Components of Vector in terms of Direction Cosines

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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

Vector-Components-in-3-Space.png

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

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