotics.

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Most popular questions from this chapter

If \(\vec{B}\) is added to \(\vec{A}\), the result is \(6.0 \hat{\mathrm{i}}+1.0 \hat{\mathrm{j}} .\) If \(\vec{B}\) is subtracted from \(\vec{A}\), the result is \(-4.0 \hat{\mathrm{i}}+7.0 \hat{\mathrm{j}}\). What is the magnitude of \(\vec{A} ?\)

A displacement vector \(\vec{r}\) in the \(x y\) plane is \(15 \mathrm{~m}\) long and directed at angle \(\theta=30^{\circ}\) in Fig. 3 -26. Determine (a) the \(x\) component and (b) the \(y\) component of the vector.

Rock faults are ruptures along which opposite faces of rock have slid past each other. In Fig. \(3-35\), points \(A\) and \(B\) coincided before the rock in the foreground slid down to the right. The net displacement \(\overrightarrow{A B}\) is along the plane of the fault. The horizontal component of \(\overrightarrow{A B}\) is the strike-slip \(A C\). The component of \(\overrightarrow{A B}\) that is directed down the plane of the fault is the dip-slip \(A D\). (a) What is the magnitude of the net displacement \(\overrightarrow{A B}\) if the strike-slip is \(22.0 \mathrm{~m}\) and the dip- slip is \(17.0 \mathrm{~m} ?\) (b) If the plane of the fault is inclined at angle \(\phi=52.0^{\circ}\) to the horizontal, what is the vertical component of \(\overrightarrow{A B}\) ?

A vector \(\vec{d}\) has a magnitude of \(2.5 \mathrm{~m}\) and points north. What are (a) the magnitude and (b) the direction of \(4.0 \vec{d}\) ? What are (c) the magnitude and (d) the direction of \(-3.0 \vec{d}\) ?

Use the definition of scalar product, \(\vec{a} \cdot \vec{b}=a b \cos \theta\), and the fact that \(\vec{a} \cdot \vec{b}=a_{x} b_{x}+a_{y} b_{y}+a_{z} b_{z}\) to cal- culate the angle between the two vectors given by \(\vec{a}=3.0 \hat{\mathrm{i}}+\) \(3.0 \hat{\mathrm{j}}+3.0 \hat{\mathrm{k}}\) and \(\vec{b}=2.0 \hat{\mathrm{i}}+1.0 \hat{\mathrm{j}}+3.0 \hat{\mathrm{k}}\)

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