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.