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If x and y are parallel, then ha = b?

  1. Jul 30, 2011 #1
    I defined the angle [itex]\theta[/itex] between [itex]x \in \mathbb{R}^k[/itex] and [itex]y \in \mathbb{R}^k[/itex] in the way that:

    [itex]\theta = \cos ^{-1}\frac{<x,y>}{||x|| \cdot ||y||}[/itex]

    where [itex]\cos ^{-1}:[-1,1] \to [0,\pi][/itex] is bijective.

    And, therefore, [itex]x[/itex] and [itex]y[/itex] are parallel in the same direction if [itex]\theta = 0[/itex] and in the opposite direction if [itex]\theta = \pi[/itex].

    Now I can't prove the following theorem within this context:

    Theorem: If [itex]x[/itex] and [itex]y[/itex] are parallel in the same direction, then [itex]x = hy[/itex] where [itex]h>0[/itex], and if in the opposite direction, then [itex]x = hy[/itex] with [itex]h<0[/itex].

    I feel very grateful if someone post the proof. Thanks.

    (Please note that [itex]<x,y> = x_1 y_1 + \cdots + x_k y_k[/itex].)

    Actually, what I've done so far for the proof is this:

    [itex]<x,y>^2 = (||x|| \cdot ||y||)^2[/itex]

    [itex]\displaystyle{ x^2 _1 y^2 _1 + \cdots + x^2 _k y^2 _k + 2\sum_{i \neq j} x_i y_i x_j y_j = x^2 _1 y^2 _1 + \cdots + x^2 _k y^2 _k + \sum_{i \neq j} x^2 _i y^2 _j }[/itex]

    [itex]\displaystyle{ 2\sum_{i \neq j} x_i y_i x_j y_j = \sum_{i \neq j} x^2 _i y^2 _j }[/itex].

    But from here I cannot progress further.
    Last edited: Jul 30, 2011
  2. jcsd
  3. Aug 3, 2011 #2
    Try computing the norm of (||y||/||x||)x-y.
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