Problem: What are the magnitude and angle of a vector whose x and y components are respectively 6 and -2?
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Solution: The magnitude of the vector is found by the Pythagorean Theorem to be `sqrt( ( 6) ^ 2 + ( -2) ^ 2) = 6.324. The angle made by the vector with the x axis is arctan ( -2/ 6) = -18.43 degrees. Since the y component of the vector is negative, the vector is closest to the negative x axis. Its angle with the positive x axis is therefore equal to the 180 degrees plus the -18.43 degrees found above. The angle is therefore -18.43 degrees.
Generalized Response: The components vx and vy correspond to the sides of a right triangle whose hypotenuse is equal to the length or magnitude of the vector, and whose angle with the positive x axis is opposite to the y component and adjacent to the x component.
The magnitude of the vector is therefore found from the Pythagorean Theorem to be v = `sqrt(vx^2 + vy^2).
The angle has a tangent equal to vy / vx, so the angle is the arctangent arcTan(vy / vx).
.
.
.
.
.
.
.
.
.
.
Figure description: The figure below depicts a vector with components vx and vy. By the Pythagorean Theorem the magnitude of the vector is `sqrt(vx^2 + vy^2). By the definition of the tangent, the angle with the positive x axis is arcTan(vy / vx).
