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Dot product

Petrus

Well-known member
Feb 21, 2013
739
Hello MHB,
I would like if someone could confirmed if I did understand correct.
In a ON-system we dont use \(\displaystyle u*v=|u||v|\cos\theta\) cause In a ON-system the \(\displaystyle \theta=\frac{\pi}{2}\) so \(\displaystyle \cos(\frac{\pi}{2})=0\)

Regards
\(\displaystyle |\pi\rangle\)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,866
Hello MHB,
I would like if someone could confirmed if I did understand correct.
In a ON-system we dont use \(\displaystyle u*v=|u||v|\cos\theta\) cause In a ON-system the \(\displaystyle \theta=\frac{\pi}{2}\) so \(\displaystyle \cos(\frac{\pi}{2})=0\)

Regards
\(\displaystyle |\pi\rangle\)
What do you mean by an ON-system?

If u and v are perpendicular to each other, their dot product is indeed zero.
 

Petrus

Well-known member
Feb 21, 2013
739
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 \(\displaystyle B=(u_1,u_2,u_3)\) be a ON base in the room and we assume that \(\displaystyle u=(a_1,b_1,c_1)\) and \(\displaystyle v=(a_2,b_2,c_2)\) in this base ten (they all u and v is vector)
\(\displaystyle u*v=a_1a_2+b_1b_2+c_1c_2\)

Regards,
\(\displaystyle |\pi\rangle\)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,866
ON=Orthogonality
If I have understand correctly Orthogonality means that it's perpendicular.
Then we say that "Let \(\displaystyle B=(u_1,u_2,u_3)\) be a ON base in the room and we assume that \(\displaystyle u=(a_1,b_1,c_1)\) and \(\displaystyle v=(a_2,b_2,c_2)\) in this base ten (they all u and v is vector)
\(\displaystyle u*v=a_1a_2+b_1b_2+c_1c_2\)

Regards,
\(\displaystyle |\pi\rangle\)
Yep. That's true.
 

Petrus

Well-known member
Feb 21, 2013
739
Yep. That's true.
Hello,
Have I also understand this correct. ON-Base means that all vector have the length 1?

Regards,
\(\displaystyle |\pi\rangle\)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,866
Hello,
Have I also understand this correct. ON-Base means that all vector have the length 1?

Regards,
\(\displaystyle |\pi\rangle\)
Yes.
ON would mean orthonormal.
Ortho means perpendicular.
Normal means length 1.
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,502
In a ON-system we dont use \(\displaystyle u*v=|u||v|\cos\theta\) cause In a ON-system the \(\displaystyle \theta=\frac{\pi}{2}\) so \(\displaystyle \cos(\frac{\pi}{2})=0\)
Just to add: If \(\displaystyle u=(a_1,b_1,c_1)\) and \(\displaystyle v=(a_2,b_2,c_2)\) in an orthonormal basis, then \(\displaystyle u\cdot v=a_1a_2+b_1b_2+c_1c_2\) is a corollary of the fact \(\displaystyle x\cdot y=|x||y|\cos\theta\) for any vectors $x$, $y$ with an angle $\theta$ between them. So it's not the case that we don't use \(\displaystyle x\cdot y=|x||y|\cos\theta\); it's just that in the case of orthonormal basis there is a special formula for the dot product.