Calculating the phase of the sum of two vectors

[cstrieder]

Hi,

suppose I have two vectors, one in the i direction and the other in j direction, i is horizontal and j vertical.

The first vector Vi = 5i and the second vector Vj = 2j.

If we sum this vectos it results in one vector that have module and phase.

I found the module by Vij = ( 5^2 + 2^2 ) ^ (1/2) = 5.3852.

But what about the phase?

I think it was atan(y/x), but in case y=0 or x=0, it let me in troubles.

 

[dx]

Your vector is 5i + 2j, so x = 5 and y = 2.

 

[cstrieder]

Hi dx,

I want to know how to calculate the angle(phase) between 5i and 2j...

 

[HallsofIvy]

The vector, with tail at (0,0) and head at (5, 2) forms a right triangle with "opposite side" 2 and "near side" 2 so tan(\theta)= 2/5, so \theta= arctan(2/5) and lies between 0 and \pi/2 since the components are both positive.

Yes, The vector xi+ yj has angle given by arctan(y/x) (with the quadrant determined by the sign of y and x). If x= 0, then obviously the vector lies along the y-axis so \theta= \pi/2 or \theta= 3\pi/2 depending on whether y is positive or negative.

 

[cstrieder]

When I have only 2j, what is the phase?

-pi/2 or pi/2

?

Why?

 

[HallsofIvy]

When I have only 2j, what is the phase?

-pi/2 or pi/2

?

Why?

\pi/2. The "phase angle" is measured counter-clockwise from the positive \vec{i} axis. \pi/2 points along the positive y-axis while -\pi/2 points along the negative y-axis.

 

[cstrieder]

Thanks for your reply HallsofIvy,

I have one more question,

If I have -5i, so the phase vector is \pi, correct?

And in case I have 5i, so the phase vector is -\pi, or zero?

 

[HallsofIvy]

NO! As I said before, the "phase" angle of 5i is \pi/2 and the "phase" angle of -5i is -\pi/2, not "\pi" or "-\pi".

 

[cstrieder]

Hi HallsofIvy,

can you provide me some referece, so I can study this by my self.

I fill I need to study.

If you can provide online reference it was better.

Thanks