Confirming Understanding of an Orthonormal System

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Discussion Overview

The discussion revolves around the understanding of an orthonormal (ON) system in the context of vector mathematics. Participants explore the properties of orthogonality and normalization, particularly in relation to the dot product of vectors in an orthonormal basis.

Discussion Character

  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that in an ON-system, the dot product formula $$u*v=|u||v|\cos\theta$$ is not applicable because the angle $$\theta$$ between orthogonal vectors is $$\frac{\pi}{2}$$, leading to a cosine value of zero.
  • Others clarify that orthogonality implies that vectors are perpendicular, and provide the formula for the dot product in an orthonormal basis as $$u*v=a_1a_2+b_1b_2+c_1c_2$$.
  • One participant questions whether an ON-base means that all vectors have a length of 1, to which others confirm this understanding, stating that "ON would mean orthonormal," where "Ortho" refers to perpendicularity and "Normal" to unit length.
  • A later reply adds that while the formula $$x\cdot y=|x||y|\cos\theta$$ is generally applicable, in the case of an orthonormal basis, there is a specific formula for the dot product that reflects the properties of orthonormal vectors.

Areas of Agreement / Disagreement

Participants generally agree on the definitions of orthogonality and normalization, but there is some contention regarding the applicability of the dot product formula in orthonormal systems, with differing interpretations of its use.

Contextual Notes

Some assumptions about the definitions of orthonormal systems and the conditions under which the dot product formula applies remain unresolved, particularly regarding the transition from general vector properties to those specific to orthonormal bases.

Petrus
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Hello MHB,
I would like if someone could confirmed if I did understand correct.
In a ON-system we don't use $$u*v=|u||v|\cos\theta$$ cause In a ON-system the $$\theta=\frac{\pi}{2}$$ so $$\cos(\frac{\pi}{2})=0$$

Regards
$$|\pi\rangle$$
 
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Petrus said:
Hello MHB,
I would like if someone could confirmed if I did understand correct.
In a ON-system we don't use $$u*v=|u||v|\cos\theta$$ cause In a ON-system the $$\theta=\frac{\pi}{2}$$ so $$\cos(\frac{\pi}{2})=0$$

Regards
$$|\pi\rangle$$

What do you mean by an ON-system?

If u and v are perpendicular to each other, their dot product is indeed zero.
 
I like Serena said:
What do you mean by an ON-system?

If u and v are perpendicular to each other, their dot product is indeed zero.
ON=Orthogonality
If I have understand correctly Orthogonality means that it's perpendicular.
Then we say that "Let $$B=(u_1,u_2,u_3)$$ be a ON base in the room and we assume that $$u=(a_1,b_1,c_1)$$ and $$v=(a_2,b_2,c_2)$$ in this base ten (they all u and v is vector)
$$u*v=a_1a_2+b_1b_2+c_1c_2$$

Regards,
$$|\pi\rangle$$
 
Petrus said:
ON=Orthogonality
If I have understand correctly Orthogonality means that it's perpendicular.
Then we say that "Let $$B=(u_1,u_2,u_3)$$ be a ON base in the room and we assume that $$u=(a_1,b_1,c_1)$$ and $$v=(a_2,b_2,c_2)$$ in this base ten (they all u and v is vector)
$$u*v=a_1a_2+b_1b_2+c_1c_2$$

Regards,
$$|\pi\rangle$$

Yep. That's true.
 
I like Serena said:
Yep. That's true.
Hello,
Have I also understand this correct. ON-Base means that all vector have the length 1?

Regards,
$$|\pi\rangle$$
 
Petrus said:
Hello,
Have I also understand this correct. ON-Base means that all vector have the length 1?

Regards,
$$|\pi\rangle$$

Yes.
ON would mean orthonormal.
Ortho means perpendicular.
Normal means length 1.
 
Petrus said:
In a ON-system we don't use $$u*v=|u||v|\cos\theta$$ cause In a ON-system the $$\theta=\frac{\pi}{2}$$ so $$\cos(\frac{\pi}{2})=0$$
Just to add: If $$u=(a_1,b_1,c_1)$$ and $$v=(a_2,b_2,c_2)$$ in an orthonormal basis, then $$u\cdot v=a_1a_2+b_1b_2+c_1c_2$$ is a corollary of the fact $$x\cdot y=|x||y|\cos\theta$$ for any vectors $x$, $y$ with an angle $\theta$ between them. So it's not the case that we don't use $$x\cdot y=|x||y|\cos\theta$$; it's just that in the case of orthonormal basis there is a special formula for the dot product.
 

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