Vector A has a magnitude of

12.3 units and points due west. Vector B points due north. (a) What
is the magnitude of B if A + B has a magnitude of 15.0 units? (b)
What is the direction of A + B relative to due west? (c) What is the
magnitude of B if A - B has a magnitude of 15.0 units? (d) What is
the direction of A - B relative to due west?

To solve the given problems step-by-step, let's define the following information:

Magnitude of vector A = 12.3 units
Direction of vector A = due west

(a) Magnitude of vector B if A + B has a magnitude of 15.0 units:
Let's assume the magnitude of vector B as x.

|A + B| = 15.0 units
|A + B| = √((12.3)^2 + x^2) [Applying Pythagoras theorem for a right triangle]

Solving this equation:
(15.0)^2 = (12.3)^2 + x^2
225 = 151.29 + x^2
x^2 = 73.71
x = √73.71
x ≈ 8.58 units

Therefore, the magnitude of vector B is approximately 8.58 units.

(b) Direction of A + B relative to due west:
Since vector A points due west and vector B points due north, A + B forms a right triangle. The direction of A + B can be found by using trigonometric functions.

tan(theta) = (opposite side / adjacent side)
tan(theta) = (Magnitude of vector B / Magnitude of vector A)

tan(theta) = (8.58 / 12.3)
theta = arctan(8.58 / 12.3)
theta ≈ 35.93 degrees

Therefore, the direction of A + B relative to due west is approximately 35.93 degrees north of due west.

(c) Magnitude of vector B if A - B has a magnitude of 15.0 units:
Let's assume the magnitude of vector B as y.

|A - B| = 15.0 units
|A - B| = √((12.3)^2 - y^2) [Applying Pythagoras theorem for a right triangle]

Solving this equation:
(15.0)^2 = (12.3)^2 - y^2
225 = 151.29 - y^2
y^2 = 151.29 - 225
y^2 = -73.71 (Since y^2 cannot be negative, there are no real solutions)

Therefore, there is no real magnitude of vector B that satisfies the given condition.

(d) Direction of A - B relative to due west:
Since there is no real magnitude of vector B that satisfies the given condition in part (c), we cannot determine the direction of A - B relative to due west.

To solve these problems, we can use vector addition and subtraction. Let's break down each question and explain how to find the answer step-by-step:

(a) To find the magnitude of vector B, given that A + B has a magnitude of 15.0 units:
1. Start by assuming the magnitude of vector B as x (unknown).
2. We know that vector A has a magnitude of 12.3 units and points due west. Therefore, the x-component of vector B is 0 (since it does not have any westward component).
3. To find the y-component of vector B, we subtract the x-component of vector A from the magnitude of the resultant vector (15.0) using the Pythagorean theorem.
- 15.0^2 = 12.3^2 + x^2
- Solve for x: x^2 = 15.0^2 - 12.3^2
- Take the positive square root: x = sqrt(15.0^2 - 12.3^2)
4. After finding the value of x, we have the magnitude of vector B.

(b) To find the direction of A + B relative to due west:
1. Since vector A points due west, it has an angle of 180 degrees.
2. The direction of vector B relative to due north is 90 degrees.
3. To find the direction of the resultant vector, we add these angles: 180 degrees + 90 degrees = 270 degrees.
4. Therefore, A + B is directed 270 degrees relative to due west.

(c) To find the magnitude of vector B, given that A - B has a magnitude of 15.0 units:
1. Start by assuming the magnitude of vector B as y (unknown).
2. We know that vector A has a magnitude of 12.3 units and points due west. Therefore, the x-component of vector B is 0 (since it does not have any westward component).
3. To find the y-component of vector B, we add the x-component of vector A to the magnitude of the resultant vector (15.0) using the Pythagorean theorem.
- 15.0^2 = 12.3^2 + y^2
- Solve for y: y^2 = 15.0^2 - 12.3^2
- Take the positive square root: y = sqrt(15.0^2 - 12.3^2)
4. After finding the value of y, we have the magnitude of vector B.

(d) To find the direction of A - B relative to due west:
1. Since vector A points due west, it has an angle of 180 degrees.
2. The direction of vector B relative to due north is 90 degrees.
3. To find the direction of the resultant vector, we subtract the angle of vector B from the angle of vector A: 180 degrees - 90 degrees = 90 degrees.
4. Therefore, A - B is directed 90 degrees relative to due west.

The two vectors are at 90°, so the magnitude of the sum is given by √(A^sup2;+B²).

(a)
√(12.3²+B²)=15
solve for B.
(b)
Angle relative to west, θ
tan(θ)=|B|/12.3
Solve for θ
(c)
A-B = A + (-B)
So (c) and (d) can be solved the same way as (a) and (b), but with the direction of B reversed (due south).