/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 You are given two vectors \(\vec... [FREE SOLUTION] | 91Ó°ÊÓ

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You are given two vectors \(\vec{A}=-3.00 \hat{\imath}+6.00 \hat{\jmath} \quad\) and \(\vec{B}=7.00 \hat{\imath}+2.00 \hat{\jmath} . \quad\) Let \(\quad\) coun- terclockwise angles be positive. (a) What angle does \(\vec{A}\) make with the \(+x\) -axis? (b) What angle does \(\boldsymbol{B}\) make with the \(+x\) -axis? (c) Vector \(\boldsymbol{C}\) is the sum of \(\vec{A}\) and \(\vec{B},\) so \(\vec{C}=\vec{A}+\vec{B} .\) What angle does \(\vec{C}\) make with the \(+x\) -axis?

Short Answer

Expert verified
The angle that vector A makes with the x-axis is \( \arctan{\frac{6.00}{-3.00}} \), angle for vector B is \( \arctan{\frac{2.00}{7.00}} \), and angle for the vector sum C is \( \arctan{\frac{8.00}{4.00}} \).

Step by step solution

01

Calculating Angle for Vector A

The angle \( \theta_A \) that the vector \( \vec{A} \) makes with the positive x-axis can be found using the formula for finding the angle: \( \theta = \arctan{\frac{y}{x}} \). In this case, 'y' is the coefficient of \( \hat{\jmath} \) which is 6.00 and 'x' is the coefficient of \( \hat{\imath} \) which is -3.00. Thus, \( \theta_A = \arctan{\frac{6.00}{-3.00}} \).
02

Calculating Angle for Vector B

Similarly, the angle \( \theta_B \) that the vector \( \vec{B} \) makes with the positive x-axis is found as \( \theta_B = \arctan{\frac{2.00}{7.00}} \).
03

Calculating Vector C

Vector \( \vec{C} \) is the sum of vectors \( \vec{A} \) and \( \vec{B} \). Its x-component is the sum of x-components of \( \vec{A} \) and \( \vec{B} \) and its y-component is the sum of y-components of \( \vec{A} \) and \( \vec{B} \). Therefore, \( \vec{C} = (-3.00 \hat{\imath} + 7.00 \hat{\imath}) + (6.00 \hat{\jmath} + 2.00 \hat{\jmath}) = 4.00 \hat{\imath} + 8.00 \hat{\jmath} \).
04

Calculating Angle for Vector C

Finally, the angle \( \theta_C \) that the vector \( \vec{C} \) makes with the positive x-axis is found as \( \theta_C = \arctan{\frac{8.00}{4.00}} \).

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Key Concepts

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

Understanding the Arctan Function
The arctan function, short for arctangent, is a trigonometric function used to determine the angle whose tangent is a given number. In simpler terms, if you know the lengths of the opposite and adjacent sides of a right triangle, you can find the angle between the hypotenuse and the adjacent side using the arctan function.
For vectors, the scenario is somewhat similar. Vectors have both a direction and magnitude and can be represented in a coordinate plane using the x and y axes. When calculating the angle that a vector makes with the positive x-axis, the arctan function becomes very helpful.
By using the formula \( \theta = \arctan{\frac{y}{x}} \), where \( y \) is the component along the y-axis and \( x \) is the component along the x-axis, you'll obtain the angles that vectors form with the x-axis in a counterclockwise direction.
  • This function outputs angles in radians typically, but these can often be converted to degrees for easier interpretation.
  • It's critical to keep mind the quadrant in which your vector lies, as this will affect the sign and magnitude of the angle calculated.
  • For instance, in the vector exercise, the angle \( \theta_A \) calculated as \( \arctan{\frac{6.00}{-3.00}} \) means that although the output might seem straightforward, breaking down the sign and quadrant is key to interpreting the direction.
Exploration of Vector Addition
Vector addition is a fundamental operation in physics and mathematics used to find the resultant vector, which combines the effects of two or more individual vectors. It is essential when solving problems involving multiple forces or movements.
The process involves algebraically adding the corresponding components of the vectors. For instance, if you have vectors represented as \( \vec{A} = -3.00 \hat{\imath} + 6.00 \hat{\jmath} \) and \( \vec{B} = 7.00 \hat{\imath} + 2.00 \hat{\jmath} \), the resultant vector can be found by:
  • Adding the x-components: \( -3.00 + 7.00 = 4.00 \)
  • Adding the y-components: \( 6.00 + 2.00 = 8.00 \)
Thus, the resultant vector \( \vec{C} \) becomes \( 4.00 \hat{\imath} + 8.00 \hat{\jmath} \).
This method forms the basis of understanding how vectors interact and is used extensively in fields such as physics, engineering, and computer graphics to determine overall motions or forces acting in a given situation. The simplicity of adding components does not undermine the complexity of the situations that vector addition can help solve.
Overview of the Coordinate System
The coordinate system is a method for describing and specifying the location of points in a plane or space. A typical Cartesian coordinate system features two axes that are perpendicular to each other: the x-axis (horizontal) and the y-axis (vertical).
Each point in this system can be expressed as a pair of coordinates \( (x, y) \), detailing the point's horizontal and vertical displacement from the origin, which is the point \( (0,0) \).
Vectors, such as \( \vec{A} \) and \( \vec{B} \) in the exercise, are represented as combinations of these coordinates. In the positive direction, moving counterclockwise is generally considered positive in angle measurements, and this assumption helps standardize the procedure for angle calculation.
  • Understanding the coordinate system is imperative for visualizing how vectors are positioned and how they relate to each other.
  • It's crucial to consider the components of each vector to accurately calculate resultant vectors and associated angles.
  • Whether in 2D or 3D space, interpreting vector problems is made significantly easier with a strong grasp on these coordinate axes.
By mastering how vectors are depicted within this system, students will be better prepared to tackle problems involving movement, force, and other vector-related phenomena.

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Most popular questions from this chapter

In the fall of 2002, scientists at Los Alamos National Laboratory determined that the critical mass of neptunium- 237 is about \(60 \mathrm{~kg}\). The critical mass of a fissionable material is the minimum amount that must be brought together to start a nuclear chain reaction. Neptunium- 237 has a density of \(19.5 \mathrm{~g} / \mathrm{cm}^{3}\). What would be the radius of a sphere of this material that has a critical mass?

The scalar product of vectors \(\vec{A}\) and \(\vec{B}\) is \(+48.0 \mathrm{~m}^{2}\). Vector \(\vec{A}\) has magnitude \(9.00 \mathrm{~m}\) and direction \(28.0^{\circ}\) west of south. If vector \(\overrightarrow{\boldsymbol{B}}\) has direction \(39.0^{\circ}\) south of east, what is the magnitude of \(\overrightarrow{\boldsymbol{B}} ?\)

Emergency Landing. A plane leaves the airport in Galisteo and flies \(170 \mathrm{~km}\) at \(68.0^{\circ}\) east of north; then it changes direction to fly \(230 \mathrm{~km}\) at \(36.0^{\circ}\) south of east, after which it makes an immediate emergency landing in a pasture. When the airport sends out a rescue crew, in which direction and how far should this crew fly to go directly to this plane?

Vector \(\vec{A}=a \hat{\imath}-b \hat{k}\) and vector \(\vec{B}=-c \hat{\jmath}+d \hat{k}\) (a) In terms of the positive scalar quantities \(a, b, c,\) and \(d,\) what are \(\vec{A} \cdot \vec{B}\) and \(\vec{A} \times \vec{B} ?(b)\) If \(c=0,\) what is the magnitude of \(\vec{A} \cdot \vec{B}\) and what are the magnitude and direction of \(\vec{A} \times \vec{B}\) ? Does your result for the direction for \(\vec{A} \times \vec{B}\) agree with the result you get if you sketch \(\vec{A}\) and \(\vec{B}\) in the \(x z\) -plane and apply the right-hand rule? The scalar product can be described as the magnitude of \(\overrightarrow{\boldsymbol{B}}\) times the component of \(\overrightarrow{\boldsymbol{A}}\) that is parallel to \(\overrightarrow{\boldsymbol{B}}\). Does this agree with your result? The magnitude of the vector product can be described as the magnitude of \(\overrightarrow{\boldsymbol{B}}\) times the component of \(\vec{A}\) that is perpendicular to \(\vec{B}\). Does this agree with your result?

Vector \(\vec{A}\) is in the direction \(34.0^{\circ}\) clockwise from the \(-y\) -axis. The \(x\) -component of \(\vec{A}\) is \(A_{x}=-16.0 \mathrm{~m}\). (a) What is the \(y\) -component of \(\vec{A} ?\) (b) What is the magnitude of \(\vec{A}\) ?

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