Problem 9 - Physics for Scientists and Engineers with Modern Physics

Question

(a) Determine the magnitude and direction of the sum of the three vectors \(\overrightarrow{\mathbf{v}}_{1}=4.0 \hat{\mathbf{i}}-8.0 \hat{\mathbf{j}}, \overrightarrow{\mathbf{v}}_{2}=\hat{\mathbf{i}}+\hat{\mathbf{j}},\) and \(\overrightarrow{\mathbf{v}}_{3}=-2.0 \hat{\mathbf{i}}+4.0 \hat{\mathbf{j}} .(b)\) Determine \(\overrightarrow{\mathbf{v}}_{1}-\overrightarrow{\mathbf{v}}_{2}+\overrightarrow{\mathbf{v}}_{3}\)

Step-by-Step Solution

Answer

(a) Magnitude: \(3\sqrt{2}\), Direction: \(315^\circ\); (b) \(1.0 \hat{\mathbf{i}} - 5.0 \hat{\mathbf{j}}\)

Step 1: Add the Three Vectors

To find the sum of the vectors \(\overrightarrow{\mathbf{v}}_{1}\), \(\overrightarrow{\mathbf{v}}_{2}\), and \(\overrightarrow{\mathbf{v}}_{3}\), add their respective components. The sum of the vectors is given by:\[\begin{align*}\overrightarrow{\mathbf{v}}_{\text{sum}} &= (4.0 + 1.0 - 2.0) \hat{\mathbf{i}} + (-8.0 + 1.0 + 4.0) \hat{\mathbf{j}} \&= 3.0 \hat{\mathbf{i}} - 3.0 \hat{\mathbf{j}}.\end{align*}\]

Step 2: Calculate the Magnitude of the Sum

The magnitude of a vector \(\overrightarrow{\mathbf{v}} = a \hat{\mathbf{i}} + b \hat{\mathbf{j}}\) is calculated using the formula:\[||\overrightarrow{\mathbf{v}}|| = \sqrt{a^2 + b^2}\]For \(\overrightarrow{\mathbf{v}}_{\text{sum}} = 3.0 \hat{\mathbf{i}} - 3.0 \hat{\mathbf{j}}\):\[||\overrightarrow{\mathbf{v}}_{\text{sum}}|| = \sqrt{(3.0)^2 + (-3.0)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}.\]

Step 3: Determine the Direction of the Sum

The direction of the vector is given by the angle \(\theta\) relative to the positive \(\hat{\mathbf{i}}\)-axis, calculated using:\[\theta = \tan^{-1}\left(\frac{b}{a}\right)\]Substitute \(a = 3.0\) and \(b = -3.0\):\[\theta = \tan^{-1}\left(\frac{-3.0}{3.0}\right) = \tan^{-1}(-1) = -45^\circ\]Since the vector lies in the fourth quadrant, we add 360° to get the positive angle: \[\theta = 360^\circ - 45^\circ = 315^\circ.\]

Step 4: Calculate the Resultant of \(\overrightarrow{\mathbf{v}}_{1} - \overrightarrow{\mathbf{v}}_{2} + \overrightarrow{\mathbf{v}}_{3}\)

Subtract \(\overrightarrow{\mathbf{v}}_{2}\) from \(\overrightarrow{\mathbf{v}}_{1}\) and then add \(\overrightarrow{\mathbf{v}}_{3}\):\[\begin{align*}\overrightarrow{\mathbf{v}}_{\text{result}} &= (4.0 \hat{\mathbf{i}} - 8.0 \hat{\mathbf{j}}) - (1.0 \hat{\mathbf{i}} + 1.0 \hat{\mathbf{j}}) + (-2.0 \hat{\mathbf{i}} + 4.0 \hat{\mathbf{j}}) \&= (4.0 - 1.0 - 2.0) \hat{\mathbf{i}} + (-8.0 - 1.0 + 4.0) \hat{\mathbf{j}} \&= 1.0 \hat{\mathbf{i}} - 5.0 \hat{\mathbf{j}}.\end{align*}\]

Key Concepts

Magnitude of a Vector

Understanding the magnitude of a vector is like knowing its size or length in a coordinate space. Imagine you have a line representing a vector. The magnitude tells you exactly how long that line is. To find this, you can use the Pythagorean theorem. For a vector represented as \( \overrightarrow{\mathbf{v}} = a \hat{\mathbf{i}} + b \hat{\mathbf{j}} \), the magnitude \( ||\overrightarrow{\mathbf{v}}|| \) is calculated using:

• \( ||\overrightarrow{\mathbf{v}}|| = \sqrt{a^2 + b^2} \)

In simple terms, you square each of the components, add them together, and then take the square root of that sum.
This formula helps you understand how to get from one point to another in the plane, with detailed knowledge of the distance covered.

Vector Direction

The direction of a vector is crucial as it shows where the vector points towards on a plane. Imagine it as the arrowhead of your vector line.
To determine the direction, you calculate the angle the vector makes with a standard direction, usually the positive x-axis (\( \hat{\mathbf{i}} \)-axis). Use:

• \( \theta = \tan^{-1}\left(\frac{b}{a}\right) \)

This gives you an angle \( \theta \).
If this angle is not within the 0° to 360° range, you adjust it depending on the vector's quadrant, according to:

• For the fourth quadrant, add 360° to \( \theta \)

This ensures the angle is a positive measure heading clockwise from the positive x-axis.

Trigonometric Functions

Trigonometric functions are like tools that help us understand relationships between angles and sides of triangles. For vectors, the most common trig function is tangent (\( \tan \)) which relates to direction.
The tangent function helps in finding vector direction, using its formula:

• \( \theta = \tan^{-1}\left(\frac{b}{a}\right) \)

This formula lets you convert between the components of a vector and the angle it makes with the x-axis.
By understanding how tangent works, you can quickly figure out the exact tilt or orientation of your vector.

Vector Components

A vector can be thought of as a combination of pieces along the x-axis and y-axis, known as components. If you imagine breaking down a slope into its flat (horizontal) and steep (vertical) parts, that's essentially what vector components are.
For the vector \( \overrightarrow{\mathbf{v}} = a \hat{\mathbf{i}} + b \hat{\mathbf{j}} \), the components are \( a \) along the x-axis and \( b \) along the y-axis. They represent how much the vector moves in each direction.
Learning to add or subtract these components is like stacking blocks. You simply add each corresponding part, like:

• \( \overrightarrow{\mathbf{v}}_{1} = 4 \hat{\mathbf{i}} - 8 \hat{\mathbf{j}} \)
• \( \overrightarrow{\mathbf{v}}_{2} = 1 \hat{\mathbf{i}} + 1 \hat{\mathbf{j}} \)

For these, you'd add \( 4 + 1 \) in the x-direction, and \( -8 + 1 \) in the y-direction. The resulting vector shows your new direction and length. Understanding these helps to break down, reconstruct, and analyze any type of motion accurately.