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 $\mathbf k^2$.

$\blacksquare$

Also see

• Dot Product of Elements of Standard Ordered Basis

Sources

• 1951: B. Hague: An Introduction to Vector Analysis (5th ed.) ... (previous) ... (next): Chapter $\text {II}$: The Products of Vectors: $2$. The Scalar Product: $(2.6)$
• 1957: D.E. Rutherford: Vector Methods (9th ed.) ... (previous) ... (next): Chapter $\text I$: Vector Algebra: $\S 2$. $(3)$
• 1970: George Arfken: Mathematical Methods for Physicists (2nd ed.) ... (previous) ... (next): Chapter $1$ Vector Analysis $1.3$ Scalar or Dot Product: $(1.22 a)$