1. Oct 2, 2007

### DoktorD

1. The problem statement, all variables and given/known data

Given:
$$\vec{A}$$=(6.0$$\hat{i}$$ - 8.0$$\hat{j}$$)
$$\vec{B}$$=(-8.0$$\hat{i}$$ + 3.0$$\hat{j}$$)
$$\vec{C}$$=(26.0$$\hat{i}$$ + 19.0$$\hat{j}$$)

If aA+bB+C=0, what are the magnitudes of a and b?

2. Relevant equations

3. The attempt at a solution

a(6.0$$\hat{i}$$ - 8.0$$\hat{j}$$) + b(-8.0$$\hat{i}$$ + 3.0$$\hat{j}$$) + (26.0$$\hat{i}$$ + 19.0$$\hat{j}$$) = 0

I know thats the set up of the equation, but I have no idea how to solve for a and b. Shouldn't there be a second equation to give the relationship between a and b or else there's an infnite number of solutions? Just by looking at it for a few periods, I saw that a = 5 and b = 7 works, but I can't figure out how to reach that answer using algebra!

2. Oct 2, 2007

### Kurdt

Staff Emeritus
Well you will get two equations. To get the zero vector both i and j components have to be zero.

3. Oct 2, 2007

### DoktorD

OK, but the thing is what method would be used here to find those two variables? What strategy would be used?

4. Oct 2, 2007

### DoktorD

Nevermind, I got it

5. Oct 2, 2007

### Kurdt

Staff Emeritus
You have:

$$a(6\mathbf{\hat{i}}-8\mathbf{\hat{j}})+b(-8\mathbf{\hat{i}}+3\mathbf{\hat{j}})+(26\mathbf{\hat{i}}+19\mathbf{\hat{j}}) = (6a-8b+26)\mathbf{\hat{i}} + (-8a+3b+19)\mathbf{\hat{j}}=0$$

For that to equal zero both i and j components must equal 0, so you have 2 simultaneous equations.