# Question: scalars from vectors

1. Sep 15, 2004

### Dorita

I saw this question posted yesterday, and now got a similar question to work out.

A = (6i-8j) cm
B = (-8i+3j) cm
C = (26i+19j) cm

aA+bB+C=0

Determine the two scalars a and b.

Ideas anyone??

Thanks

Dora

Last edited: Sep 15, 2004
2. Sep 15, 2004

### K.J.Healey

C=0 ?? But you jsut said C=26i+19j . Is this a typo? or did you mean aA+bB-C=0

in which case aA+bB=C
seems pretty straightforward to me. Split it up into the vector components, and youll have 2 equations with 2 unknowns, easily solveable.

3. Sep 15, 2004

### Dorita

Sorry. That was a tipo. I made a mistake.

aA+bB+C=0 not aA+bB=C=0 not

4. Sep 15, 2004

### K.J.Healey

Well, what have you done so far? How have you approached it?

5. Sep 15, 2004

### Dorita

I used the equation a^2 + b^2 = c^2 and the coodinates (6,-8) and (-8,3) to determine that the magnitude of A is 0.5cm and that the magnitude of B is 0.7cm. But I don't know if that is what is meant by "determine the two scalars a and b". I'm asuming scalars in this question is the scalar quantity or "magnitude".

Last edited: Sep 15, 2004
6. Sep 15, 2004

### HallsofIvy

Staff Emeritus
The "equation a^2+ b^2= c^2" doesn't even make sense here. You are given vectors A, B, C, not numbers a, b, c (and you certainly don't have any number c).

Do you know how to add vectors and multiply vectors by a number? That should have been ther first thing you learned!

If A= 6i+8j, then aA= (6a)i+ (8a)j.

If B= -8i+ 3j, then bB= (-8b)i+ (3b)j

aA+ bB = (6a- 8b)i+ (8a+ 3b)j and that must be equal to C= 26i+ 19j.

Okay, have you learned that two vectors are equal only if the respective components are equal?

To have aA+ bB= C, you must have (6a- 8b)i+ (8a+ 3b)j= 26i+ 19j and so
6a- 8b= 26 and -8a+ 3b= 19.

Can you solve those two equations for a and b?