# Why does the area of two base vectors form a square and not a triangle?

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: 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?: 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.

• Are you talking about the determinant ? – krirkrirk Jul 5 '17 at 19:41
• Note that base vectors not necessarily are orthogonal and of unit length. They can therefore span some other parallelogram than a square. – md2perpe Jul 5 '17 at 20:20

• 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? – sangstar Jul 6 '17 at 1:41