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I have scoured all the answers on this website but I still cannot understand why $a\cdot b = |a||b|\cos\theta$, if the dot product is interpreted as the amount of one vector, say $a$, in the same direction as the other, say $b$, then why do we scale it by multiplying it with the magnitude of the vector on which it projects? why do we need $|b|$?

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  • $\begingroup$ Don't you like your products to be bilinear? $\endgroup$ – Angina Seng Jul 13 '20 at 4:55
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    $\begingroup$ @AnginaSeng What does that mean? Sounds exciting for some reason lol $\endgroup$ – Sarhaan Gulati Jul 13 '20 at 4:59
  • $\begingroup$ Sometimes it’s good to look for a proof of that result $\endgroup$ – Fakemistake Jul 13 '20 at 12:36
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To understand how the dot product is defined, it's better to first look at why the dot product is defined. The idea of the dot product is to have some operation which takes in two vectors (say $a$ and $b$), and returns a single (meaningful) value. As you suggest, if we were merely interested in the length of the projection of $a$ onto $b$ (or, as you put it, the amount of one vector goes in the same direction as the other), then we would indeed want to use $|a|\cos\theta$. But there are several problems with this ''definition''.

Indeed, let's say we did define such a product: say $a\star b=|a|\cos\theta$. Observe this ''product'' has the following problems:

  • We do not have $a\star b=b\star a$. In other words, the product is not commutative.

  • We do not have $a\star (b+c)=a\star b+a\star c$, it is not distributive.

  • We do not have $a\star (xb)=x(a\star b)$ for $x\in \mathbb{R}$, it does not respect scalar multiplication.

That is why defining $a\cdot b=|a||b|\cos\theta$ is a much more natural definition. It satisfies the properties we want a usual product to satisfy, and hence tells us more about the vectors $a$ and $b$.

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  • $\begingroup$ Yes, but what do you interpret that scalar value as then? $\endgroup$ – Sarhaan Gulati Jul 13 '20 at 5:17
  • $\begingroup$ You could interpret it as the length of the projection of $a$ onto $b$, scaled by the length of $b$ itself. $\endgroup$ – Romain S Jul 13 '20 at 5:18
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    $\begingroup$ And that scaling is important to preserve the properties that u mentioned $\endgroup$ – Sarhaan Gulati Jul 13 '20 at 5:19
  • $\begingroup$ Exactly! That scaling is what allows us to have all of the above properties, and hence obtain a ''nice'' and useful mathematical tool. $\endgroup$ – Romain S Jul 13 '20 at 5:20
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    $\begingroup$ Sweet man thanks! $\endgroup$ – Sarhaan Gulati Jul 13 '20 at 5:24

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