# Vector components with unknowns, ultimately adding to zero.

1. Sep 25, 2009

### Lassitude

1. The problem statement, all variables and given/known data
If Avec = (6.00 ihat -8.00 jhat) units, Bvec = (-8.00 ihat + 3.00 jhat) units, and Cvec = (26.0 ihat + 22.0 jhat) units, determine a and b such that a Avec + b Bvec + Cvec= 0.

a=
b=

2. Relevant equations
Avec + Bvec = Rvec
Rvec = (Ax + Bx)i-hat + (Ay + By)j-hat

3. The attempt at a solution
I tried this two ways. The first way was to isolate one of the variables in terms of the i-hat, and then plug it into the equation for the other variable.
So, for instance, doing aAx + bBx + Cx = 0, putting in the numbers to get 6a + (-8b) + 26 = 0. If I proceed algebraically I wind up with values for a and b which make i-hat or j-hat =0, but my understanding of these types of problems is not sufficient to really know how to proceed from there. Would it be assumed to be a given that the value for a in terms of i-hat and the value for a in terms of j-hat which make their separate equations zero must be synonymous with one-another?

It's the two dimensions of the vectors which are tripping me up, now.

Last edited: Sep 25, 2009
2. Sep 25, 2009

### Delphi51

Too many symbols! As one student told me (high school physics teacher): "We don't have time to write all that stuff!"
It says aA + bB + C = 0 so you have
a(6i-8j) + b(-8i+3j) + 26i+22j = 0
Expand that out so you have all the i's together and all the j's together.
Note that the total of the i's must be zero and the total of the j's must also be zero.
Thus you have two equations involving just a and b, and so you can find the two unknowns.

3. Sep 25, 2009

### Delphi51

My apologies! I did not go back and read your solution after finding my own.
Check that: I get a = (8b -26) /6

4. Sep 25, 2009

### Lassitude

Yes, I messed up my signs there.

The problem I'm faced with, now that my own stupid mistake is out of the way, is how I do something with the two separate values of a which I get for both proving the i-hat and j-hat statements true.

In the case of the i-hats, a=(8b-26)/6 whereas with the j-hats, a=(-3b-22)/-8. I'm just not sure how to unify the two into a single value for a and a single value for b which together cause the statement to be true.

Or is it enough to simply find the a and b for the i-hats, let's say, and assume therefore that if it makes them zero, the j-hats must be made zero as well?

I really wish I had done OAC physics.

5. Sep 25, 2009

### Lassitude

Would anyone be able to help me out with this? I hate to be a pest, but it's the last question to a homework assignment and I've been struggling with it since yesterday. The assignment is due at noon EST today (Friday) and I'd really like to figure out an answer before that deadline.

6. Sep 25, 2009

### TwoTruths

This is a case of solving a system of linear equations with n variables and n equations. Just a note, for any n > 3 or maybe 4, use matrices. Much easier. Anyways.

What you've done is isolate the a variable in both equations. What you need to do now is to set both equal to each other, because a is a constant, so this must be true. You want to find the b such that those two equations for a yield the same value.

Solving any system of linear equations with n variables and n equations can be described generally like this:

Take the ith equation and isolate one variable. Take this expression and substitute it into the (i+1)st equation. When you are left with only one variable in the kth equation, solve for this variable. Substitute the value for this variable into the (k-1)st equation.

In your case, you should take the 1st (i) equation and isolate one variable (such as a, as you have done for the i-hat component). Substitute this expression for a into the 2nd (i+1) equation. You see that you are now left with one variable in this equation (b). Solve for it. This equation is the kth equation above. So take a step backgrounds to the 1st equation (k-1), and substitute this new value for b. Solve for a. You have solved your problem.

Ignore the i's and k's if you don't want to think about it generally. You can use either method (set the two expressions for a equal to each other and solve for b) or you can use the general method (solve for a, plug into 2nd equation, solve for b value, plug into 1st equation, solve for a value).

7. Sep 25, 2009

### Lassitude

I'm not sure what constitutes (i+1) in this problem. Which equation is that?

8. Sep 25, 2009

### TwoTruths

Basically, number the equations (in this case, eq 1 is the i-hat components, and eq 2 is the j-hat components). So if you look at the general form, it says isolate a variable in the ith equation, substitute into the (i+1)st equation, and repeat until only one variable remains.

Start with i=1. So by the general form, we want to take the 1st equation and solve for a variable, say $$a$$. You've done so. This is your new equation 1. Then, the general form says to substitute this expression for $$a$$ into the (i+1)st equation, or the (1+1)st equation, or the 2nd equation. So whenever you see an $$a$$ in the 2nd equation, you replace it with the expression you found previously. You can see that you only have one variable now.

Going by the general form, we now define a variable k to be the number of the equation we're on. We're on the 2nd right now, right? So k=2. The general form says to solve the kth equation to find a value for the one variable. Do so.

Next, it says to substitute this new value into the (k-1)st equation, or (2-1)st equation, or 1st equation. Put the value you found for b into the first equation, after you've messed around with it (remember we defined a new equation 1 above). Doing so, you can see that only the variable $$a$$ is in the equation. Solve for it to find its value.

Since you have found all the wanted variables, you can stop.

It probably seems confusing at first, but it's better to know the general way to solve a system of linear equations without matrices than to try and remember specific ways to solve 2 equations, 3 equations, etc.

9. Sep 25, 2009

### Lassitude

Yeah I just suck at basic arithmetic. Thanks a lot for the help, this shouldn't have been anything like as difficult as it was!

10. Sep 25, 2009

### Delphi51

In case it is just another tiny error, for the j's I got -8a + 3b + 22 = 0.
Subbing in the "a" value from the i's this is -8(8b-26)/6 + 3b + 22 = 0.
I reduced the 8/6 to 4/3, then multiplied the whole thing by 3 to clear the fractions.
This gave -4(8b-26) + 9b + 66 = 0
It simplified to b = 170/41.
The final step is subbing this value in for b in a = (8b-26)/6

11. Sep 25, 2009

### Lassitude

The equation, solved, wound up being a = 127/23, b = 170/23 if I recall correctly.

I made a couple very basic mathematical errors which turned this simple problem into something that I thought warranted a thread. But, again, thanks for the help. I'm at least pleased that my initial plan of attack was correct, even if my execution sucked.