Angle between Vector Quantities in terms of Direction Cosines

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Theorem

Let $\mathbf a$ and $\mathbf b$ be vector quantities embedded in Cartesian $3$-space

Let $\theta$ be the angle between $\mathbf a$ and $\mathbf b$.

Then:

$\cos \theta = \lambda_a \lambda_b + \mu_a \mu_b + \nu_a \nu_b$

where $\lambda_a$, $\mu_a$ and $\nu_a$ are the direction cosines of $\mathbf a$ with respect to the $x$-axis, $y$-axis and $z$-axis respectively, and similarly for $\mathbf b$.


Proof

Let $\mathbf r$ be an arbitrary vector quantity embedded in a Cartesian $3$-space.

From Components of Vector in terms of Direction Cosines:

\(\ds x\) \(=\) \(\ds r \lambda_r\)
\(\ds y\) \(=\) \(\ds r \mu_r\)
\(\ds z\) \(=\) \(\ds r \nu_r\)

where:

$x$, $y$ and $z$ denote the components of $\mathbf r$ in the $\mathbf i$, $\mathbf j$ and $\mathbf k$ directions respectively.
$r$ denotes the magnitude of $\mathbf r$.

Hence:

\(\ds \mathbf a \cdot \mathbf b\) \(=\) \(\ds a b \cos \theta\) Definition of Dot Product
\(\ds \) \(=\) \(\ds \paren {a \lambda_a} \paren {b \lambda_b} + \paren {a \mu_a} \paren {b \mu_b} + \paren {a \nu_a} \paren {b \nu_b}\) Definition of Dot Product
\(\ds \leadsto \ \ \) \(\ds \cos \theta\) \(=\) \(\ds \lambda_a \lambda_b + \mu_a \mu_b + \nu_a \nu_b\) dividing by $a b$

$\blacksquare$


Sources

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