High School Confused about dot product multiplication

[Hill]

Given a gradient of a function, $\vec \nabla f$, and a vector $\vec v$, the dot product, $\vec \nabla f \cdot \vec v$, is the directional derivative of $f$ in $\vec v$.

 

[Steve4Physics]

I could be wrong. But don't all intuitive examples involving both physics and dot product that are also simple, involve forces?

You could be right. But presumably it depends upon what is classified as intuitive/simple.

For example Gauss's law is fairly intuituve/simple after studying basic electromagnetism. It involves the dot product of the electric field vector and the 'area vector'. (But in fact force is hidden in there - because the electric field vector depends on electric force.)

Edit: Aha. @PhDeezNutz beat me to it.

 

[Mark44]

I meant some kind of word problem example using a specific real world application to illustrate how the dot product works.

Suppose a company sells five different products, call them A, B, C, D, and E. Suppose also that during a month, there were, respectively 25, 30, 15, 20, and 40 units sold, with per-unit revenues of $100, $120, $85, $140, and $135.
If the number sold is $\overrightarrow{N_{sold}} = <25, 30, 15, 20, 40>$ and the per-unit revenue is $\vec R = <100, 120, 85, 140, 135>$, then the total revenue is given by $\overrightarrow{N_{sold}} \cdot \vec R$. The units of this number would be dollars.

 

[PeroK]

I meant some kind of word problem example using a specific real world application to illustrate how the dot product works.

There must be a parable that covers your situation. Mathematics is a mountain. You have to start climbing to make progress. You can only sit at the bottom looking up for so long. You must have made a decision at some point to attempt this mountain, so you ought to get on with it.