Question

I have a question about adding vectors. I know that if the the angle is the 2nd or 3rd quadrant, you must add 180 degrees, but how do I know what quadrant it falls in?

A = 15m@35* and B = 7m@120*
I got 17.1 for the magnitude and 59.07 for the angle but am unsure if I need to add 180*...

Answer 1

Therefore the final vector for A + B = 8.95m@(35+51.18)degrees
=8.95m@(86.18)degrees

Answer 2

Great picture - thanks! I drew something similar (length of 15 and 7), but am unsure where the origin is located. Also I don't think that's how I attempted to solve it (which it appears I did wrong anyway).

Is this a viable method? I guess I need more help than I thought, where did I go astray?

A = < 15cos(35*) , 15sin(35*) >

B = < 7cos(120*) , 7sin(120*) >
B = < 7(-1/2) , 7(sqrt3/2) >
B = < -7/2 , 7sqrt3/2 >

A+B = < 15cos(35*)-7/2 , 15sin(35*)+7sqrt3/2 >
A+B = <8.79 , 14.67 >
A+B = sqrt[(8,79)^2 + (14.67)^2] = 17.1 magnitude

tan(theta) = y/x
tan(theta) = 14.67/8.79
theta = tan^-1(14.67/8.79)
theta = 59.07*

17.1m @ 59.07*

Answer 3

Oh, wow, I'm sorry, I have made a mistake with the 5m length. Your answer is correct and your working is brilliant. I used a way which involves trigonometry and since you gave me the vectors with angles and magnitudes, I though that was the way you were being taught. The method that you use uses vector components in the form:

vector = x i + y j
So 'i' is the horizontal component and 'j' is the vertical component.

The way you know which quadrant it is in is by using the sign of the 'i' and 'j' components. See the diagram below.



If you were in quadrant 3 and did the "tan" rule for the angle, without taking into account the negatives, you would need to take away 180 degrees to get the angle from the positive x-axis. This is because tan gives you the result that is to the negative x-axis as below.



This is similar with quadrant 4, except you have to do 180 - {the angle from tan}.

As a summary:
Quadrant 1 - positive i and j - result from tan
Quadrant 2 - negative i, positive j - 180-[result from tan]
Quadrant 3 - negative i and j - [result from tan]-180
Quadrant 4 - negative j, positive i - negative[result from tan]

Hope this helps :)

Answer 4

Thanks!