- #1
julypraise
- 104
- 0
I defined the angle \theta between x \in \mathbb{R}^k and y \in \mathbb{R}^k in the way that:
\theta = \cos ^{-1}\frac{<x,y>}{||x|| \cdot ||y||}
where \cos ^{-1}:[-1,1] \to [0,\pi] is bijective.
And, therefore, x and y are parallel in the same direction if \theta = 0 and in the opposite direction if \theta = \pi.
Now I can't prove the following theorem within this context:
Theorem: If x and y are parallel in the same direction, then x = hy where h>0, and if in the opposite direction, then x = hy with h<0.
I feel very grateful if someone post the proof. Thanks.
(Please note that <x,y> = x_1 y_1 + \cdots + x_k y_k.)
Actually, what I've done so far for the proof is this:
<x,y>^2 = (||x|| \cdot ||y||)^2
\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 }
\displaystyle{ 2\sum_{i \neq j} x_i y_i x_j y_j = \sum_{i \neq j} x^2 _i y^2 _j }.
But from here I cannot progress further.
\theta = \cos ^{-1}\frac{<x,y>}{||x|| \cdot ||y||}
where \cos ^{-1}:[-1,1] \to [0,\pi] is bijective.
And, therefore, x and y are parallel in the same direction if \theta = 0 and in the opposite direction if \theta = \pi.
Now I can't prove the following theorem within this context:
Theorem: If x and y are parallel in the same direction, then x = hy where h>0, and if in the opposite direction, then x = hy with h<0.
I feel very grateful if someone post the proof. Thanks.
(Please note that <x,y> = x_1 y_1 + \cdots + x_k y_k.)
Actually, what I've done so far for the proof is this:
<x,y>^2 = (||x|| \cdot ||y||)^2
\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 }
\displaystyle{ 2\sum_{i \neq j} x_i y_i x_j y_j = \sum_{i \neq j} x^2 _i y^2 _j }.
But from here I cannot progress further.
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