MHB Confirming Understanding of an Orthonormal System

[Petrus]

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
$$|\pi\rangle$$

 

[I like Serena]

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.

 

[Petrus]

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,
$$|\pi\rangle$$

 

[I like Serena]

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.

 

[Petrus]

Yep. That's true.

Hello,
Have I also understand this correct. ON-Base means that all vector have the length 1?

Regards,
$$|\pi\rangle$$

 

[I like Serena]

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.

 

[Evgeny.Makarov]

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.