Vector has a magnitude of 136 units and points 35.0 ° north of west. Vector points 69.0 ° east of north. Vector points 19.0 ° west of south. These three vectors add to give a resultant vector that is zero. Using components, find the magnitudes of (a) vector and (b) vector .
I have drawn a diagram of the vectors but besides that im stumped
add up the north components (y axis)
add up the east components (x axis)
take the square root of the sum of the squares for the magnitude
Tan theta = total y/total x
but that theta is the math theta, counterclockwise from x axis
for compass angle -- > 90 - theta
Sorry, I made that too hard.
the two unknown vectors result in a vector that is equal and opposite to the first one
so
B + C is 136 units 35 deg south of east
|B|^2 + |C|^2 = 136 ^2
Bx + Cx = 136 cos 35
By + Cy = -136 sin 35
Bx = |B| sin 69
By = |B| cos 69
Cx = -|C| sin 19
Cy = -|C| cos 19
To solve this problem, let's break down each vector into its x and y components, and then sum up the components to find the resultant vector.
Vector A:
Given that Vector A has a magnitude of 136 units and points 35.0° north of west, we can draw a diagram and break it down into its x and y components.
The x component of vector A (Ax) can be found using trigonometry:
Ax = magnitude of A * cos(angle)
Ax = 136 * cos(35°)
Ax ≈ 111.104 units
The y component of vector A (Ay) can also be found using trigonometry:
Ay = magnitude of A * sin(angle)
Ay = 136 * sin(35°)
Ay ≈ 77.468 units
Vector B:
Given that Vector B points 69.0° east of north, we can follow a similar process to find its x and y components.
The x component of vector B (Bx) can be found using trigonometry:
Bx = magnitude of B * cos(angle)
Since vector B does not have a given magnitude, we'll represent it with a variable, which we'll call B.
Bx = B * cos(69°)
The y component of vector B (By) can be found using trigonometry:
By = magnitude of B * sin(angle)
By = B * sin(69°)
Vector C:
Given that Vector C points 19.0° west of south, we can again find its components.
The x component of vector C (Cx) can be found using trigonometry:
Cx = magnitude of C * cos(angle)
Since vector C does not have a given magnitude, we'll represent it with a variable, which we'll call C.
Cx = C * cos(19°)
The y component of vector C (Cy) can be found using trigonometry:
Cy = magnitude of C * sin(angle)
Cy = C * sin(19°)
Now, let's sum up the x and y components of the three vectors to find the x and y components of the resultant vector, which is zero.
Sum of x components: Ax + Bx + Cx = 0
111.104 + B * cos(69°) + C * cos(19°) = 0
Sum of y components: Ay + By + Cy = 0
77.468 + B * sin(69°) + C * sin(19°) = 0
Now we have two equations with two unknowns (B and C). We can solve these equations simultaneously to find the values of B and C.
Once we have the values of B and C, we can calculate the magnitudes of vector B and vector C using the Pythagorean theorem:
Magnitude of B: √(Bx^2 + By^2)
Magnitude of C: √(Cx^2 + Cy^2)
Solving these equations will give you the magnitudes of vector B and vector C, as requested.