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Anonymous Sat 07 Sep 16:59:07 2019 No.10952945 View Reply Original Report

Quoted By: >>10952967 >>10952986 >>10953353 >>10953443

Anyone have a good explanation as to why and how this cosine version of the dot product relates to the sum of the component products?
I know how to use the dot product but it makes no sense to me.

Why is the dot product called a projection when it doesn’t give projection but a projection multiplied by a number?

What is the mathematical reasoning for making this definition? I’ve seen the reasoning for the cross product, the sine definition is the area of the parallelogram but the dot product eludes me

Anonymous

Anonymous Sat 07 Sep 2019 17:04:34 No.10952967 Report

Quoted By: >>10953331

>>10952945

it's a right triangle. cos(theta) is length of adjacent/ length of hypotenuse. multiplying by |A| cancels the denominator so you get the length of A projected onto B.

Anonymous

Anonymous Sat 07 Sep 2019 17:10:44 No.10952986 Report

Quoted By: >>10953331

>>10952945
My text book first computed the cosine using basic geometry and coordinates and then defined the dot product.
Yours didn't?

Anonymous

Anonymous Sat 07 Sep 2019 18:48:36 No.10953331 Report

Quoted By:

>>10952967
Yeah I get the trig, I understand |A|cos(theta) is the projection but then why even multiply by |B|? By multiplying by |B| the value is no longer the projection. It’s a multiple of the projection
>>10952986
No, it just stated that it is defined that way. What was your textbooks definition/proof?

Anonymous

Anonymous Sat 07 Sep 2019 18:53:29 No.10953353 Report

Quoted By:

>>10952945
because the length of the line being projected matters..

Anonymous

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Anonymous Sat 07 Sep 2019 19:27:07 No.10953443 Report

Quoted By: >>10953448 >>10953461 >>10953566 >>10953646

>>10952945
Okay, if I was teaching the dot product to someone from scratch, this is the short of it:

1) given law of cosines (be sure to be able to prove this to yourself)
$c^2 = a^2 + b^2 - 2 a b \cos \theta$
2) define dot product as
$\vec{a} \cdot \vec{b} \equiv a b \cos \theta$
3) solve law of cosines for dot product
$ \begin{align} \vec{a} \cdot \vec{b} &= \frac{1}{2}(a^2 + b^2 - c^2) \\ &= \frac{1}{2}\left (a^2 + b^2 - \sum_i (b_i - a_i)^2 \right ) \\ &= \frac{1}{2}\left (a^2 + b^2 - \sum_i (b_i^2 - 2 a_i b_i - a_i^2) \right ) \\ &= \frac{1}{2}\left (a^2 + b^2 - \sum_i b_i^2 + 2 \sum_i a_i b_i - \sum_i a_i^2) \right ) \\ &= \frac{1}{2}\left (a^2 + b^2 - b^2 + 2 \sum_i a_i b_i - a^2) \right ) \\ &= \sum_i a_i b_i \end{align}$
4) boom, done

Vectors are geometrically inspired, so when working out definitions, understanding the geometry is key. Once you understand 0D through 3D well enough, you see the pattern, and distill what it means to be a vectorspace down to about 10 axioms, which allows you to extend the math to ND easily. Similarly, with the dot product and the two forms you now know (cosine and sum of components), you can boil down what it means to be an "inner product" to 3 or 4 axioms. In this way you can start to consider wacky vectorspaces (like the space of all matrices of a certain size, or countably infinite and uncountably infinite dimensional vectorspaces) and inner products (like using integrals).

Anonymous

Anonymous Sat 07 Sep 2019 19:28:22 No.10953448 Report

Quoted By:

>>10953443
Great post.

Anonymous

Anonymous Sat 07 Sep 2019 19:33:35 No.10953461 Report

Quoted By:

>>10953443
based

Anonymous

Anonymous Sat 07 Sep 2019 20:05:46 No.10953566 Report

Quoted By: >>10953646

>>10953443
I'll add, this gives the algebraic connection, but it doesn't shed much light on the geometric interpretation of what an inner product does. The projection interpretation you gave is perfectly fine. You seem to be wondering why you'd ever want to do something that wasn't a projection, where one of the vectors didn't have unit length. Knowing some applications here helps.

For instance, in basic physics, you learn work is "force times a displacement". However, the force has to be in the direction of the displacement, otherwise no work is done. So really, work is force dot displacement. Earth's gravity isn't doing any work on you when you are sitting still or orbiting it in a circle, only when you fall.

Another example would be trying to determine the volume rate of air flowing through a flat hoop in a uniform wind. If you represent the area of the hoop as a vector normal to the hoop with a length equal to the area [m^2], and knew the velocity vector of the air [m/s], then dotting them yields the volume rate [m^3/s]. If the hoop area vector points into the wind, you get some non-zero number. If the hoop area vector points orthogonal to the wind, you get zero air flow.

There are countless other (probably better) examples.

Anonymous

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Anonymous Sat 07 Sep 2019 20:35:56 No.10953646 Report

Quoted By:

>>10953443
This makes much more sense now and I finally get it now. Thanks anon
>>10953566
This was also really helpful too. Thanks anon as well.

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