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By the naming of scalar we know that it scales. When It is multiplied with a vector it changes the vector's magnitude because it only has magnitude but no direction. So then why a negative scalar rotates a vector's direction by 180 degree? For example, if the scalar is -5 then it will scale the magnitude 5 times and also rotate the vector by 180 degree. Thus it changes the vector's direction. Doesn't it imply that scalars have also direction??

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    $\begingroup$ this is called inversion through the origin, rather than rotation $\endgroup$ – Andrew Steane Sep 19 '20 at 14:14
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    $\begingroup$ The directions of scalars are called signs $\endgroup$ – Hagen von Eitzen Sep 19 '20 at 14:46
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Vector spaces are defines on a field where those scalers comes from and in your case real numbers are the field and if you multiply any vector, scalars scales the components of the vector on the axises defined. If you multiply these components with any scalar that will scale all the three components. Basically scaling doesn't rotates the vector it simply scales the vector along a line such that, assume you have a vector where tail is on the origin and when you scale this vector with all the real numbers your vector's tip will draw a line where you can simply think as scaling pulls or pushes the vector along the direction of the vector. Also, turning back to field definition where scalars come from, in your case real numbers are the scalars and also real numbers are a vector space in itself and each element is both vector and scalar so, in one perspective scalars have direction but on the other hand when multiplying it to vectors in three dimensions you are doing matrix product where the scalar placed on the right of the vector. So, you can think scaling as a matrix product which means a scalar is a transformation matrix (standard matrix) but it's geometrical interpretation may differ for different vector spaces.

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look at this:

enter image description here

the blue arrow has the components $[x~,y]~$ the red one $[-x~,-y]~$, both with the magnitude $\sqrt{x^2+y^2}$

thus you have to rotate the blue arrow by $2\,\pi$ radiant to obtain the red one.

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