B What is the reality of the scalar product?

[hugo_faurand]

Hi everyone !

I would like to know the real meaning of scalar product. So, I know scalar product is defined as :

||a||.||b||.cos(a;b) = k

But what k is ?

(Sorry for my english, I am french).

Regards :)

 

[scottdave]

It is the product of the magnitudes, which are pointing in the same direction. First one vector is projected onto the other. Think of casting a shadow down of one vector onto the other one, to get the portion which is in the same direction.

One practical use is in calculations of work performed by a force. Suppose you are pulling a wagon with a force. But you are pulling at an angle 30° up from the horizontal surface. You take the cosine of the angle then multiply the magnitude of the force by the distance traveled (magnitude of distance vector), to find the amount of work done.

This video is about tensors and vectors. I think the whole thing is worth watching, but the part around 3:17 talks about projecting vectors onto another vector.

 

[jedishrfu]

Basically its a useful product where you can get:
- the projection of A on B by dividing it by the magnitude of B or
- the projection of B on A by dividing it by the magnitude of A or
- the angle between A and B by dividing by the magnitudes of A and B

This is useful when you are say computing the effect of a force A along a certain direction B (represented as a unit vector in the B direction). The effect of the force is the projection of A on B.

Since B is a unit vector its magnitude is 1 and so the A . B product is the magnitude of the A projection along B.

https://en.wikipedia.org/wiki/Dot_product

In contrast, the vector product A x B geometrically can represent the area of a parallelogram partially bounded by A and B.

 

[fresh_42]

Hi everyone !

I would like to know the real meaning of scalar product. So, I know scalar product is defined as :

||a||.||b||.cos(a;b) = k

But what k is ?

(Sorry for my english, I am french).

Regards :)

This depends on whom you ask. A geometer, an algebraist, a topologist, a physicist or a functional analyst may all give a different answer to this question corresponding to what they use it for. In the end it's a convenient tool to measure something. Perhaps only the algebraist would give a different answer. To him it's simply an operation which some structures carry and others don't.

 

[scottdave]

I liked how Dr. Fleisch described projecting a vector as casting a shadow of one vector onto another.
I know what a vector projection is, but I don't recall seeing it presented like that. It made me think "Aha"

 

[hugo_faurand]

Thanks for your answers! But, as @fresh_42 says, I would like to have more precisions on algebric meaning.

 

[fresh_42]

Thanks for your answers! But, as @fresh_42 says, I would like to have more precisions on algebric meaning.

I'm not sure how far an answer can go on "B" level, which wasn't already given. If you're interested in the geometrical aspect, this: https://arxiv.org/pdf/1205.5935.pdf might be worth reading. Another example is physics itself. Lie theory plays a crucial role in large part of physics and within Lie theory certain objects, whose most important property is to allow a kind of scalar product. Those products are really often used, resp. more generally speaking, bilinear forms, which associate a scalar to two elements of the structure. If it is not degenerated and positive definite as in the case of a scalar product, it is especially useful, as it allows geometry (length and angles) and to some extend a kind of division on the structure considered. But it's not only Lie theory. Quantum field theory heavily relies on Hilbert spaces, where the elements are functions and which have a scalar product.

In more detail:

Not degenerated means, $a\cdot b = 0$ for all $b$ implies $a=0$ and positive definite $a\cdot a \geq 0$ with equality only in the case $a=0$. The usual scalar product has these properties. So if we have them (plus linearity in the arguments, i.e. the distributive law), we can build transformations like $w \longmapsto w - \frac{2(w \cdot v)}{(v \cdot v)} v$ and do geometry by investigating the scalar $\frac{2(w \cdot v)}{(v \cdot v)}$ which can be seen as a normalized angle, a slope. You might be surprised how far this can get you. It was the beginning of an entire classification in Lie theory and the physicist actually use this classification.

This is a short glimpse on how scalar products can be used. They are simply incredibly useful; especially on structures which otherwise don't have methods of measuring. And to measure quantities is beside philosophy the only way we try to understand everything around us.

 

[scottdave]

Another way to calculate the scalar product (also called dot product) of two vectors is to multiply the components, then add them.
For example if vector u = ai + bj + ck and vector v = di + ej + fk, then the scalar product of u and v is a*d + b*e + c*f.

 

[jedishrfu]

Another way to calculate the scalar product (also called dot product) of two vectors is to multiply the components, then add them.
For example if vector u = ai + bj + ck and vector v = di + ej + fk, then the scalar product of u and v is a*d + b*e + c*f.

What's interesting here is that each vector is represented as the sum of three vectors which are all orthogonal to each other aka the basis vectors. The coefficients of u and v are actually the projections of u and v on each basis vector I,j and k i.e. a=u.i and b=u.j ...

Expanding the u.v = ai.di + ai.ej + ai.fk + bj.di + bj.ej + bj.fk + ... = ai.di + bj.ej + ck.fk since the ij, jk and ik dot products are zero because they are orthogonal.

 

[jedishrfu]

Theres a cool series of videos on YouTube by 3brown1blue called the Essence of Linear Algebra that could check out to get more insight into this and other vector operations.

 

[Svein]

The scalar product is the fundamental operator in Hilbert space (https://en.wikipedia.org/wiki/Hilbert_space).