Question

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

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

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 \( \overrightarrow{\mathbf{B}} = (15, 0) \).

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

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

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

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

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

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

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

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

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

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

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

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

Key concepts

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.