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Sir $P=F/A$ And since $F$ and $A$ both are vectors but $P$ is scalar. So doesn't it violates that "Division is NOT defined for vectors"?

BioPhysicist
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Vinay Sheokand
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2 Answers2

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Pressure is actually $$P=\frac{F_\bot}{A}$$ where $F_\bot$ is the force component perpendicular to the surface in question, and $A$ is the area of the surface.

Therefore, there is no "division by a vector" here. Certainly, the area vector is used in various areas of physics; this is not one of those areas (pun always intended).

I suppose if you wanted a definition based on vectors you can exploit the use of projections: $$P=\frac{\mathbf F\cdot\mathbf A}{||\mathbf A||^2}$$

since the area vector, by definition, is perpendicular to the surface in question.

BioPhysicist
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division is not defined for vectors. However, it is obvious for vectors that are scalar multiples of each other. Just because we can apply division to a subset of vectors doesn't mean it's defined for the entire set

Jim
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  • What do you mean by "it is obvious for vectors that are scalar multiples of each other"? – Riemann'sPointyNose Aug 04 '20 at 23:13
  • @Riemann'sPointyNose If A =c B , then you can do some fairly obvious division (so obvious that people don't always notice when they do it) to say that A / B =c – Jim Aug 06 '20 at 12:05
  • Hmm... It's not really meaningful to say ${\frac{\mathbf{A}}{\mathbf{B}}=c}$. I mean, sure - it would stand to reason that this would be a nice candidate result for such an operation, but the operation remains undefined. You cannot just say ${\frac{\mathbf{A}}{\mathbf{B}}=c}$ without defining what ${\frac{\mathbf{A}}{\mathbf{B}}}$ means in this context. You could of course define the division - but this isn't what we typically do. So I don't think this is the answer to this question. – Riemann'sPointyNose Aug 08 '20 at 14:20
  • @Riemann'sPointyNose I'm fairly certain that the main point of the answer I provided is exactly as you say, that the main operation is undefined even though that it stands to reason in this one case that one could do this. – Jim Aug 11 '20 at 12:58
  • Furthermore, you cannot say that in the special case of collinear vectors that we do not do such division. If given (1,0,1) and (2,0,2), we know one is twice the other through easy division in our heads. Just the same as when given 36 and 18, we do 36/18 in our head to get 2, but the division is done nonetheless. One could not "pull the two out" of the vector without a mathematical way of doing that division. Object permanence is a thing; just because you skip a step doesn't mean the step doesn't exist. – Jim Aug 11 '20 at 12:58
  • Yes, one could define this to be the way they handle such an expression. But it is not at all "obvious" and "easy" - you still have to define it. Things in Mathematics don't have to behave whatsoever how you think they should, and there are plenty of examples of other operations you can define where this is the case. I don't think at all it stands to reason and is "easy" and "obvious" this is how we should handle such a division. Division between numbers is a well defined operation - yes, ${\frac{36}{18}=2}$. But the vector division you refer to has not been well-defined. – Riemann'sPointyNose Aug 11 '20 at 15:15
  • I am also curious as to what you mean by "One could not "pull the two out" of the vector without a mathematical way of doing that division"? Because we can, in fact, pull out that $2$ (so long as scalar multiplication is defined well, which it is) without defining any division. Not every operation has to be invertible whatsoever. The point I'm trying to make is I doubt this is the correct way to understand OP's question. It's too Mathematically sloppy – Riemann'sPointyNose Aug 11 '20 at 15:20
  • @Riemann'sPointyNose Having scalar multiplication defined for vectors does not allow you to pull the two out when given only the two vectors. As you say, not every action is invertible. If given the two vectors only, there must be a mathematically defined way of finding the 2 that doesn't boil down to simply "we tried trial and error on every number until scalar multiplication resulted in a win". – Jim Aug 11 '20 at 17:44
  • I'm not opposed to saying that the operation is not defined for vectors in general. I said that explicitly in my answer. But when everyone simply does collinear vector division in their head without realizing it, I feel justified in calling it obvious. As for the mathematical sloppiness, it's you that have read too far into this. I was pointing out the mathematical sloppiness of the OP by saying "Yes, you can do division this way when they are scalar multiples and that's fine shorthand but that doesn't mean you can define vector division" – Jim Aug 11 '20 at 17:49
  • Then you decided I was saying the exact opposite and got mad at me for saying the same thing as what you wanted but in more colloquial terms – Jim Aug 11 '20 at 17:50