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
- 1957: D.E. Rutherford: Vector Methods (9th ed.) ... (previous) ... (next): Chapter $\text I$: Vector Algebra: $\S 2$. $(5)$
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): direction angles
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): direction angles