Vector A has a magnitude of 13.0 units and points due west. Vector B points due north.

(a) what is the magnitude of B if A+B has a magnitude of 16.8 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 16.8 units?
(d) What is the direction of A-B relative to due west?

a. X = A = -13 Units.

Y = B = ?
Z= A+B = 16.8 Units = Resultant.

X^2 + Y^2 = Z^2
(-13)^2 + Y^2 = 16.8^2
Y^2 = 16.8^2 - (-13)^2 = 282.24 - 169 =
113.24
Y = 10.64 Units = B.

b. Tan Cr = Y/X = 10.64/-13 = -0.81846
Cr = -39.3o = Reference angle.
C = -39.3 + 180 = 140.7o CCW = 39.3o N of W.

c. B = Y = 10.64

d. X = -13. Y = -10.64.
Tan Cr = Y/X = -10.64/-13 = 0.81846
Cr = 39.30o = Reference angle.

C = 39.3 + 180 = 219.3o CCW. = 39.3o S of W.

To solve these questions, we can use vector addition and subtraction rules. Let's go step by step:

(a) To find the magnitude of vector B, we can use the given information that A+B has a magnitude of 16.8 units. Since vector A has a magnitude of 13.0 units, we can write the equation as:

|A + B| = 16.8

Now, we know that A points due west and B points due north, which means they are perpendicular to each other. Using the Pythagorean theorem, we can solve for the magnitude of vector B:

|A + B| = √(13^2 + B^2) = 16.8

Simplifying the equation:

169 + B^2 = 16.8^2
B^2 = 16.8^2 - 169
B^2 = 282.24 - 169
B^2 = 113.24
B ≈ √113.24
B ≈ 10.65

Therefore, the magnitude of vector B is approximately 10.65 units.

(b) To find the direction of A+B relative to due west, we can use trigonometry. Since vector A points due west and vector B points due north, their resultant will form a right-angled triangle with the angle relative to due west. We can use the inverse tangent function to find this angle:

θ = arctan(B/A)
θ = arctan(10.65/13.0)

Using a calculator, we find:
θ ≈ arctan(0.8208)
θ ≈ 39.67 degrees

Therefore, the direction of A+B relative to due west is approximately 39.67 degrees north of west.

(c) To find the magnitude of vector B when A-B has a magnitude of 16.8 units, we follow a similar procedure as in part (a). Using the given information, we can write the equation as:

|A - B| = 16.8

Again, since vector A points due west and vector B points due north, they are perpendicular to each other. Using the Pythagorean theorem, we can solve for the magnitude of vector B:

|A - B| = √(13^2 + B^2) = 16.8

Simplifying the equation:

169 + B^2 = 16.8^2
B^2 = 16.8^2 - 169
B^2 = 282.24 - 169
B^2 = 113.24
B ≈ √113.24
B ≈ 10.65

Therefore, the magnitude of vector B is approximately 10.65 units, which is the same as in part (a).

(d) To find the direction of A-B relative to due west, we once again use trigonometry. Since vector A points due west and vector B points due north, their resultant will form a right-angled triangle with the angle relative to due west. We can use the inverse tangent function to find this angle:

θ = arctan(B/A)
θ = arctan(10.65/13.0)

Using a calculator, we find:
θ ≈ arctan(0.8208)
θ ≈ 39.67 degrees

Therefore, the direction of A-B relative to due west is approximately 39.67 degrees north of west. This is the same direction as in part (b).

To solve these vector addition and subtraction problems, we can use the Pythagorean theorem and trigonometry.

(a) To find the magnitude of vector B, we can use the equation A+B = 16.8, where A is the magnitude of vector A. Let's call the magnitude of vector B as B.

Using the Pythagorean theorem, we can express vector A and vector B in terms of their horizontal and vertical components. Since vector A points due west, its horizontal component is -A, and its vertical component is 0. Vector B points due north, so its horizontal component is 0, and its vertical component is B.

To find the magnitude of vector A + vector B, we can use the formula magnitude = √(horizontal component^2 + vertical component^2). Using this formula, we can set up the equation as follows:

√((-A)^2 + 0^2) + √(0^2 + B^2) = 16.8

Simplifying the equation:

√(A^2 + B^2) = 16.8

Since A = 13.0 (given), the equation becomes:

√(13.0^2 + B^2) = 16.8

Squaring both sides:

169 + B^2 = 16.8^2

Simplifying further:

B^2 = 16.8^2 - 169

B^2 = 282.24 - 169

B^2 = 113.24

Taking the square root of both sides:

B ≈ √113.24

B ≈ 10.64

Therefore, the magnitude of vector B is approximately 10.64 units.

(b) To find the direction of vector A + vector B relative to due west, we need to calculate the angle between the resultant vector (A + B) and the west direction.

Since vector A points due west and vector B points due north, the angle between them is 90 degrees (a right angle). Therefore, the direction of vector A + vector B relative to due west is 90 degrees clockwise from due west.

(c) To find the magnitude of vector B, we can use the equation A - B = 16.8.

Using the same approach as before, we can express vector A and vector B in terms of their horizontal and vertical components. Since vector A points due west, its horizontal component is -A. Vector B points due north, so its horizontal component is 0, and its vertical component is B.

Applying the formula for magnitude:

√((-A)^2 + 0^2) - √(0^2 + B^2) = 16.8

Simplifying the equation:

√(A^2 + B^2) - √(B^2) = 16.8

Since A = 13.0 (given), the equation becomes:

√(13.0^2 + B^2) - √(B^2) = 16.8

Squaring both sides:

(13.0^2 + B^2) - 2√(13.0^2 + B^2)√(B^2) + B^2 = 16.8^2

Simplifying further:

169 + B^2 - 2B√(169 + B^2) + B^2 = 282.24

2B^2 - 2B√(169 + B^2) + 169 - 282.24 = 0

Now we have a quadratic equation in terms of B. We can solve this equation by using the quadratic formula or factoring. However, the exact value of B cannot be obtained without further information on A. Therefore, we cannot determine the magnitude of vector B with the given information.

(d) Similarly, we cannot determine the direction of vector A - vector B relative to due west without knowing the exact values of A and B.