Dot Product of Elements of Standard Ordered Basis

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

Then:
 * $\mathbf i \cdot \mathbf j = \mathbf j \cdot \mathbf k = \mathbf k \cdot \mathbf i = 0$

where $\cdot$ denotes the dot product.

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

By definition of rectangular coordinate system, the coordinate axes are perpendicular to each other.

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:
 * $\mathbf i$ is perpendicular to $\mathbf j$
 * $\mathbf j$ is perpendicular to $\mathbf k$
 * $\mathbf k$ is perpendicular to $\mathbf i$

The result follows from Dot Product of Perpendicular Vectors.

Also see

 * Self-Product of Standard Ordered Basis Element equals 1