Understanding why the definition $\cos(\widehat{vw})=\frac{v\cdot w}{\|v\| \|w\|} $ works so well.

Question

After perusing several geometry / vector space posts on the forum to better understand formal notions of angles and rotations, while avoiding circularity, I came across a definition that I really like from How do you formally prove that rotation is a linear transformation? :

$$\cos(\widehat{vw})=\frac{v\cdot w}{\|v\| \|w\|} $$

After playing around with this definition, I noticed that one of the properties this definition (obviously quite purposefully) possesses is that holding $w$ constant, one can vary the length (but not 'direction') of $v$ without changing the value of the expression (which adheres to our intuitive notion of an 'angle' between two vectors).

I am going through the Chapter 4 appendices of Spivak's Calculus, so I have not yet really even transitioned into any notion of derivative/rates of change, etc. I was wondering if an analytical justification for my observation can be stated without using calculus.

More or less, I am wondering if I can gain insight into why the value of the expression is independent of the length of $v$.

Perhaps the very nature of this question cannot be answered without a nod to calculus, in which case, my apologies for the poorly posed question.

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Edit 1:

It seems like the idea here is that, treating the dot product as a function, $\cdot(v,w)$ and $\sqrt{\cdot(v,v)}\sqrt{\cdot (w,w)}$ change equally while varying the length of $v$.

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Edit 2:

After some thinking, one can reduce my argument to:

Prove $\cos \big ( \widehat{(\alpha v)w})=\cos(\widehat{vw})$ for any scalar $\alpha$.

Applying the definition: $\cos \big ( \widehat{(\alpha v)w})=\dfrac{(\alpha v) \cdot w}{\lVert \alpha v \rVert \lVert w \rVert}$

From properties of norms and dot products, we know the following:

$(\alpha v) \cdot w=\alpha(v \cdot w)$

$\lVert \alpha v \rVert = \alpha \lVert v \rVert$

Applying these two properties, we get:

$\dfrac{(\alpha v) \cdot w}{\lVert \alpha v \rVert \lVert w \rVert} = \dfrac{\alpha (v \cdot w)}{\alpha \lVert v \rVert \lVert w \rVert}$

Canceling out the $\alpha$'s, the answer follows.

Using the same approach, we can further generalize this to:

$\cos \big ( \widehat{(\alpha v)(\beta w})=\cos(\widehat{vw})$ for any scalars $\alpha$ and $\beta$.

Answer 1

If you are willing to believe that two vectors are orthogonal to each other iff their dot product is $0,$ then consider this:

By computing the dot product, one can see that

• $\dfrac{u\cdot v}{\|v\|\|v\|} v$ and $u - \dfrac{u\cdot v}{\|v\|\|v\|} v$ are orthogonal to each other, and

• the first of the two vectors above is in the direction of $v,$ and

• the sum of those two vectors is $u.$

Thus we have a right triangle in which the adjacent side is $\dfrac{u\cdot v}{\|v\|\|v\|} v$ and the hypotenuse is $u$. The cosine is the length of the adjacent side divide by the length of the hypotenuse.