Vector has a magnitude of 113 units and points 34.0 ° north of west. Vector points 63.0 ° east of north. Vector points 17.0 ° 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 .
A is 20 degrees and B is 60 degrees. I don't even know how to start
You sure are cagey with the names of your vectors.
I'll call them u,v,w.
Just draw a diagram. It should be clear that the x- and y-components of vector u are
x: -113 cos34°
y: 113 sin34°
Do the others the same way. For magnitudes, recall that
|<x,y>| = √(x^2+y^2)
To solve this problem, we need to find the components of the given vectors and use them to calculate the magnitudes of vector A and vector B.
Let's start by finding the components of each vector:
For vector A:
Given magnitude: 113 units
Direction: 34.0° north of west
To find the components, we can use trigonometry. Recall that the horizontal component is given by the magnitude multiplied by the cosine of the angle, and the vertical component is given by the magnitude multiplied by the sine of the angle.
Horizontal component of vector A:
A_x = 113 * cos(34.0°)
Vertical component of vector A:
A_y = 113 * sin(34.0°)
Similarly, we can find the components for vector B and vector C using their magnitudes and directions.
Now, let's calculate the components and magnitudes for vectors A and B:
For vector A:
A_x = 113 * cos(34.0°)
A_y = 113 * sin(34.0°)
For vector B:
B_x = 0 (since it is pointing directly north)
B_y = 113 (since it is pointing directly east)
To calculate the magnitude of a vector using its components, we can use the Pythagorean theorem. Recall that the magnitude of a vector (V) is given by:
|V| = sqrt(V_x^2 + V_y^2)
Now, let's calculate the magnitude of each vector:
Magnitude of vector A:
|A| = sqrt(A_x^2 + A_y^2)
Magnitude of vector B:
|B| = sqrt(B_x^2 + B_y^2)
Plug in the values we calculated into these formulas, and you will find the magnitudes of vector A and vector B.
Remember to convert the angles to radians if necessary before using trigonometric functions.