You are given a vector in the xy plane that has a magnitude of 100.0 units and a y component of -50.0 units

(b) Assuming the x component is known to be positive, specify the vector which, if you add it to the original one would give a resultant vector that is 85.0 units long and points entirely in the -x direction.
Magnitude ?

Direction ?

Correction:

Direction = 16.24 deg NORTH of WEST.

Y = ver = -50 (given),

X^2 + (-50)^2 = (100)^2,
X^2 = 10000 - 2500 = 7500,

X = hor = sqrt(7500) = 86.6,

Vector to be added:
-50 + Y = 0,

Y = 50 added.

86.6 + X = -85,

X = hor = -85 - 86.6 = -171.6 added.

tanA = Y/X = 50/171.6 = 0.2914,
A = 16.24 deg.

Magnitude = y / sinA = 50/sin16.24
= 178.7

Direction = 16.24 deg above WEST.

Well, if we want the resultant vector to be entirely in the -x direction, that means the x component of the resultant vector should be negative.

Since the magnitude of the original vector is 100.0 units and the y component is -50.0 units, we can use Pythagoras' theorem to find the x component.

The equation is: magnitude^2 = x^2 + y^2

Plugging in the values: 85^2 = x^2 + (-50)^2

Simplifying: 7225 = x^2 + 2500

Rearranging: x^2 = 7225 - 2500

x^2 = 4725

Taking the square root of both sides: x ≈ √4725

Calculating: x ≈ 68.74

So, the x component of the resultant vector would be approximately -68.74 units.

Now, let's calculate the magnitude of the resultant vector using Pythagoras' theorem again:

magnitude^2 = x^2 + y^2

magnitude^2 = (-68.74)^2 + (-50)^2

magnitude^2 = 4725 + 2500

magnitude^2 = 7225

Taking the square root of both sides: magnitude ≈ √7225

Calculating: magnitude ≈ 85.00

Therefore, the magnitude of the vector that, when added to the original vector, gives a resultant vector that is 85.0 units long and points entirely in the -x direction is 85.0 units.

As for the direction, it would be directly opposite to the positive x-axis, so we can say it points in the direction of "Left" or "West".

To find the magnitude and direction of the vector that, when added to the original vector, gives a resultant vector that is 85.0 units long and points in the negative x-direction, we can use vector addition and trigonometry.

Let's break down the information given:

Original Vector:
Magnitude = 100.0 units
y-component = -50.0 units

Resultant Vector:
Magnitude = 85.0 units
Direction = -x direction (negative x-direction)

To find the magnitude of the vector that needs to be added, we can use the Pythagorean theorem. The magnitude of the resultant vector is the sum of the original vector's magnitude and the magnitude of the vector we are looking for.

Let the magnitude of the vector we are looking for be represented as 'a'.

Using the Pythagorean theorem:
(100.0 + a)^2 + (-50.0)^2 = 85.0^2

Expanding the equation:
10000 + 200a + a^2 + 2500 = 7225

Combining like terms:
a^2 + 200a + 12500 - 7225 = 0

Simplifying:
a^2 + 200a - 5275 = 0

Now, we can solve this quadratic equation using the quadratic formula. The quadratic formula is given by:
a = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = 1, b = 200, and c = -5275.

Plugging in these values:
a = (-200 ± √(200^2 - 4 * 1 * -5275)) / (2 * 1)

Simplifying further, we get:
a = (-200 ± √(40000 + 21100)) / 2
a = (-200 ± √(61100)) / 2

Calculating the square root of 61100:
√(61100) ≈ 247.49

Now, we can substitute this value into our equation:
a = (-200 ± 247.49) / 2

Simplifying:
a_1 = (-200 + 247.49) / 2 ≈ 23.75
a_2 = (-200 - 247.49) / 2 ≈ -223.75

Since we assume the x-component is positive, the magnitude of the vector that needs to be added to the original vector is approximately 23.75 units.

Now, let's find the direction of the vector. We are told that the vector points in the negative x-direction, which means it has an angle of 180 degrees with respect to the positive x-axis.

Therefore, the direction of the vector that needs to be added is 180 degrees.

To summarize:
Magnitude = approximately 23.75 units
Direction = 180 degrees