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For the displacement vectors \(\vec{a}=(3.0 \mathrm{~m}) \hat{\mathrm{i}}+(4.0 \mathrm{~m}) \hat{\mathrm{j}}\) and \(\vec{b}=\) \((5.0 \mathrm{~m}) \hat{\mathrm{i}}+(-2.0 \mathrm{~m}) \hat{\mathrm{j}}\), give \(\vec{a}+\vec{b}\) in (a) unit-vector notation, and as (b) a magnitude and (c) an angle (relative to \(\hat{i}\) ). Now give \(\vec{b}-\vec{a}\) in (d) unit-vector notation, and as (e) a magnitude and (f) an angle.

Short Answer

Expert verified
\(\vec{a} + \vec{b} = 8.0\hat{\mathrm{i}} + 2.0\hat{\mathrm{j}}\), magnitude \(= 8.25\, \mathrm{m}\), angle \(= 14.0^\circ\); \(\vec{b} - \vec{a} = 2.0\hat{\mathrm{i}} - 6.0\hat{\mathrm{j}}\), magnitude \(= 6.32\, \mathrm{m}\), angle \(= -71.6^\circ\).

Step by step solution

01

Calculate \(\vec{a} + \vec{b}\) in Unit-Vector Notation

To find \(\vec{a} + \vec{b}\), we add the corresponding components of vectors \(\vec{a}\) and \(\vec{b}\).Given:\[ \vec{a} = 3.0 \hat{\mathrm{i}} + 4.0 \hat{\mathrm{j}} \]\[ \vec{b} = 5.0 \hat{\mathrm{i}} - 2.0 \hat{\mathrm{j}} \]Adding the \(\hat{\mathrm{i}}\) components:\[ 3.0 + 5.0 = 8.0 \hat{\mathrm{i}} \]Adding the \(\hat{\mathrm{j}}\) components:\[ 4.0 + (-2.0) = 2.0 \hat{\mathrm{j}} \]Thus, \(\vec{a} + \vec{b} = 8.0 \hat{\mathrm{i}} + 2.0 \hat{\mathrm{j}}\).
02

Calculate the Magnitude of \(\vec{a} + \vec{b}\)

The magnitude \(|\vec{a} + \vec{b}|\) is found using the Pythagorean theorem:\[ |\vec{a} + \vec{b}| = \sqrt{(8.0)^2 + (2.0)^2} \]Calculate the squares:\[ 8.0^2 = 64.0 \]\[ 2.0^2 = 4.0 \]Add them:\[ 64.0 + 4.0 = 68.0 \]Take the square root:\[ |\vec{a} + \vec{b}| = \sqrt{68.0} \approx 8.25\, \mathrm{m} \]
03

Calculate the Angle of \(\vec{a} + \vec{b}\) Relative to \(\hat{\mathrm{i}}\)

The angle \(\theta\) with respect to \(\hat{\mathrm{i}}\) is given by:\[ \theta = \text{tan}^{-1}\left(\frac{\text{component along } \hat{\mathrm{j}}}{\text{component along } \hat{\mathrm{i}}}\right) = \text{tan}^{-1}\left(\frac{2.0}{8.0}\right) \]Calculate the fraction:\[ \frac{2.0}{8.0} = 0.25 \]Now, compute the angle:\[ \theta = \text{tan}^{-1}(0.25) \approx 14.0^\circ \]
04

Calculate \(\vec{b} - \vec{a}\) in Unit-Vector Notation

To find \(\vec{b} - \vec{a}\), subtract the components of \(\vec{a}\) from \(\vec{b}\).Subtract the \(\hat{\mathrm{i}}\) components:\[ 5.0 - 3.0 = 2.0 \hat{\mathrm{i}} \]Subtract the \(\hat{\mathrm{j}}\) components:\[ -2.0 - 4.0 = -6.0 \hat{\mathrm{j}} \]Thus, \(\vec{b} - \vec{a} = 2.0 \hat{\mathrm{i}} - 6.0 \hat{\mathrm{j}}\).
05

Calculate the Magnitude of \(\vec{b} - \vec{a}\)

The magnitude \(|\vec{b} - \vec{a}|\) is:\[ |\vec{b} - \vec{a}| = \sqrt{(2.0)^2 + (-6.0)^2} \]Calculate the squares:\[ 2.0^2 = 4.0 \]\[ (-6.0)^2 = 36.0 \]Add them:\[ 4.0 + 36.0 = 40.0 \]Take the square root:\[ |\vec{b} - \vec{a}| = \sqrt{40.0} \approx 6.32\, \mathrm{m} \]
06

Calculate the Angle of \(\vec{b} - \vec{a}\) Relative to \(\hat{\mathrm{i}}\)

The angle \(\phi\) with respect to \(\hat{\mathrm{i}}\) is:\[ \phi = \text{tan}^{-1}\left(\frac{\text{component along } \hat{\mathrm{j}}}{\text{component along } \hat{\mathrm{i}}}\right) = \text{tan}^{-1}\left(\frac{-6.0}{2.0}\right) \]Calculate:\[ \frac{-6.0}{2.0} = -3.0 \]Compute the angle:\[ \phi = \text{tan}^{-1}(-3.0) \approx -71.6^\circ \]

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Key Concepts

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

Unit-Vector Notation
Vectors can be represented in unit-vector notation, which is a standardized way of expressing vectors in terms of their components along the coordinate axes.
In two-dimensional space, this involves the unit vectors
  • \(\hat{\mathrm{i}}\) for the x-axis
  • \(\hat{\mathrm{j}}\) for the y-axis
For example, a vector with x-component 3 meters and y-component 4 meters can be expressed as \(3\hat{\mathrm{i}} + 4\hat{\mathrm{j}}\).
When adding vectors like \(\vec{a}\) and \(\vec{b}\), the rule is straightforward: simply add the corresponding components.
This means you add the \(\hat{\mathrm{i}}\) parts together and the \(\hat{\mathrm{j}}\) parts together.
The result is a new vector, which is a combination of the two initial vectors' components.
Magnitude of a Vector
The magnitude of a vector is its length or size, representing how much of something there is in terms of direction.
It is always a non-negative number. For a vector in two-dimensional space, the magnitude is the distance from the origin, calculated using the Pythagorean theorem:
\[ |\vec{v}| = \sqrt{(x)^2 + (y)^2} \] where \(x\) and \(y\) are the components of the vector.
For example, the magnitude of the vector \(8\hat{\mathrm{i}} + 2\hat{\mathrm{j}}\) is \(\sqrt{8^2 + 2^2} = \sqrt{68} \approx 8.25\,\mathrm{m}\).
This calculation uses the vector's components to determine its overall length in space, crucial for understanding its real-world impact in physical terms.
Angle of a Vector
The angle of a vector illustrates its direction relative to a reference axis, commonly the positive x-axis represented by \(\hat{\mathrm{i}}\).
This angle is determined using trigonometry, specifically the inverse tangent function (tan\(^{-1}\)):
\[ \theta = \text{tan}^{-1}\left(\frac{\text{component along } \hat{\mathrm{j}}}{\text{component along } \hat{\mathrm{i}}}\right) \] For example, for a vector \(8\hat{\mathrm{i}} + 2\hat{\mathrm{j}}\), the calculation is \(\theta = \text{tan}^{-1}(\frac{2}{8}) \approx 14.0^\circ\).
This angle shows how much the vector deviates from the horizontal axis, providing insight into its directional flow.
The proper computation of this angle is vital for analyzing vectors' directional tendencies in physics and engineering contexts.
Pythagorean Theorem
The Pythagorean theorem is one of the most fundamental concepts in geometry and vector mathematics.
It states that in a right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides:
\[ c^2 = a^2 + b^2 \] In vector terms, if you have a vector \(\vec{v}\) with components \(x\) and \(y\), then the magnitude (hypotenuse) is calculated as:
\[ |\vec{v}| = \sqrt{x^2 + y^2} \]
This method enables you to find a vector's overall length based on its component parts.
It's useful for calculating distances in physics problems involving two-dimensional planes, ensuring precise measurements of vector quantities.
Trigonometry in Vectors
Trigonometry is a powerful tool in understanding vector quantities because it links angles to the ratios of the sides in right triangles.
In vector mathematics, trigonometric functions like sine, cosine, and tangent help describe the relationships between a vector's magnitude and direction.
Key formulas include:
  • \( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \)
  • \( \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} \)
  • \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \)
These relationships enable you to solve for missing vector components or angles, making trigonometry indispensable in vector addition, subtraction, and other vector operations.
Its application is widespread in fields like physics, engineering, and computer graphics, where precise vector calculations are essential.

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Most popular questions from this chapter

Starting from an oasis, a camel walks \(25 \mathrm{~km}\) in a direction \(30^{\circ}\) south of west and then walks \(30 \mathrm{~km}\) toward the north to a second oasis. What is the direction from the first oasis to the second oasis?

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An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) while returning to base camp. He was supposed to travel due north for \(4.8 \mathrm{~km}\), but when the snow clears, he discovers that he actually traveled \(7.8 \mathrm{~km}\) at \(50^{\circ}\) north of due east. (a) How far and (b) in what direction must he now travel to reach base camp?

Find the (a) \(x\), (b) \(y\), and (c) z components of the sum \(\vec{r}\) of the displacements \(\vec{c}\) and \(\vec{d}\) whose components in meters are \(c_{x}=7.4, c_{y}=-3.8, c_{z}=-6.1 ; d_{x}=4.4, d_{y}=-2.0, d_{z}=3.3\).

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