Dot Product of Elements of Standard Ordered Basis
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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.
$\blacksquare$
Also see
Sources
- 1951: B. Hague: An Introduction to Vector Analysis (5th ed.) ... (previous) ... (next): Chapter $\text {II}$: The Products of Vectors: $2$. The Scalar Product: $(2.5)$
- 1957: D.E. Rutherford: Vector Methods (9th ed.) ... (previous) ... (next): Chapter $\text I$: Vector Algebra: $\S 2$. $(2)$
- 1970: George Arfken: Mathematical Methods for Physicists (2nd ed.) ... (previous) ... (next): Chapter $1$ Vector Analysis $1.3$ Scalar or Dot Product: $(1.22 b)$