Self-Product of Standard Ordered Basis Element equals 1

Theorem
Let $\tuple {\mathbf i, \mathbf j, \mathbf k}$ be the standard ordered basis of Cartesian $3$-space $S$.

Then:
 * $\mathbf i^2 = \mathbf j^2 = \mathbf k^2 = 1$

where $\mathbf i^2$ and so on denotes the square of a vector quantity:
 * $\mathbf i^2 := \mathbf i \cdot \mathbf i$

Proof
By definition, the Cartesian $3$-space is a frame of reference consisting of a rectangular coordinate system.

By definition of Component of Vector in $3$-space, the vectors $\mathbf i$, $\mathbf j$ and $\mathbf k$ are the unit vectors in the direction of the $x$-axis, $y$-axis and $z$-axis respectively.

Hence $\mathbf i^2$ is the square of a unit vector:


 * $\mathbf i^2 = \norm {\mathbf i}^2 = 1^2 = 1$

and the same for $\mathbf j^2$ and $\mathbf k^2$.