# Why do we solve i and j components of a vector using trig?

1. Jun 29, 2015

### ALRedEye

1. The problem statement, all variables and given/known data
I'm having trouble understanding why we solve vector components (i and j, or the horizontal and vertical legs) like a right triangle?
An example would be a 5-4-3 triangle. If 5 N was the force vector I am solving for then I would end up with 4 N in the horizontal direction and 3 N in the vertical direction. The part I don't understand is how the sum of the legs can equal more than the original force? To me it seems like i^2 + j^2 > (the original force vector)^2 and i + j = (the original force vector) makes more sense.
When I try and relate this to the real world I'm thinking maybe that 5 N force is me pushing on a box at a -53 degree angle. So that means the box is moving at 4 N along the floor and 3 N into the floor, but I'm misunderstanding how these components relate to each other since the total of the two components is 7 N and I'm only pushing at 5 N.
I'd really appreciate some help wrapping my head around this!
(Sorry if I posted to the wrong category)

2. Relevant equations

3. The attempt at a solution

2. Jun 29, 2015

### HallsofIvy

Staff Emeritus
What, exactly do you mean by "the sum of the legs"? The sum of the horizontal and vertical force vectors are equal to the total force- that's the whole point.

Assuming you mean the dot product here, (3i).(3i)= 9 and (4i).(4i)= 16. 9+ 16= 25= 5^2. Where did you get the idea that it was larger than
(the original force vector)^2

Did you not mean to write 3i+ 4j= (the original force vector)?

You mean there would be a 4N force pushing the box and a 3N force pressing it into the floor, right?

Your mistake is thinking that "the total of the two components is 7 N". You do not add the magnitudes of two vectors- you add the vectors themselves. These two vectors, of magnitude 3 and 4, add to a vector of magnitude 5. That is because the magnitude of a vector of the form ai+ bj is $\sqrt{a^2+ b^2}$.

3. Jun 29, 2015

### jbriggs444

Do you understand that if you walk 3 miles north and 4 miles west that you can get back where you started by walking 5 miles diagonally to the southeast?

4. Jun 29, 2015

### ALRedEye

That makes more sense. Thanks