# Impossible vector problem. Help!

1. ### relativitydude

70
Most of the time, one uses vectors to find an overall magnitude acting on something, but I need to go in reverse.

Say I know there are a slew of charges and the center one feels a force of some magnitude and there is one undefined vector (magnitude and angle unknown) I would think if you know the overall vector, one could calculate the missing vector in terms of magnitude and angle.

I want to solve for this unknown vector but I have no idea how. It seems impossible.

2. ### HallsofIvy

40,918
Staff Emeritus
Essentially, you are saying that you have the equation
$$x_1+ x_2+ ...+ x_n= F$$ where $$x_1,... x_{n-1}$$ are the known vector forces, $$x_n$$ is the one unknown vector and F is the known resultant. Solve that exactly the way you would any equation: $$x_n= F- x_1- x_2- ...- x_{n-1}$$. Essentially that "subtraction" on the right is just like addition of vectors except that you reverse the direction of $$x_1,... x_{n-1}$$. You might find the calculation easier as $$-x_n= x_1+ x_2+ ...+ x_{n-1}- F$$. That is, you just reverse the direction of F, add the vectors and then reverse the direction of the result to find $$x_n$$.

### Staff: Mentor

Consider the vector equation:
$$\vec{F}_{net} = \vec{F}_{known} + \vec{F}_{unknown}$$

Where F(net) is the net force at the center and F(known) is the sum of the known forces from each charge. To solve for the unknown vector, F(unknown), just subtract.

(Looks like Halls beat me to it.)

4. ### relativitydude

70
Overall Magnitude = sqrt( (mag1*cos(A) + mag2*cos(B) + mag3*cos(C))^2 + (mag1*sin(A) + mag2*sin(B) + mag3*sin(C))^2)

I dont think it's that easy since the angle and magnitude go hand in hand. I need to solve for both mag1 and angle A. I'm sorry if I forgot to really point that out.

I thought it would be something more involved along the lines of langrange multipliers.

### Staff: Mentor

Instead of using overall magnitude, find the vector components. If you write the vectors in terms of their components, adding and subtracting will be a breeze. Once you find the components of the unknown vector, then you can determine its magnitude and angle.

6. ### relativitydude

70
Yes, using the horizontal and vertical components would be really easy but I only know the overall magnitude :(

### Staff: Mentor

If all you know is the magnitude, then you can't solve the problem. I suspect that you know both the magnitude and the angle. Use those to find the components.

8. ### marlon

Working with vectors is essential in both mathematics and physics. Here is an intro and some exercises to test your knowledge...courtesy of PF

Here You Go

regards
marlon

9. ### relativitydude

70
Thanks for all the help. I guess not having an overall angle made the problem, well, impossible. Including it brought me quickly to the answer.

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