1. Two points in a plane have polar coordinates (2.5 m, 30 degrees) and (3.8 m, 120 deg). Determine (a) the cartesian coordinates of both points and (b) the distance between the points.

2. Vector A has magnitude 29 units and points along +y. When vector B is added to the vector A, the resultant ponits along -x with a magnitude of 14 units. Find the magnitude and directions of B.

3. Repeat #2, for the same vector A, but for the case where the resultant of A and B points along -y with a magnitude of 14 units .

1a. 2.5m[30o]

X = 2.5*cos30 = 2.17 m.
Y = 2.5*sin30 = 1.25 m.
P1(2.17,1.25).

3.8m[120o]
X = 3.8*cos120 = -1.9 m.
Y = 3.8*sin120 = 3.29 m.
P2(-1.9,3.29).

1b. (2.17,1.25), (-1.9,3.29)

d^2=(-1.9-2.17)^2 + (3.29-1.25)^2= 20.73
d = 4.55 m.

2. A + B = -14
29i + B = -14
B = -14 -29i = 32.2[244.2o]

3. A + B = -14i
29i + B = -14i
B = -43i = 43[270o].

To solve the given problems, we need to use some formulas and concepts from coordinate systems and vectors. I will guide you step by step on how to approach each question.

1. Two points in polar coordinates are given. We need to find their Cartesian coordinates and the distance between them.

(a) To convert polar coordinates (r, θ) to Cartesian coordinates (x, y), we use the following formulas:
x = r * cos(θ)
y = r * sin(θ)

For the first point (2.5 m, 30 degrees):
x1 = 2.5 * cos(30) ≈ 2.165 m
y1 = 2.5 * sin(30) ≈ 1.25 m

For the second point (3.8 m, 120 degrees):
x2 = 3.8 * cos(120) ≈ -1.9 m
y2 = 3.8 * sin(120) ≈ 3.292 m

The Cartesian coordinates of the first point are approximately (2.165 m, 1.25 m), and the Cartesian coordinates of the second point are approximately (-1.9 m, 3.292 m).

(b) To find the distance between two points in Cartesian coordinates (x1, y1) and (x2, y2), we use the distance formula:
distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)

In this case, the distance between the two points is approximately:
distance = sqrt((-1.9 - 2.165)^2 + (3.292 - 1.25)^2)
distance ≈ 3.667 m

Therefore, the distance between the two points is approximately 3.667 m.

2. We are given the magnitude and direction of vector A and the magnitude and direction of the resultant vector when vector B is added to A. We need to find the magnitude and direction of vector B.

Let's assume the given information as follows:
Magnitude of vector A = 29 units
Direction of vector A = +y
Magnitude of the resultant vector = 14 units
Direction of the resultant vector = -x

First, we need to find the components of vector A and the resultant vector. Since vector A points along +y, its x-component would be 0. The y-component would be equal to the magnitude of vector A.

Components of vector A:
Ax = 0
Ay = 29 units

Components of the resultant vector (R):
Rx = -14 units
Ry = 0

To find the components of vector B, we can subtract the components of vector A from the components of the resultant vector:
Bx = Rx - Ax = -14 - 0 = -14 units
By = Ry - Ay = 0 - 29 = -29 units

The magnitude of vector B can be found using the Pythagorean theorem:
Magnitude of vector B = sqrt(Bx^2 + By^2)
Magnitude of vector B = sqrt((-14)^2 + (-29)^2)
Magnitude of vector B ≈ 32.8 units

The direction of vector B can be found using the inverse tangent function:
Direction of vector B = atan(By / Bx)
Direction of vector B = atan((-29) / (-14))
Direction of vector B ≈ 64.7 degrees (This is the angle with respect to the positive x-axis.)

Therefore, the magnitude of vector B is approximately 32.8 units, and its direction is approximately 64.7 degrees relative to the positive x-axis.

3. We are given the same vector A as before, but now the resultant of vector A and vector B points along -y with a magnitude of 14 units. We need to find the magnitude and direction of vector B.

Since the resultant vector points along -y, its x-component would be 0 and the y-component would be -14 units.

Components of vector A (same as before):
Ax = 0
Ay = 29 units

Components of the resultant vector (R):
Rx = 0
Ry = -14 units

Again, to find the components of vector B, we can subtract the components of vector A from the components of the resultant vector:
Bx = Rx - Ax = 0 - 0 = 0
By = Ry - Ay = -14 - 29 = -43 units

The magnitude of vector B can be found using the Pythagorean theorem:
Magnitude of vector B = sqrt(Bx^2 + By^2)
Magnitude of vector B = sqrt((0)^2 + (-43)^2)
Magnitude of vector B ≈ 43 units

The direction of vector B can be found using the inverse tangent function:
Direction of vector B = atan(By / Bx)
Direction of vector B = atan((-43) / 0)
Direction of vector B = -90 degrees (This is the direction along the negative y-axis.)

Therefore, the magnitude of vector B is approximately 43 units, and its direction is along the negative y-axis.