Lagrange's Formula/Corollary
Jump to navigation
Jump to search
Theorem
Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be vectors in a vector space $\mathbf V$ of $3$ dimensions.
Then:
- $\paren {\mathbf a \times \mathbf b} \times \mathbf c = \paren {\mathbf a \cdot \mathbf c} \mathbf b - \paren {\mathbf b \cdot \mathbf c} \mathbf a$
Proof
\(\ds \mathbf c \times \paren {\mathbf a \times \mathbf b}\) | \(=\) | \(\ds \paren {\mathbf c \cdot \mathbf b} \mathbf a - \paren {\mathbf c \cdot \mathbf a} \mathbf b\) | Lagrange's Formula | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {\mathbf a \times \mathbf b} \times \mathbf c\) | \(=\) | \(\ds -\paren {\paren {\mathbf c \cdot \mathbf b} \mathbf a - \paren {\mathbf c \cdot \mathbf a} \mathbf b}\) | Vector Cross Product is Anticommutative | ||||||||||
\(\ds \) | \(=\) | \(\ds -\paren {\mathbf b \cdot \mathbf c} \mathbf a + \paren {\mathbf a \cdot \mathbf c} \mathbf b\) | Dot Product Operator is Commutative | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\mathbf a \cdot \mathbf c} \mathbf b - \paren {\mathbf b \cdot \mathbf c} \mathbf a\) | Real Addition is Commutative |
$\blacksquare$
Source of Name
This entry was named for Joseph Louis Lagrange.
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 22$: Miscellaneous Formulas involving Dot and Cross Products: $22.19$