Question

Given two vectors, \(\overrightarrow{\mathbf{a}}=4.0 \hat{\mathbf{i}}-3.0 \hat{\mathbf{j}}\) and \(\overrightarrow{\mathbf{b}}=6.0 \hat{\mathbf{i}}+8.0 \hat{\mathbf{j}}\), find the magnitudes and directions (with the \(+x\) axis) of ( \(a\) ) \(\overrightarrow{\mathbf{a}},(b) \overrightarrow{\mathbf{b}},(c) \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}},(d) \overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{a}}\), and \((e) \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}\)

Short answer

The magnitudes and directions of the vectors are: \(\overrightarrow{\mathbf{a}}\) (5, -36.87°), \(\overrightarrow{\mathbf{b}}\) (10, 53.13°), \(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}\) (11.18, 26.57°), \(\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{a}}\) (11.18, 80.54°), and \(\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}\) (11.18, -80.54°).

Step-by-step solution

Understand vector notation

A vector notated as \(a\hat{i} + b\hat{j}\) means it has \(a\) units in the x direction (denoted by \(\hat{i}\)) and \(b\) units in the y direction (denoted by \(\hat{j}\)). The given vectors are: \(\overrightarrow{\mathbf{a}}=4.0 \hat{\mathbf{i}}-3.0 \hat{\mathbf{j}}\) and \(\overrightarrow{\mathbf{b}}=6.0 \hat{\mathbf{i}}+8.0 \hat{\mathbf{j}}\).

Compute the magnitudes

The magnitude of a vector \(\overrightarrow{\mathbf{v}}=a\hat{i} + b\hat{j}\) can be found using the formula \(\|v\| = \sqrt{a^2 + b^2}\). Using this, the magnitude of vector \(\overrightarrow{\mathbf{a}}\) is \(\sqrt{4^2 + (-3)^2} =5 \) and the magnitude of \(\overrightarrow{\mathbf{b}}\) is \(\sqrt{6^2 + 8^2}=10\).

Compute the directions

The direction (angle) \(\Theta\) of a vector \(\overrightarrow{\mathbf{v}}=a\hat{i} + b\hat{j}\) can be found using the formula \( \Theta = \tan^{-1}(b/a) \). Using this, the direction of \(\overrightarrow{\mathbf{a}}\) is \(\tan^{-1}(-3/4) = -36.87^\circ \) and of \(\overrightarrow{\mathbf{b}}\) is \(\tan^{-1}(8/6) = 53.13^\circ \). The negative angle indicates that the vector is pointing downwards from the positive x-axis.

Add and subtract vectors

Vectors can be added or subtracted by adding or subtracting their corresponding component. The vectors \( \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}} = (4+6)\hat{\mathbf{i}} + (-3+8)\hat{\mathbf{j}} = 10\hat{\mathbf{i}}+5\hat{\mathbf{j}} \) and \( \overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{a}} = (6-4)\hat{\mathbf{i}} + (8-(-3))\hat{\mathbf{j}} = 2\hat{\mathbf{i}}+11\hat{\mathbf{j}} \) and \( \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}} = (4-6)\hat{\mathbf{i}} + ((-3)-8)\hat{\mathbf{j}} = -2\hat{\mathbf{i}}-11\hat{\mathbf{j}} \).

Find magnitudes and directions of the result vectors

Using the same procedures applied in Steps 2 and 3, the magnitude and direction of the vectors found in step 4 will be found. \(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}} = \sqrt{10^2 + 5^2} = 11.18, \angle = \tan^{-1}(5/10) = 26.57^\circ \) . \(\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{a}} =\sqrt{2^2 + 11^2} = 11.18, \angle = \tan^{-1}(11/2) = 80.54^\circ \) . \(\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}} =\sqrt{(-2)^2 + (-11)^2} = 11.18, \angle = \tan^{-1}(-11/(-2)) = -80.54^\circ \). The negative angle indicates that the vector is pointing downwards from the positive x-axis for \(\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}\).

Key concepts

Vector Addition and Subtraction

Understanding vector addition and subtraction is key to working with multiple forces, velocities, or other vector quantities. In our exercise, we add and subtract vectors by combining their horizontal and vertical components separately.

For example, when we combine vector \(\overrightarrow{\mathbf{a}}\) with vector \(\overrightarrow{\mathbf{b}}\), we add their corresponding components: the \(\hat{\mathbf{i}}\) (x) components together and the \(\hat{\mathbf{j}}\) (y) components together. This straightforward method is used because vectors obey the commutative law of addition, which allows us to add them in any order. The resultant vector has its own magnitude and direction, which may represent a new force, speed, or another physical vector quantity in a given problem.

Vector Components

The components of a vector are essentially the projections of the vector along the axes of a coordinate system. For the vectors in our exercise, \(\overrightarrow{\mathbf{a}}=4.0\hat{\mathbf{i}}-3.0\hat{\mathbf{j}}\) and \(\overrightarrow{\mathbf{b}}=6.0\hat{\mathbf{i}}+8.0\hat{\mathbf{j}}\), the \(\hat{\mathbf{i}}\) and \(\hat{\mathbf{j}}\) represent the vector's x and y components, respectively.

It's crucial to note that the x component (along \(\hat{\mathbf{i}}\)) indicates how far the vector goes in the horizontal direction, and the y component (along \(\hat{\mathbf{j}}\)) illustrates how far it goes in the vertical direction. In real-life applications, breaking a vector into its components allows us to analyze and solve problems that involve forces or motions in two-dimensional space more easily.

Magnitude Calculation

The magnitude of a vector represents its 'size' or 'length'. To calculate the magnitude of vectors such as \(\overrightarrow{\mathbf{a}}\) and \(\overrightarrow{\mathbf{b}}\) in our exercise, we use the Pythagorean theorem applied to the vector's components. This is done by squaring each component, adding the squares together, and then taking the square root of that sum.

In mathematical terms, for a vector \(\overrightarrow{\mathbf{v}}=a\hat{\mathbf{i}} + b\hat{\mathbf{j}}\), its magnitude is \(||v|| = \sqrt{a^2 + b^2}\). The calculation of magnitude is essential when you need to know the 'strength' of a force, the length of a displacement, or the speed of a velocity vector.

Direction Angle of Vectors

The direction angle of a vector is the angle it makes with the positive x-axis and its essential for understanding in which direction the vector is 'pointing'. In our problem, the direction angles for vectors \(\overrightarrow{\mathbf{a}}\) and \(\overrightarrow{\mathbf{b}}\) are found using the arctangent function: \(\Theta = \tan^{-1}(b/a)\).

The angle is measured counterclockwise from the positive x-axis, which is standard convention. However, if a vector's y-component is negative, the direction angle will be negative as well, indicating that the vector points below the x-axis. This concept helps us visualize vector interactions and is extremely useful in fields such as physics and engineering, where the direction of a force or velocity is just as important as its magnitude.