Before starting this problem, review Conceptual Example 7. The force vector A has a magnitude of 79.0 newtons (N) and points due east. The force vector B has a magnitude of 107 N and points 75.0 ° north of east. Find the (a) magnitude and (b) direction of A - B. (Give the direction as a positive angle with respect to due east). Find the (c) magnitude and (d) direction of B - A. (Give the direction as a positive angle with respect to due west).

To solve this problem, we can break down the vectors A and B into their x and y components and then subtract them.

(a) To find the magnitude of A - B:
- First, let's find the x and y components of vector A:
- Since vector A points due east, its x component is equal to its magnitude (79.0 N) and its y component is 0.

- Next, let's find the x and y components of vector B:
- To find the x component of vector B, we use the cosine function:
- cos(75.0°) = Adjacent/Hypotenuse = x-component of B/107 N
- Solving for the x-component of B:
- x-component of B = cos(75.0°) * 107 N

- To find the y component of vector B, we use the sine function:
- sin(75.0°) = Opposite/Hypotenuse = y-component of B/107 N
- Solving for the y-component of B:
- y-component of B = sin(75.0°) * 107 N

- Now, let's subtract the x and y components of B from A:
- x-component of A - B = x-component of A - x-component of B
- y-component of A - B = y-component of A - y-component of B

- Finally, we can find the magnitude of A - B using the Pythagorean theorem:
- magnitude of A - B = sqrt((x-component of A - B)^2 + (y-component of A - B)^2)

(b) To find the direction of A - B:
- The direction of a vector can be found using the arctan function:
- direction of A - B = atan((y-component of A - B)/(x-component of A - B))

(c) To find the magnitude of B - A:
- We can use the same process as before, but with the x and y components of B and A switched.

(d) To find the direction of B - A:
- Again, we can use the same process as before, but with the x and y components of B and A switched.

Please note that we need specific values for the x and y components of vector B to continue with the calculations.

To solve this problem, we need to break down the given information and use vector addition and subtraction techniques.

Conceptual Example 7 likely explains how to add and subtract vectors using their components, so it's important to review that before attempting this problem.

Now let's solve the problem step by step:

(a) To find the magnitude of A - B, we need to subtract the components of vector B from A.

The given magnitude of vector A is 79.0 N, pointing due east. Therefore, the components of vector A are (79.0, 0) N.

The given magnitude of vector B is 107 N, and it points 75.0° north of east. To determine the components of vector B, we can use trigonometry.

The horizontal component of B can be found by multiplying the magnitude by the cosine of the angle: 107 N * cos(75.0°) = 107 N * 0.25882 = 27.737 N.

The vertical component of B can be found by multiplying the magnitude by the sine of the angle: 107 N * sin(75.0°) = 107 N * 0.96593 = 103.539 N.

Now we can subtract the components:

Magnitude of A - B = sqrt((79.0 - 27.737)^2 + (0 - 103.539)^2) = sqrt(3132.294) = 55.98 N (rounded to two decimal places).

(b) To find the direction of A - B, we need to determine the angle it makes with due east.

The angle can be found using arctan(dy/dx), where dy is the change in the y-component and dx is the change in the x-component.

Angle of A - B = arctan((103.539 - 0) / (27.737 - 79.0)) = arctan(-103.539 / -51.263) = arctan(2.019) = 63.50° (rounded to two decimal places). Since the vector A - B points west of due east, the angle is positive.

(c) To find the magnitude of B - A, we need to subtract the components of vector A from B.

Using the components we found earlier:

Magnitude of B - A = sqrt((27.737 - 79.0)^2 + (103.539 - 0)^2) = sqrt(3132.294) = 55.98 N (rounded to two decimal places) - same magnitude as in part (a).

(d) To find the direction of B - A, we need to determine the angle it makes with due west.

The angle can be found using arctan(dy/dx), where dy is the change in the y-component and dx is the change in the x-component.

Angle of B - A = arctan((103.539 - 0) / (79.0 - 27.737)) = arctan(103.539 / 51.263) = arctan(2.019) = 63.50° (rounded to two decimal places). Since the vector B - A points east of due west, the angle is positive.

So, the answers to the given questions are:
(a) The magnitude of A - B is 55.98 N.
(b) The direction of A - B is 63.50° with respect to due east.
(c) The magnitude of B - A is also 55.98 N.
(d) The direction of B - A is also 63.50° with respect to due west.

OOPS!!

Ignore data shown after part b.
It is incorrect.

Given: A = 79N, East; B = 107N @ 75deg.

a. X = hor. = 79 - 107cos75,
X = 79 - 27.7 = 51.3N.

Y = ver. = 0 - 107sin75,
Y = 0 - 103.4 = -103.4N.
M^2 = X^2 + Y^2,
M^2 = (51.3)^2 + (-103.4)^2,
M^2 = 2631.7 + 10691.6 = 13323,
M = 115.4N Magnitude of A-B.

b. tanA = Y/X = -103.4 / 51.3 = -2.015,
A = -63.6Deg. = 63.6deg South of East = Direction of A-B.

Use the above procedure for B-A.

Mag.= X/cosA = 106.7 / cos44.1 = 148.6N.

b. A = 79N,
B = 107N@75deg. = 107cos75 + i107sin75,
B = 27.7 + i103.4.

A - B = A to B = (27.7 + i103.4) -79,
A to B = -51.3 + i103.4
tan(theta) = 103.4 / -51.3 = -2.0156,
Theta = -63.6deg = 63.6deg South of East.

c.