From my understanding from university physics from Pearson (Page ~20):

Dot product multiplies "like" unit vector terms and cross product multiplies "unlike" unit vector terms. So, why does the cross product retain its unit vector (ñ), but dot product does not?


$A\cdot B = AB\cos(\theta) =$ magnitude of $A$ * The component of $B$ that is parallel to $A$.

So shouldn't the unit vector be going in the direction of $A$? Do mathematicians drop unit vectors when stuff only goes in one direction and then call it a scalar?

  • $\begingroup$ Don't blame the mathematicians for this. $\endgroup$ – Eric Towers Jul 8 '16 at 1:39

The cross product is a rather special beast, only for $3$-dimensional space. A more general version of it is the exterior product (wedge product), which maps into a different vector space (the exterior square), which happens in the case of dimension $3$ to be identifiable with the original space. Dot products, on the other hand, can be defined in any finite number of dimensions.

  • $\begingroup$ You can define a "cross product" on $\mathbb R^n$ but it takes $n - 1$ vectors as inputs. Oh well. $\endgroup$ – Hoot Jul 7 '16 at 23:55

Hint: Suppose dot products were vectors and not scalars. What direction should they have?


Asker: If the dot product was a vector, then it should be in the direction that was multiplied by the like components. Thus, for Newton's second law:

F = mr = xi + yj + zk

(F)•j = (xi + yj + zk )•j

(F_y)*j = (yj)*j

Cancel out j from both sides, which is our "residue" unit vector:

F_y = yj

  • Chris
  • 1
    $\begingroup$ The thing is, if you're taking the dot product of two vectors, call them $a$ and $b$, it's algebraically the same whether you choose to view the dot product as "$b$ times (projection of $a$ onto $b$)" or "$a$ times (projection of $b$ onto $a$)". With this in mind, it wouldn't make sense to say that the dot product has the direction of $a$ instead of that of $b$, or vice versa. This is why we leave it as a scalar. $\endgroup$ – Asker Jul 7 '16 at 23:40

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