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Vector Proof

  1. Apr 8, 2005 #1
    Hi, i was required to show that

    -1 < [tex] \frac{a.b}{\|{a}\|\|{b}\|}} [/tex] > -1

    I did this by using the cosine rule which is [tex] c^2 = a^2 + b^2 - 2a.b\cos{\vartheta} [/tex]

    How ever our teacher did it by a scharts proof which i don't quite understand, :mad: , Now my question is why can't i prove it using the cosine rule and could somebody explain the schwartz proof a bit better?
  2. jcsd
  3. Apr 8, 2005 #2


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    What is going on with your inequality. It looks like you have -1 < something > -1. Why not just say that -1 < something. Also, how are you defining a.b? Is it the regular dot product a.b = |a||b|cos(theta)? If so, then this just says cos(theta) > -1 which is false, it may be -1 (if theta = pi).
  4. Apr 8, 2005 #3
    Has your teacher defined the angle between two vectors to be the arccosine of your expression, or is he using the inequality 1 > [tex] \frac{a.b}{\|{a}\|\|{b}\|}} [/tex] > -1 to motivate such a definition for general vector spaces (not just Euclidean space) ? In the latter case, it is understandable that he does not equate it with what he is trying to motivate yet.
  5. Apr 8, 2005 #4
    yes the inequality is suppose to be -1< something < 1

    Also i'm suppose to prove that the expression [tex] \frac{a.b}{\|{a}\|\|{b}\|}} [/tex]
    lies within the domain [-1,1]

    I did this using the cosine rule because cos(theta) is within that domain, how do i do it with the schwartz proof?
  6. Apr 8, 2005 #5


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    Schwarz's Inequality states that (a.b)² < (a.a)(b.b) does it not? And you've defined the norm by |a|² = a.a, right? You're really not asking the question right because you haven't shown your work, you haven't shown what you know, etc. Anyways, assuming that you have the above inequality (Schwarz's) and definition (of norm) to work with, isn't it obvious?
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