Can a vector with zero magnitude have one or more components that are nonzero?

Answer 1

You can always divide such a zero-magnitude vector into N number of non-zero components, such that they all cancel out and yield zero for your magnitude.

If you choose, however, an orthogonal basis (all vectors are 90 degrees from each other) for your components, then you can never have two components cancel (since they'll be at right angles), so it's thus impossible to have a zero-magnitude vector.

Bottom line: Yes, if you choose a non-orthogonal basis for the expansion of vector components.

Answer 2

It's not possible in Euclidean space or, in general, any space where the metric tensor is positive definite. In fact, this is precisely what "positive definite" means. But it's possible in other spaces. For example, in Minkowski space, used in special relativity, the magnitude of a 4-vector is given by: x² + y² + z² - t² or: t² - x² - y² - z² (you'll see both in the literature; they just differ by a sign). Clearly it's possible for this to be zero if t² = x² + y² + z². Such a vector is called "light-like" (as opposed to "space-like" or "time-like"), because it represents a displacement which lies directly on the light cone. The first link gives an overview of Minkowski space. The second is a gentle introduction to geometric algebra, which goes into this topic in a little more detail without the language of tensor calculus. (Note: The second link didn't work when I tried it; I'll check again later and edit if necessary.)

Answer 3

yes definitely if your referring to general vector components. one component vector can be going 4 at 90 degrees and another can be going at 4 at 270 degrees. they will cancel each other out. however if your talking about x and y then no. also i'm not even sure a vector can technically even have zero magnitude.

Answer 4

No. It may be the vector sum of 2 or more non-zero vectors, but the magnitude of a vector with components x, y, z.....w is given by
|v| = √(x² + y² + z² + .......w²) so for |v|| to be 0, all of the x, y, z, .....w terms must be 0.

Doug

Answer 5

It could have direction and no speed, I suppose.

Answer 6

yes