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Chapter 3: Problem 53

A vector is given by \(\mathbf{R}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}} .\) Find \((\mathbf{a})\) the magnitudes of the \(x, y,\) and \(z\) components, \((\mathrm{b})\) the magnitude of \(\mathbf{R},\) and \((c)\) the angles between \(\mathbf{R}\) and the \(x, y,\) and \(z\) axes.

Short Answer

Expert verified
The magnitudes of the x, y, and z components are 2, 1, and 3, respectively. The magnitude of \(\mathbf{R}\) is \(\sqrt{14}\). The angles with the x, y, and z axes are \(\cos^{-1}(2/\sqrt{14})\), \(\cos^{-1}(1/\sqrt{14})\), and \(\cos^{-1}(3/\sqrt{14})\) respectively.

Step by step solution

01

Part A - Finding the magnitudes of components

For the vector \(\mathbf{R} = 2\hat{\mathbf{i}} + \hat{\mathbf{j}} + 3\hat{\mathbf{k}}\), the magnitudes of the components are simply the coefficients of the unit vectors. Thus, the magnitude of the x component is 2, the magnitude of the y component is 1, and the magnitude of the z component is 3.
02

Part B - Calculating the magnitude of \(\mathbf{R}\)

To find the magnitude of the vector \(\mathbf{R}\), we can use the Pythagorean theorem in three dimensions: \[\lVert \mathbf{R} \rVert = \sqrt{(2)^2 + (1)^2 + (3)^2} = \sqrt{4 + 1 + 9} = \sqrt{14}.\]
03

Part C - Determining the angles with the axes

The angles \(\theta_x\), \(\theta_y\), and \(\theta_z\) between \(\mathbf{R}\) and the x, y, and z axes can be calculated using the dot product formula: \[\cos(\theta_x) = \frac{2}{\sqrt{14}}, \quad \cos(\theta_y) = \frac{1}{\sqrt{14}}, \quad \cos(\theta_z) = \frac{3}{\sqrt{14}}.\] Take the inverse cosine to find the angles: \[\theta_x = \cos^{-1}\left(\frac{2}{\sqrt{14}}\right), \quad \theta_y = \cos^{-1}\left(\frac{1}{\sqrt{14}}\right), \quad \theta_z = \cos^{-1}\left(\frac{3}{\sqrt{14}}\right).\]

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Key Concepts

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

Magnitude of a Vector
The concept of the magnitude of a vector is fundamental in understanding vector properties. It represents the 'size' or 'length' of the vector. Imagine drawing a straight line from one point to another; the vector's magnitude would be the length of that line.

When we have a vector, such as \(\mathbf{R} = 2\hat{\mathbf{i}} + \hat{\mathbf{j}} + 3\hat{\mathbf{k}}\), we're looking at a vector in three-dimensional space. To calculate its magnitude, we use a formula analogous to the Pythagorean theorem, \(\lVert \mathbf{R} \rVert = \sqrt{(2)^2 + (1)^2 + (3)^2}\), which simplifies to \(\sqrt{14}\). This magnitude gives us an idea of how far the point described by the vector is from the origin in a spatial sense.
Vector Components
Vector components are essentially projections of a vector along the axes of a coordinate system. In simpler terms, they answer the question: 'How much does this vector go in the direction of each axis?'

For the vector \(\mathbf{R}\) from our exercise, its components along the \(x\), \(y\), and \(z\) axes are given as 2, 1, and 3, respectively. These are the vector magnitudes in the direction of each standard basis vector, \(\hat{\mathbf{i}}\), \(\hat{\mathbf{j}}\), and \(\hat{\mathbf{k}}\). Conceptually, you can imagine breaking down a diagonal movement into separate steps along the street grid - one step east (\(x\)), one step north (\(y\)), and one step up (\
Angles Between Vectors and Axes
Understanding the angles between a vector and the coordinate axes helps in visualizing the vector's orientation in space. These angles tell us how steeply a vector inclines from each axis.

Let's simplify this: if a vector is aimed directly along an axis, the angle with that axis would be 0 degrees. Conversely, if it points directly away, that angle would be 180 degrees. Any other orientation results in an angle somewhere between 0 and 180 degrees. To find these angles for \(\mathbf{R}\), we rely on the dot product and trigonometric functions. For example, for the x-axis, we calculate \(\theta_x\) using \(\cos(\theta_x) = \frac{2}{\sqrt{14}}\) and similarly for the other axes. Taking the inverse cosine of each gives us the angle \(\theta_x\), \(\theta_y\), and \(\theta_z\), respectively.

These angles are essential for tasks ranging from physics problems, where we analyze force directions, to computer graphics, where they help determine an object's orientation in a 3D environment.

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Most popular questions from this chapter

Instructions for finding a buried treasure include the following: Go 75.0 paces at \(240^{\circ},\) turn to \(135^{\circ}\) and walk 125 paces, then travel 100 paces at \(160^{\circ} .\) The angles are measured counterclockwise from an axis pointing to the east, the \(+x\) direction. Determine the resultant displacement from the starting point.

Find the magnitude and direction of the resultant of three displacements having rectangular components 2.00 ) \(\mathrm{m},(-5.00,3.00) \mathrm{m},\) and \((6.00,1.00) \mathrm{m}\)

Consider the three displacement vectors \(\mathbf{A}=(3 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}) \mathrm{m},\) \(\mathbf{B}=(\hat{\mathbf{i}}-4 \hat{\mathbf{j}}) \mathrm{m},\) and \(\mathbf{C}=(-2 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}) \mathrm{m} .\) Use the component method to determine (a) the magnitude and direc- tion of the vector \(\mathbf{D}=\mathbf{A}+\mathbf{B}+\mathbf{C},\) (b) the magnitude and direction of \(\mathbf{E}=-\mathbf{A}-\mathbf{B}+\mathbf{C}\)

Consider a game in which \(N\) children position themselves at equal distances around the circumference of a circle. At the center of the circle is a rubber tire. Each child holds a rope attached to the tire and, at a signal, pulls on his rope. All children exert forces of the same magnitude \(F\) . In the case \(N=2,\) it is easy to see that the net force on the tire will be zero, because the two oppositely directed force vectors add to zero. Similarly, if \(N=4,6,\) or any even integer, the resultant force on the tire must be zero, because the forces exerted by each pair of oppositely positioned children will cancel. When an odd number of children are around the circle, it is not so obvious whether the total force on the central tire will be zero. (a) Calculate the net force on the tire in the case \(N=3,\) by adding the components of the three force vectors. Choose the \(x\) axis to lie along one of the ropes. (b) What If? Determine the net force for the general case where \(N\) is any integer, odd or even, greater than one. Proceed as follows: Assume that the total force is not zero. Then it must point in some particular direction. Let every child move one position clockwise. Give a reason that the total force must then have a direction turned clockwise by \(360^{\circ} / N .\) Argue that the total force must nevertheless be the same as before. Explain that the contradiction proves that the magnitude of the force is zero. This problem illustrates a widely useful technique of proving a result "by symmetry"-by using a bit of the mathematics of group theory. The particular situation is actually encountered in physics and chemistry when an array of electric charges (ions) exerts electric forces on an atom at a central position in a molecule or in a crystal.

Arbitrarily define the "instantaneous vector height" of a person as the displacement vector from the point halfway between his or her feet to the top of the head. Make an order-of-magnitude estimate of the total vector height of all the people in a city of population 100000 (a) at 10 o'clock on a Tuesday morning, and (b) at 5 o'clock on a Saturday morning. Explain your reasoning.
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