A vector has an x-component of -2.5m and a y-component of 4.2m. Express the vector in magnitude-angle form.?

Question

A vector has an x-component of -2.5m and a y-component of 4.2m. Express the vector in magnitude-angle form.?

Answer 1

Either with complex numbers

-2.5m + j 4.2m = 4.89m <121°

or magnitude = √ [ (-2.5)^2 + (4.2)^2 ] = 4.89m

& angle = arctan (+4.2/-2.5) = -59.2°

We have a negative x and a positive y.  This puts us in quadrant II, so we have to add 180° to our angle to get the angle relative to the positive x-axis.

angle = 180° + -59.2° = 121°

Ans: 4.89m <121°

Answer 2

wat?

Answer 3

Letting r, represent the resultant magnitude of the vector we

have:

       r = absolute value of  x + yi = squirt x^2 + y^2

       r = squirt ( -2.5m)^2 + (4.2m)^2

       r = squirt 23.89m^2

       r = 4.89m ( approx, 4.9m )

      tan(theta) = y/x = 4.2m/-2.5m = -1.68

        

      angle theta = arctan ( -1.68 ) = -59.2 degrees

     the resulting angle is in the second quadrant because

x = -2.5m and y = 4,2 also sin(theta) is positive and cos (theta) is negative.

      180 degrees - 59.2 degrees = 120.7 degrees

  or approx. 121.0 degrees.

   Therefore: ( r, theta ) = ( 4.9m, 121 degrees )