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It is known that a dot product is a scalar, but a cross product is a vector.

Dot product: $$ A \cdot B \in \mathbb{R} $$

$$ A \cdot B = |A||B| \cos\theta $$

$$ A \cdot B = A_xB_x + A_yB_y + A_zB_z $$

Cross product: $$ A \times B \in \mathbb{R}^3 $$

$$ \| A \times B \| = |A||B|\sin\theta $$

$$ A \times B = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \ A_x & A_y & A_z \\ \ B_x & B_y & B_z \\ \end{vmatrix} $$

$$ A \times B = (A_yB_z-A_zB_y)\mathbf{i} + (A_zB_x - A_xB_z)\mathbf{j} + (A_xB_y - A_yB_x)\mathbf{k} $$

It is known that the dot product results in a scalar, and the cross product results in a vector. Why does one have direction, and the other does not, and why are they calculated differently?

Glorfindel
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Larry
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  • You just said $A \times B$ is a vector, so is $A \times B = |A||B|sin\theta$ true ? – Ahmad Bazzi Sep 12 '18 at 01:15
  • While this is a pretty broad question, I think the suggested duplicate (full disclosure: I have a vested interest in the duplicate) goes a long way towards answering at least one possible root question here. – Steven Stadnicki Sep 12 '18 at 01:16
  • @AhmadBazzi: it is the magnitude, so it is true. – Larry Sep 12 '18 at 01:27
  • so $\vert A \times B \vert = |A||B|sin\theta$ – Ahmad Bazzi Sep 12 '18 at 01:27
  • @Larry also please check this question here https://math.stackexchange.com/questions/62318/origin-of-the-dot-and-cross-product – Ahmad Bazzi Sep 12 '18 at 01:28
  • @Ahmad Bazzi: the question is really helpful. I saw Steven Stadnicki's answer, and I think it is very detailed. I think your statement is right. – Larry Sep 12 '18 at 01:38
  • this stupid stack exchange! it will not let me correct a small change (a change from a minus sign to a plus sign) without forcing me to make big changes. and then when i tried to, it changed it back to the original. i understand the need to review (i have only 206 rep), but these restrictions are stupid. – robert bristow-johnson Sep 12 '18 at 01:45
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    @ rober bristow-johnson I think I just made the correction, thanks – Larry Sep 12 '18 at 01:49
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    i can't add an answer, evidently the question is closed. $$ $$ one thing that motivates mathematical definitions are a need in the physical sciences. work (or energy), a scaler, is the dot product of force and displacement, both vectors. torque, a vector, is the cross product of force and and "moment arm", both vectors. – robert bristow-johnson Sep 12 '18 at 02:08
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    the last change was to make the matrix symbol into determinent. but since SE requires 6 characters to change, i had to add some "xxxxx" to it. – robert bristow-johnson Sep 12 '18 at 04:41

2 Answers2

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The only thing I think they have in common is that they take two vectors as input and have the word "product" in their name. They don't work in the same dimensions, produce the same types of values or have a meaningful interpretation in terms of the other.

CyclotomicField
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  • both operations have product-like properties. it's not merely that they have "product" in the terminology. – robert bristow-johnson Sep 14 '18 at 18:40
  • @robertbristow-johnson I'm not familiar with "product-like" properties. What exactly defines "product-like" mathematical objects? – CyclotomicField Sep 15 '18 at 16:53
  • the magnitudes of the cross product vector and the dot product are both proportional to the product of the magnitudes of the two argument vectors. and under different conditions, the magnitudes of the results of the cross or dot products are equal to the product of the magnitudes of the two argument vectors. – robert bristow-johnson Sep 16 '18 at 21:46
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You a;lready know one produces a scalar, and the other a vector.

Dot product is defined between any two vectors of same (but arbitrary) sizes. However cross-product is defined only for 3D-vectors.

Your definition is presented in a way that expects one to know the angle between the vectors to calculate the dot product. Actually it is the other way: one can calculate the dot product given all the components of the vector, and use that product value to determine the angle between those vectors.

In a theoretical view point dot product makes available a natural isomorphism to the dual vector space. And cross-product gives a map from the second exterior power of the 3-dimensional vectors space to itself, which is an isomorphism. These are the theoretical advantages.

P Vanchinathan
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  • Can you expand on the isomorphism between ∧²ℝ³ and ℝ³? That one exists is clear, as dim ∧ⁿ V = (dim V choose n), which is 3 = dim ℝ³ in this case. It's probably not unique, but is there one that is natural in the sense of cross products? – Jos van Nieuwman Oct 20 '20 at 20:41