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Question: scalars from vectors

  1. Sep 15, 2004 #1
    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


    Determine the two scalars a and b.

    Ideas anyone??


    Last edited: Sep 15, 2004
  2. jcsd
  3. Sep 15, 2004 #2
    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.
  4. Sep 15, 2004 #3
    Sorry. That was a tipo. I made a mistake.

    aA+bB+C=0 not aA+bB=C=0 not
  5. Sep 15, 2004 #4
    Well, what have you done so far? How have you approached it?
  6. Sep 15, 2004 #5
    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
  7. Sep 15, 2004 #6


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    Staff Emeritus
    Science Advisor

    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?

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