Components of Vector in terms of Direction Cosines
Theorem
Let $\mathbf r$ be a vector quantity embedded in a Cartesian $3$-space.
Let $\mathbf i$, $\mathbf j$ and $\mathbf k$ be the unit vectors in the positive directions of the $x$-axis, $y$-axis and $z$-axis respectively.
Let $\cos \alpha$, $\cos \beta$ and $\cos \gamma$ be the direction cosines of $\mathbf r$ with respect to the $x$-axis, $y$-axis and $z$-axis respectively.
Let $x$, $y$ and $z$ be the components of $\mathbf r$ in the $\mathbf i$, $\mathbf j$ and $\mathbf k$ directions respectively.
Let $r$ denote the magnitude of $\mathbf r$, that is:
- $r := \size {\mathbf r}$
Then:
\(\ds x\) | \(=\) | \(\ds r \cos \alpha\) | ||||||||||||
\(\ds y\) | \(=\) | \(\ds r \cos \beta\) | ||||||||||||
\(\ds z\) | \(=\) | \(\ds r \cos \gamma\) |
Proof
By definition, the direction cosines are the cosines of the angles that $\mathbf r$ makes with the coordinate axes.
By definition of the components of $\mathbf r$:
- $\mathbf r = x \mathbf i + y \mathbf j + z \mathbf k$
Thus:
- $\mathbf r = r \cos \alpha \mathbf i + r \cos \beta \mathbf j + r \cos \gamma \mathbf k$
and the result follows.
$\blacksquare$
Also presented as
This relationship can also be seen presented as:
Then:
\(\ds \cos \alpha\) | \(=\) | \(\ds \dfrac x r\) | ||||||||||||
\(\ds \cos \beta\) | \(=\) | \(\ds \dfrac y r\) | ||||||||||||
\(\ds \cos \gamma\) | \(=\) | \(\ds \dfrac z r\) |
Sources
- 1921: C.E. Weatherburn: Elementary Vector Analysis ... (previous) ... (next): Chapter $\text I$. Addition and Subtraction of Vectors. Centroids: Components of a Vector: $7$. The unit vectors $\mathbf i$, $\mathbf j$, $\mathbf k$
- 1970: George Arfken: Mathematical Methods for Physicists (2nd ed.) ... (previous) ... (next): Chapter $1$ Vector Analysis $1.1$ Definitions, Elementary Approach: $(1.4)$