Vector A has magnitude 8.0 units at an angle of 60o from the x-axis. Vector B has magnitude 6.0 at an angle of -30o from the x-axis. Find the magnitude and direction of vector C = A + B
I tried actually drawing it to see if it would help but it didn't I'm pretty sure the magnitude is found by using tang and the distance is found by using the Pythagorean theorem.
Sketch this...it is a 3 4 5 triangle.
To find the magnitude and direction of vector C = A + B, we can break down each vector into its x and y components.
For vector A:
- Magnitude: given as 8.0 units.
- Angle: given as 60° from the x-axis.
To find the x-component of A:
- x-component = magnitude * cos(angle) = 8.0 * cos(60°).
To find the y-component of A:
- y-component = magnitude * sin(angle) = 8.0 * sin(60°).
Similarly, for vector B:
- Magnitude: given as 6.0 units.
- Angle: given as -30° from the x-axis.
To find the x-component of B:
- x-component = magnitude * cos(angle) = 6.0 * cos(-30°).
To find the y-component of B:
- y-component = magnitude * sin(angle) = 6.0 * sin(-30°).
Once we have the x and y components of vectors A and B, we can add them to find the x and y components of vector C as follows:
- x-component of C = x-component of A + x-component of B.
- y-component of C = y-component of A + y-component of B.
To find the magnitude of vector C using the Pythagorean Theorem:
- magnitude of C = sqrt((x-component of C)^2 + (y-component of C)^2).
To find the angle of vector C relative to the x-axis:
- angle of C = atan2(y-component of C, x-component of C).
Now let's calculate the values.
For vector A:
- x-component of A = 8.0 * cos(60°) = 8.0 * 0.5 = 4.0 units (positive x-direction).
- y-component of A = 8.0 * sin(60°) = 8.0 * sqrt(3)/2 = 6.93 units (positive y-direction).
For vector B:
- x-component of B = 6.0 * cos(-30°) = 6.0 * √(3)/2 = 6.0 * 0.866 = 5.20 units (positive x-direction).
- y-component of B = 6.0 * sin(-30°) = 6.0 * (-0.5) = -3.0 units (negative y-direction).
Adding the x-components of A and B:
- x-component of C = 4.0 + 5.20 = 9.20 units (positive x-direction).
Adding the y-components of A and B:
- y-component of C = 6.93 - 3.0 = 3.93 units (positive y-direction).
Using the Pythagorean Theorem to find the magnitude of C:
- magnitude of C = sqrt((9.20)^2 + (3.93)^2) = sqrt(84.64 + 15.40) = sqrt(100.04) = 10.0 units.
Using the atan2 function to find the direction of C:
- angle of C = atan2(3.93, 9.20) = 21.26°.
Therefore, the magnitude of vector C is 10.0 units, and its direction is 21.26° from the x-axis.