# Calculating the phase of the sum of two vectors

## Main Question or Discussion Point

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.

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

dx
Homework Helper
Gold Member
Your vector is 5i + 2j, so x = 5 and y = 2.

Hi dx,

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

HallsofIvy
Homework Helper
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.

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

-pi/2 or pi/2

?

Why?

HallsofIvy
Homework Helper
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.

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
Homework Helper
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 "$-\pi$".

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