MHB 1-Forms .... Interpretation by Bachman ....

[Math Amateur]

I am reading David Bachman's book: "A Geometric Approach to Differential Forms" (Second Edition) ...

I need some help with some remarks Bachman makes near the start of his section on 1-forms ...

The relevant section reads as follows:View attachment 8604In the above text from Bachman we read the following:

" ... ... Recall the geometric interpretation of the dot product: You project $$\langle -1, 2 \rangle$$ onto $$\langle 2, 3 \rangle$$ and then multiply by $$\mid \langle 2, 3 \rangle \mid = \sqrt{13}$$. ... ... "My question is as follows:

Is there any reason Bachman has chosen the projection of $$\langle -1, 2 \rangle$$ onto $$\langle 2, 3 \rangle$$ ... ... ?

Could he just have easily chosen the projection of $$\langle 2, 3 \rangle$$ onto $$\langle -1, 2 \rangle$$ ... ... ?Can someone please clarify ...

Peter==================================================================================

It may help MHB readers of the above post to have access to Bachman's Section 3.1 ... ... so I am providing the same .. ... as follows:View attachment 8605
View attachment 8606Hope that helps ...

Peter

 

[Evgeny.Makarov]

My question is as follows:

Is there any reason Bachman has chosen the projection of $$\langle -1, 2 \rangle$$ onto $$\langle 2, 3 \rangle$$ ... ... ?

Could he just have easily chosen the projection of $$\langle 2, 3 \rangle$$ onto $$\langle -1, 2 \rangle$$ ... ... ?

In this place, yes. The scalar projection $\text{proj}_vu$ of vector $u$ on vector $v$ is $\dfrac{u\cdot v}{|v|}$, from where $u\cdot v=(\text{proj}_vu)\cdot|v|$. The equality $u\cdot v=(\text{proj}_uv)\cdot|u|$ is equally valid. But the conclusion "Evaluating 1-form on a vector is the same as projecting [that vector] onto some line..." assumes that we project the vector we feed to the form, i.e., $\langle dx,dy\rangle$.

 

[Math Amateur]

In this place, yes. The scalar projection $\text{proj}_vu$ of vector $u$ on vector $v$ is $\dfrac{u\cdot v}{|v|}$, from where $u\cdot v=(\text{proj}_vu)\cdot|v|$. The equality $u\cdot v=(\text{proj}_uv)\cdot|u|$ is equally valid. But the conclusion "Evaluating 1-form on a vector is the same as projecting [that vector] onto some line..." assumes that we project the vector we feed to the form, i.e., $\langle dx,dy\rangle$.

Thanks Evgeny

Peter