An air traffic controller observes two airplanes approaching the airport. The displacement from the control tower to plane 1 is given by the vector A , which has a magnitude of 220 km and points in a direction 32 degrees north of west. The displacement from the control tower to plane 2 is given by the vector B , which has a magnitude of 140 km and points 65 degrees east of north.sketch the vectors A, B and D=A-B.Notice that vector D is the displacement from plane 2 to plane 1. Find the direction and magnitude of vector D.
Student of upng can you help me please
To sketch the vectors A, B, and D, first, draw a coordinate system with the control tower at the origin.
Vector A has a magnitude of 220 km and points 32 degrees north of west. To draw it, start at the origin and draw a line in the direction 32 degrees north of west for a length of 220 km.
Vector B has a magnitude of 140 km and points 65 degrees east of north. To draw it, start at the origin and draw a line in the direction 65 degrees east of north for a length of 140 km.
To find vector D, which is the displacement from plane 2 to plane 1, subtract vector B from vector A. To do this, place the tail of vector B at the head of vector A. The resulting vector D starts from the tail of B and ends at the head of A.
Now, to find the magnitude and direction of vector D:
Magnitude of vector D can be found by finding the difference between the magnitudes of vectors A and B:
|D| = |A| - |B|
= 220 km - 140 km
= 80 km
The direction of vector D can be found by finding the angle it makes with the positive x-axis (east) in a counterclockwise direction:
θ = arctan(Dy / Dx)
Here, Dx is the x-component of vector D, and Dy is the y-component of vector D.
By calculating the x and y components of vector D, we get:
Dx = Ax - Bx
= 220 km * cos(32°) - 140 km * sin(65°)
= 98.753 km
Dy = Ay - By
= 220 km * sin(32°) + 140 km * cos(65°)
= 222.373 km
Now, substitute the values in the formula to find the direction of vector D:
θ = arctan(222.373 km / 98.753 km)
= 66.5°
So, the direction of vector D is 66.5 degrees counterclockwise from the positive x-axis (east), which means it points in the direction 66.5 degrees north of east.
Therefore, the magnitude of vector D is 80 km, and its direction is 66.5 degrees north of east.
To sketch the vectors A and B, we need to create a coordinate system with the control tower as the origin. The direction is labeled with respect to two reference directions - north and west.
First, draw a line in the northwest direction representing vector A. The length of this line will be 220 km. The angle 32 degrees north of west tells us that vector A points in a direction between west and north.
Next, draw a line in the east-north direction representing vector B. The length of this line will be 140 km. The angle 65 degrees east of north tells us that vector B points in a direction between north and east.
Now, to find vector D, subtract vector B from vector A by subtracting the corresponding components. We subtract the x-components and the y-components separately. The x-component of vector A is 220 km * cos(32 degrees) in the negative x-direction (west). The y-component of vector A is 220 km * sin(32 degrees) in the positive y-direction (north). Similarly, the x-component of vector B is 140 km * sin(65 degrees) in the positive x-direction (east), and the y-component of vector B is 140 km * cos(65 degrees) in the positive y-direction. Subtracting the corresponding components, we get the x-component and y-component of vector D.
To find the magnitude and direction of vector D, we can use the Pythagorean theorem and trigonometry. The magnitude is given by the square root of the sum of squares of the x and y components. The direction is given by the angle between vector D and the positive x-axis.
Once we have the x and y components of vector D, we can calculate the magnitude as follows:
Magnitude of D = sqrt((xD^2) + (yD^2))
To find the direction, we can use the inverse tangent (arctan) function:
Direction of D = arctan(yD / xD)
Performing these calculations will give us the direction and magnitude of vector D, which represents the displacement from plane 2 to plane 1.