vector help?

Question

vector A has a magnitute of 5.7 and points at an angle of 125 deg from the x-axis. vector B has a magnitute of 3.8 and points in the same direction as the x axis.

A+B = ([Ax + Bx]), ([Ay+By])
.........= ([5.7cos(180-125) + 3.8cos(0)] , [ 5.7sin(180-125)+ 3.8sin(0)]
.........= (7.1,4.7)

The magnitute of A+B is:
||A+B|| = sqrt( 7.1^2 + 4.7^2)
............= 8.5

did i do this right? and how do i find the angle for the new vector?

Answer 1

If I am picturing this correctly, Vector A is in the second quadrant (up and to the left), so its x-component will be 5.7 cos(125º) [or -5.7 cos (180-125)] = -3.27. Added to Vector B, the resultant will have an x-component of -3.27 + 3.80 = 0.53.

The y-component of the resultant is 4.67. The only difference in our results is that I have used one more significant figure, possibly unwisely, since 5.7 and 3.8 are given only to the nearest tenth, not the nearest hundredth.

The angle between the resultant and the x-axis is
arctan (4.67/0.54) = arctan(8.65) = 83.4º

Answer 2

Subtracting the angle from 180 gives you the right reference angle but this is not necessary. If you do it this way you need to remember what the correct sign will be (which you didn't by the way). For instance, in your case, Ax will be negative since the vector is in the 2nd quadrant. So it should be -5.7*cos(180-125).

It is easier to just do this as 5.7cos(125). The cos and sin functions automatically give you the correct sign if your angle is measured from the x-axis.

To find the angle of A + B, use the tangent function. tan(theta) = y / x
so theta = arctan(y/x)
The inverse trig functions DO only give the reference angle. So you have to figure out the actual angle based on what quadrant your vector is. The signs of x and y determine the quadrant.

Answer 3

A+B = ([Ax + Bx]), ([Ay+By])
.........= ([5.7cos(180-125) + 3.8cos(0)] , [ 5.7sin(180-125)+ 3.8sin(0)]
.........= (0.53,4.67)

The magnitute of A+B is:
||A+B|| = sqrt( 0.53^2 + 4.67^2)
............= 4.70