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I've created two (poorly) drawn graphics to illustrate my point.

Two unit vectors, $\widehat i$ and $\hat j$, in their standard orientation of \begin{bmatrix} 1 & 0 \\ 0 & 1\\ \end{bmatrix}

Form an enclosed area like this:

enter image description here

Please note that this is merely a graphic to illustrate my point and salient details of the vector's size and coordinate system have been omitted.

Why, however, is the enclosed area not this, as it seems more intuitive to me?:

enter image description here

The vectors in both graphics are of the same magnitudes and only visually aren't due to my lack of artistic skills, but in this one I have half the enclosed area. This seems more intuitive to me because it's now a shape using the two length projections of $\hat i$ and $\hat j$ instead of using all the enclosed area. Is it convention? Or is there something else to it? Let me know if my question is unclear and I'll try to clarify, but I'm essentially asking why we establish that the enclosed area of the matrix I've stated above forms a square and not a triangle, when I imagine you would need $4$ lengths to create a square, and $2$ to create a triangle. What contradicts my logic is that by that logic the area of a triangle would be $base \ \dot \ \ height$, but I can't explain why beyond that point, as visually it would make sense to me.

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  • $\begingroup$ Are you talking about the determinant ? $\endgroup$ – krirkrirk Jul 5 '17 at 19:41
  • $\begingroup$ Note that base vectors not necessarily are orthogonal and of unit length. They can therefore span some other parallelogram than a square. $\endgroup$ – md2perpe Jul 5 '17 at 20:20
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The basis vectors tell you where in the space you can go. Let us say you want to go one unit up and one unit to the right. In this situation, the shaded triangle area does not cover this point.

Another way to think of it is to take a vector and move it along the path of the other vector. Doing this and shading any point that is touched tells you where you can go (assuming you are limited to the length of the vectors). If you do this with your drawn vectors you will see it shades out a square.

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  • $\begingroup$ Oh, so any superposition, like anywhere along the $i$ vector the $j$ vector can sit from head to tail, creates a square? Because an object can be in any point in that square space? $\endgroup$ – sangstar Jul 6 '17 at 1:41
  • $\begingroup$ Yes. If I were you, I would look into learning some Linear Algebra, and really grab the intuition, so that you can truly appreciate vectors. $\endgroup$ – David Jul 6 '17 at 12:56

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