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Vector \(\overrightarrow{\mathbf{A}}\) points in the negative \(y\) direction and has a magnitude of \(5 \mathrm{~km}\). Vector \(\overrightarrow{\mathbf{B}}\) has a magnitude of \(15 \mathrm{~km}\) and points in the positive \(x\) direction. Use components to find the magnitude of (a) \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}\), (b) \(\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}\), and (c) \(\overrightarrow{\mathbf{B}}-\overrightarrow{\mathbf{A}}\).

Short Answer

Expert verified
All magnitudes \( |\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}| \), \( |\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}| \), and \( |\overrightarrow{\mathbf{B}}-\overrightarrow{\mathbf{A}}| \) equal \( 5\sqrt{10} \approx 15.81 \text{ km} \).

Step by step solution

01

Represent Vectors in Component Form

For vector \( \overrightarrow{\mathbf{A}} \), since it is pointing in the negative \( y \) direction with a magnitude of 5 km, its components are \( \overrightarrow{\mathbf{A}} = (0, -5) \). For vector \( \overrightarrow{\mathbf{B}} \), since it points in the positive \( x \) direction with a magnitude of 15 km, its components are \( \overrightarrow{\mathbf{B}} = (15, 0) \).
02

Calculate Components of \( \overrightarrow{\mathbf{A}} + \overrightarrow{\mathbf{B}} \)

To add vectors \( \overrightarrow{\mathbf{A}} \) and \( \overrightarrow{\mathbf{B}} \), sum the respective components: \[\overrightarrow{\mathbf{A}} + \overrightarrow{\mathbf{B}} = (0 + 15, -5 + 0) = (15, -5)\].
03

Calculate Magnitude of \( \overrightarrow{\mathbf{A}} + \overrightarrow{\mathbf{B}} \)

Use the Pythagorean theorem to find the magnitude of the resultant vector: \[|\overrightarrow{\mathbf{A}} + \overrightarrow{\mathbf{B}}| = \sqrt{15^2 + (-5)^2} = \sqrt{225 + 25} = \sqrt{250} = 5\sqrt{10} \approx 15.81 \, \text{km}\].
04

Calculate Components of \( \overrightarrow{\mathbf{A}} - \overrightarrow{\mathbf{B}} \)

To subtract vectors, subtract the components of \( \overrightarrow{\mathbf{B}} \) from \( \overrightarrow{\mathbf{A}} \): \[\overrightarrow{\mathbf{A}} - \overrightarrow{\mathbf{B}} = (0 - 15, -5 - 0) = (-15, -5)\].
05

Calculate Magnitude of \( \overrightarrow{\mathbf{A}} - \overrightarrow{\mathbf{B}} \)

Calculate the magnitude of \( \overrightarrow{\mathbf{A}} - \overrightarrow{\mathbf{B}} \) with the Pythagorean theorem: \[|\overrightarrow{\mathbf{A}} - \overrightarrow{\mathbf{B}}| = \sqrt{(-15)^2 + (-5)^2} = \sqrt{225 + 25} = \sqrt{250} = 5\sqrt{10} \approx 15.81 \, \text{km}\].
06

Calculate Components of \( \overrightarrow{\mathbf{B}} - \overrightarrow{\mathbf{A}} \)

Now, subtract \( \overrightarrow{\mathbf{A}} \) from \( \overrightarrow{\mathbf{B}} \): \[\overrightarrow{\mathbf{B}} - \overrightarrow{\mathbf{A}} = (15 - 0, 0 - (-5)) = (15, 5)\].
07

Calculate Magnitude of \( \overrightarrow{\mathbf{B}} - \overrightarrow{\mathbf{A}} \)

Find the magnitude of \( \overrightarrow{\mathbf{B}} - \overrightarrow{\mathbf{A}} \) using the Pythagorean theorem: \[|\overrightarrow{\mathbf{B}} - \overrightarrow{\mathbf{A}}| = \sqrt{15^2 + 5^2} = \sqrt{225 + 25} = \sqrt{250} = 5\sqrt{10} \approx 15.81 \, \text{km}\].

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Key Concepts

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

Vector Magnitude
Vectors have both direction and magnitude. The magnitude of a vector is essentially its "length" and is always a positive number. It tells you how far a vector extends in space without considering the direction in which it points. The magnitude of a vector \(\overrightarrow{\mathbf{V}}\) with components \( (x, y)\) can be calculated using the formula: \[|\overrightarrow{\mathbf{V}}| = \sqrt{x^2 + y^2}\].
For example, if we have a vector with components \( (3, 4)\), its magnitude would be \( \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5\).
This measure is especially useful in physics and engineering to understand the intensity or speed associated with the vector, such as in vectors representing forces or velocities. The magnitude helps in comparing different vectors to see which one is "stronger" or longer.
Vector Components
Vectors can be broken down into simpler parts known as components. These components represent the projections of the vector along the coordinate axes, usually the \(x\) and \(y\) axes in a two-dimensional plane. Breaking down a vector into its components makes it easier to analyze, especially when performing operations like addition and subtraction.
To determine the components of a vector, you look at how much it "moves" horizontally and vertically. For a vector pointing due east with a magnitude of 10 units, its components would be \( (10, 0)\) because it only moves in the \(x\) direction. Similarly, a vector pointing due south with the same magnitude would have components \((0, -10)\).
When vectors are represented by their components, vector addition becomes a simple task. We add corresponding components to obtain the resultant vector. For instance, adding \( (10, 0)\) and \( (0, -10)\) results in \( (10, -10)\).
Understanding vector components is fundamental for applying them in real-world scenarios, such as calculating net forces in physics or displacement in navigation.
Pythagorean Theorem
The Pythagorean Theorem is a crucial tool in geometry and vector analysis. It connects the lengths of the sides of a right triangle, stating that the square of the hypotenuse (the side opposite the right angle) equals the sum of the squares of the other two sides. In mathematical terms, for a right triangle with sides \(a\) and \(b\), and hypotenuse \(c\), the theorem is written as: \[c^2 = a^2 + b^2\].
In vectors, the Pythagorean Theorem helps calculate the magnitude of a vector when given its components. The components form a right triangle, with the vector acting as the hypotenuse. The theorem is thus central in calculating vector magnitudes during addition or subtraction processes.
For example, if a vector's components are \( (x, y)\), using the Pythagorean Theorem, the magnitude is \( |\overrightarrow{\mathbf{V}}| = \sqrt{x^2 + y^2}\), showing how the theorem seamlessly integrates with vector calculations.
The Pythagorean Theorem provides a straightforward way to solve numerous problems involving right triangles, making it indispensable in both academic learning and practical applications, like determining distances and calculating force magnitudes.

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Most popular questions from this chapter

A road that rises \(1 \mathrm{ft}\) for every \(100 \mathrm{ft}\) traveled horizontally is said to have a 1\% grade. Portions of the Lewiston grade, near Lewiston, Idaho, have a \(6 \%\) grade. At what angle is this road inclined above the horizontal?

When running around a track on a windy day, you feel the wind is stronger when you are running into it than when you run with it. Explain in terms of relative velocity.

A diver runs horizontally off the end of a \(3.0-\mathrm{m}\)-high diving board with an initial speed of \(1.8 \mathrm{~m} / \mathrm{s}\). (a) Given that the diver's initial position is \(x_{\mathrm{i}}=0\) and \(y_{\mathrm{i}}=3.0 \mathrm{~m}\), find her \(x\) and \(y\) positions at the times \(t=0.25 \mathrm{~s}\), \(t=0.50 \mathrm{~s}\), and \(t=0.75 \mathrm{~s}\). (b) Plot the results from part (a), and sketch the corresponding parabolic path.

Suppose that \(\overrightarrow{\mathbf{A}}\) and \(\overrightarrow{\mathbf{B}}\) have nonzero magnitudes. Is it possible for \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}\) to be zero?

An air traffic controller observes two airplanes approaching the airport. The displacement from the control tower to plane 1 is given by the vector \(\overrightarrow{\mathbf{A}}\), which has a magnitude of \(220 \mathrm{~km}\) and points in a direction \(32^{\circ}\) north of west. The displacement from the control tower to plane 2 is given by the vector \(\overrightarrow{\mathbf{B}}\), which has a magnitude of \(140 \mathrm{~km}\) and points \(65^{\circ}\) east of north. (a) Sketch the vectors \(\overrightarrow{\mathbf{A}},-\overrightarrow{\mathbf{B}}\), and \(\overrightarrow{\mathbf{D}}=\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}\). Notice that \(\overrightarrow{\mathbf{D}}\) is the displacement from plane 2 to plane 1 . (b) Use components to find the magnitude and the direction of the vector \(\overrightarrow{\mathrm{D}}\).

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