# Dot product

#### Petrus

##### Well-known member
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
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
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
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
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
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
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.