Visualization of span of 3 vectors?

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If I visualize 3 vectors in 3d space, is the span of the 3 vectors a 3d region within the 3 vectors? I need help visualizing so I can understand the concept better.

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Start of with the three vectors you know and assume they are all different from the zero vector (thus I don't want to span a point).

We start with taking the first vector. He can point in any direction. If the second vectors points in the same direction as the first one, we span a line. Also assume that the third vector also lies in the same direction, then we still span a line. You see, if all those three vectors lie in the same direction, you can only span a line.

If the second vector does not point in the same direction as the first one, we already span a plane.

Now you can take a third vector. If this vector point in the same direction as the plane, you still span the plane. If this vector does not point in the direction of the plane and thus uses your last dimension properly, you span a 3D space.

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The span $S = <u_1, u_2, u_3>$ consists of all possible linear combinations

$$ \alpha_1 u_1 + \alpha_2 u_2 + \alpha_3 u_3 $$

where the coefficients $\alpha_i$ are arbitrary real numbers, including negative, zero and positive numbers, e.g. +5002, -136 and 0.

The span $S$ will be either a point, a line, a plane in 3D space or the full 3D space, each containing the origin.

This depends on the choice of vectors $u_i$ and how many of them are linear independent.

Note: I assume that your vector space is $\mathbb{R}^3$.

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Imagine being in a spaceship in $\mathbb{R}^3$. Start at $\mathbf{0}$. The vectors in question are directions that your ship can travel in (this corresponds to adding that vector on). You can travel any distance, forwards or backwards, along any of these directions (this corresponds to multiplying the vector by a constant $c$ or $-c$).

The span of these vectors is the set of all points your spaceship can get to this way.

It so happens that this set will either be a line, a plane, or the whole space. It will never be, for instance, some cube in $\mathbb{R}^3$. (Why not?)