Derivation of the angle between two vectors from law of cosine

[saaz]

Hello!

I was looking at the way to derive the formulae of the angle between to vectors from the law of cosine.

I want to get this, where the dot product of two vectors is equal to the product of the modulus of the two vectors and the cosine of the angle between them:

[math]uv = |u||v|Cos(A)[/math]
If I start with the law of cosine and simplify, I get this:

[math](u-v)^2 = u^2+v^2-2uvCos(A)[/math]
[math]u^2-2uv+v^2 = u^2+v^2-2uvCos(A)[/math]
[math]uv= uvCos(A)[/math]
At this point, I don't get why, on the expression before the equality, the product of the two vectors is a dot product, while on the expression after the equality is the product of the length of the vectors.

Thanks in advance!

 

[Harry_the_cat]

If \(\displaystyle u\) and \(\displaystyle v\) are vectors, how do you define \(\displaystyle u^2\) and \(\displaystyle v^2\)?

 

[blamocur]

Hello!

I was looking at the way to derive the formulae of the angle between to vectors from the law of cosine.

I want to get this, where the dot product of two vectors is equal to the product of the modulus of the two vectors and the cosine of the angle between them:

[math]uv = |u||v|Cos(A)[/math]
If I start with the law of cosine and simplify, I get this:

[math](u-v)^2 = u^2+v^2-2uvCos(A)[/math]
[math]u^2-2uv+v^2 = u^2+v^2-2uvCos(A)[/math]
[math]uv= uvCos(A)[/math]
At this point, I don't get why, on the expression before the equality, the product of the two vectors is a dot product, while on the expression after the equality is the product of the length of the vectors.

Thanks in advance!

What exactly is your question? What do you consider as a given and what are you trying to derive?

Also, I suggest being more careful with the notation. You use [imath]uv[/imath] both for dot product and product of magnitudes, whereas using, for example, [imath]u\cdot v[/imath] and [imath]|u||v|[/imath] would be less confusing. And, to partially answer @Harry_the_cat's question, [imath]|u|^2[/imath] looks less ambiguous than [imath]u^2[/imath].

 

[Dr.Peterson]

Hello!

I was looking at the way to derive the formulae of the angle between to vectors from the law of cosine.

I want to get this, where the dot product of two vectors is equal to the product of the modulus of the two vectors and the cosine of the angle between them:

[math]uv = |u||v|Cos(A)[/math]
If I start with the law of cosine and simplify, I get this:

[math](u-v)^2 = u^2+v^2-2uvCos(A)[/math]
[math]u^2-2uv+v^2 = u^2+v^2-2uvCos(A)[/math]
[math]uv= uvCos(A)[/math]
At this point, I don't get why, on the expression before the equality, the product of the two vectors is a dot product, while on the expression after the equality is the product of the length of the vectors.

Thanks in advance!

If you are trying to derive [imath]u\cdot v = |u||v|\cos(A)[/imath] from the Law of Cosines, you will also need to somehow involve a definition of the dot product. What definition are you using? I would expect you to be starting from the definition in terms of components.

Your application of the LOC should be written as [imath]|u-v|^2 = |u|^2+|v|^2-2|u||v|\cos(A)[/imath]. What you wrote next doesn't follow from it.

Please tell us more about what you have learned, and what you are trying to do.

 

[saaz]

If \(\displaystyle u\) and \(\displaystyle v\) are vectors, how do you define \(\displaystyle u^2\) and \(\displaystyle v^2\)?

Mmh, they would be squares of the modules of the respective vectors... I tried to treat them either as: [math]x^2+y^2[/math] or using the cosine as [math]x =Cos(ϴ)u[/math] and rearranging to obtain the final equation, but It seems I get nowhere.

What exactly is your question? What do you consider as a given and what are you trying to derive?

Also, I suggest being more careful with the notation. You use [imath]uv[/imath] both for dot product and product of magnitudes, whereas using, for example, [imath]u\cdot v[/imath] and [imath]|u||v|[/imath] would be less confusing. And, to partially answer @Harry_the_cat's question, [imath]|u|^2[/imath] looks less ambiguous than [imath]u^2[/imath].

My question is to understand this equation: [math]u⋅v=∣u∣∣v∣Cos(A)[/math]. The book suggests that I could derive it from the law of cosine [math]a^2 =u^ 2 +v^ 2 −2uvCos(A)[/math], but once I get to the final step of my first post I don't see how to proceed.

If you are trying to derive [imath]u\cdot v = |u||v|\cos(A)[/imath] from the Law of Cosines, you will also need to somehow involve a definition of the dot product. What definition are you using? I would expect you to be starting from the definition in terms of components.

Your application of the LOC should be written as [imath]|u-v|^2 = |u|^2+|v|^2-2|u||v|\cos(A)[/imath]. What you wrote next doesn't follow from it.

Please tell us more about what you have learned, and what you are trying to do.

The way I know the dot product is: [math]u⋅v=x1x2+y1y2[/math]. About the way that I wrote the LOC, I used the same format of the book, without absolute values, although I think it doesn't make a real difference since the magnitude of a vector is always positive, albeit I trust you that with absolute values is more correct.

Ok, so maybe I should write down a sketch of an obtuse angle, treat all sides as vectors, and use all the components of each vector. Maybe this way it will pop out the cosine of the angle between them?

 

[saaz]

Oh yes, actually if I transform everything into basic components the left side simplifies in the dot product. I think I was hoping there was a more simple or straightforward way to get it from where I ended to be...

Anyway, thanks again for your answers! This forum is great

 

[Dr.Peterson]

About the way that I wrote the LOC, I used the same format of the book, without absolute values, although I think it doesn't make a real difference since the magnitude of a vector is always positive, albeit I trust you that with absolute values is more correct.

It absolutely does make a difference. (Pun not originally intended, but recognized.)

In the statement of the Law of Cosines, the book is taking a, b, and c to be lengths of sides. In the description of the dot product, they are vectors (and typically are written either in a different font, or with arrows above them, or something like that). It is imperative not to confuse vectors with numbers.

Oh yes, actually if I transform everything into basic components the left side simplifies in the dot product. I think I was hoping there was a more simple or straightforward way to get it from where I ended to be...

I'd like to see the details of what you did. It sounds right.

The derivation of the fact you are interested in can be done in several different ways. A couple different perspectives are shown in my blog here. (The last answer uses the Law of Cosines.)

 

[saaz]

It absolutely does make a difference. (Pun not originally intended, but recognized.)

In the statement of the Law of Cosines, the book is taking a, b, and c to be lengths of sides. In the description of the dot product, they are vectors (and typically are written either in a different font, or with arrows above them, or something like that). It is imperative not to confuse vectors with numbers.

I'd like to see the details of what you did. It sounds right.

The derivation of the fact you are interested in can be done in several different ways. A couple different perspectives are shown in my blog here. (The last answer uses the Law of Cosines.)

I read the article, nice, thanks! I actually found there a way to derive it that I like more.

So, here is what I have done. The top half of the page is the method that I like from your blog; now I get why the cosine of the angle between pop out. While the bottom half is the way to derive it from the LOC. I rewrote it quickly working only with the left side to show that most of the terms cancel out with the modules of the terms on the right side (not showed). I hope is clear enough.

 

[Dr.Peterson]

I read the article, nice, thanks! I actually found there a way to derive it that I like more.

So, here is what I have done. The top half of the page is the method that I like from your blog; now I get why the cosine of the angle between pop out. While the bottom half is the way to derive it from the LOC. I rewrote it quickly working only with the left side to show that most of the terms cancel out with the modules of the terms on the right side (not showed). I hope is clear enough.

I can't follow the thinking in the bottom version, though I can see it hidden between the lines. I think you have used "=" where it doesn't belong (at the start of the last two lines), and omitted all the work using the law of cosines. For a proof, you'd need to show a lot more work, since a proof is communication.