Question

Tower 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 \(\vec{D}\) is the displacement from plane 2 to plane 1 . (b) Use components to find the magnitude and the direction of the vector \(\overrightarrow{\mathbf{D}}\).

Short answer

\(\overrightarrow{\mathbf{D}}\) has a magnitude of 318.9 km and points 169.6° north of west.

Step-by-step solution

Sketch the Vectors

To begin, draw the vectors for visualization. Vector \( \overrightarrow{\mathbf{A}} \) is 220 km, pointing 32° north of west. Start from the origin, draw a vector 32° above the negative x-axis (westward). Vector \( \overrightarrow{\mathbf{B}} \) is 140 km, making an angle of 65° east of north. Draw this vector 65° to the right of the y-axis (northward). Finally, \( -\overrightarrow{\mathbf{B}} \) is the same vector \( \overrightarrow{\mathbf{B}} \) but in the opposite direction.

Determine Components of \( \overrightarrow{\mathbf{A}} \)

The vector \( \overrightarrow{\mathbf{A}} \) has components: \[A_x = -220 \cos(32°), \quad A_y = 220 \sin(32°)\]Calculate: \( A_x = -220 \cos(32°) \approx -186.7 \) km and \( A_y = 220 \sin(32°) \approx 116.7 \) km.

Determine Components of \( \overrightarrow{\mathbf{B}} \)

Vector \( \overrightarrow{\mathbf{B}} \) components: \[B_x = 140 \sin(65°), \quad B_y = 140 \cos(65°)\]Calculate: \( B_x = 140 \sin(65°) \approx 127.0 \) km and \( B_y = 140 \cos(65°) \approx 59.2 \) km.

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

To find \( -\overrightarrow{\mathbf{B}} \), reverse the signs of \( \overrightarrow{\mathbf{B}} \) components: \[(-B_x) = -127.0 \text{ km}, \quad (-B_y) = -59.2 \text{ km}\]

Find Components of \( \overrightarrow{\mathbf{D}} \)

Now, find vector \( \overrightarrow{\mathbf{D}} = \overrightarrow{\mathbf{A}} - \overrightarrow{\mathbf{B}} \) by subtracting the components:\[D_x = A_x - B_x = -186.7 - 127.0 = -313.7 \text{ km}\]\[D_y = A_y - B_y = 116.7 - 59.2 = 57.5 \text{ km}\]

Calculate Magnitude of \( \overrightarrow{\mathbf{D}} \)

Use the Pythagorean theorem to find the magnitude of \( \overrightarrow{\mathbf{D}} \):\[|\overrightarrow{\mathbf{D}}| = \sqrt{(-313.7)^2 + (57.5)^2} \approx 318.9 \text{ km}\]

Determine Direction of \( \overrightarrow{\mathbf{D}} \)

Calculate the angle \( \theta \) north of west:\[\theta = \tan^{-1}\left( \frac{D_y}{D_x} \right) = \tan^{-1}\left( \frac{57.5}{-313.7} \right) \approx -10.4°\]Since the angle is negative, add 180° to find the angle below the negative x-axis. The direction is approximately 169.6° north of west.

Key concepts

Understanding Displacement Vectors

Displacement vectors are essential in describing the shift of an object from one point to another. They are not merely measurements of distance; they include direction as part of their definition. In problems involving airplanes, displacement vectors represent the route from the control tower to each plane.
For vector \( \overrightarrow{\mathbf{A}} \), the displacement is described with a magnitude of 220 km, pointing 32° north of west. This tells us the plane's approximate position in relation to the tower. The vector's direction is crucial because it provides more information than merely stating the plane is 220 km away.
Vector \( \overrightarrow{\mathbf{B}} \) similarly describes another plane, with a magnitude of 140 km at a direction 65° east of north. Understanding these vectors helps air traffic controllers manage safe distances and avoid collisions in the airspace.

Breaking Down Vector Components

Vector components are the building blocks used to analyze vectors in more detail. Any vector in a plane can be broken down into its horizontal (x) and vertical (y) components using trigonometry. This breakdown allows for easier calculations when dealing with multiple vectors.
For vector \( \overrightarrow{\mathbf{A}} \):

• The x-component is determined by \( A_x = -220 \cos(32°) \) and calculates to approximately -186.7 km. The negative sign indicates that this component is in the opposite direction on the horizontal axis.
• The y-component is \( A_y = 220 \sin(32°) \), resulting in approximately 116.7 km, reflecting the vertical direction.

These components allow for straightforward manipulations, such as adding or subtracting vectors, which is essential in vector calculus. Similarly, \( \overrightarrow{\mathbf{B}} \) can be broken down into its components to further solve vector problems.

Mastering Magnitude Calculation

Magnitude calculation is a vital step in vector analysis, providing us with the vector's total length regardless of its direction. Using the Pythagorean theorem, we calculate the magnitude of a resultant vector like \( \overrightarrow{\mathbf{D}} = \overrightarrow{\mathbf{A}} - \overrightarrow{\mathbf{B}} \).
The components of vector \( \overrightarrow{\mathbf{D}} \) have already been calculated:

• \( D_x = -313.7 \text{ km} \)
• \( D_y = 57.5 \text{ km} \)

The magnitude \(|\overrightarrow{\mathbf{D}}|\) is then found using:\[|\overrightarrow{\mathbf{D}}| = \sqrt{(-313.7)^2 + (57.5)^2} \approx 318.9 \text{ km}\]Calculating magnitude helps us understand how far apart the planes are based solely on the shift in their position vectors.

Determining Direction

Direction determination involves figuring out where a vector is pointing relative to a given reference, typically a compass direction or an axis.
For vector \( \overrightarrow{\mathbf{D}} \), knowing the direction helps understand its implication in the real world, such as which way to travel to get from one plane to another. We use the arctangent function to find the angle:\[\theta = \tan^{-1}\left( \frac{57.5}{-313.7} \right) \approx -10.4°\]
This angle, being negative, initially suggests a direction below the x-axis. By adding 180°, we adjust it to have: 169.6° north of west. This calculation helps in comprehending how the resultant displacement vector aligns in space to guide decisions based on navigation or control.