To simply present the question I will be using the two vectors (velocity and displacment), and the scalar time. As you may know, $$\vec{s} = \vec{v}\cdot t $$ This makes perfect sense as multiplying a vector by a scalar gives a vector with a different magnitude. But my question is when we want to calculate the time taken given that we know the velocity at which an object was travling and the total displacment it was displaced, we will have to equate the following equation $$\vec{s}/\vec{v}$$ But what does that mean? we have two operations on vectors regarding multiplication (i.e the dot product and the scalar product), but $$\vec{s}/\vec{v}$$ doesnt fit in any of the two. The first intuitive thing that came to mind is to think of it like a vector product of two vectors assuming that $$\frac{1}{\vec{v}}$$ is a vector. But here I face two problems, 1. I dont know if $$\frac{1}{\vec{v}}$$ (v being a velocity vector) gives a vector or, in other words if a recprocal of a vector gives another vector. I tried to google this but I couldnt get an answer that I can understand. I only have a high school level of mathematics so please try consider that when you answer the question if possible. The second thing is when I learn this at school I have not applied the vector product to find the answer I would just manipulate the numbers as they are just scalars. But what is the right way?
So to just put my question in short words Does the recprocal of a vector give a vector? And can a product of a vector (a) and the recporcal of another vector(b) treated as a vector product (is a/b = scalar product) or is there another way to calcaulate that?
Best Answer
In mathematics vector space is an additive abelian(commutative) group and satisfies vector distribution over scalar addition, scalar distribution over vector addition and other two properties.
Scalar is a field(a ring). When we talk about velocity in physics then as we can see it satisfies all the properties of vector space so it is a vector. But, a vector does not guarantee that it has an inverse like field( see all the properties of vector space it does not say that if field of dimension n contains one more property apart from vector properties, it cannot be a vector space) as you are saying.
I recommend taking vectors as scalars or most simple would be breaking vector into individual dimensions(components) and then in individual dimension you are free to take each component of the vector as scalar. Try this.