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The vector triple-product identity in $R^{3}$

Statement

For all $a, b, c \in R^{3}$ , the Euclidean cross product satisfies

$a \times (b \times c) = b ⟨ a, c ⟩ - c ⟨ a, b ⟩ .$

Statement references

The vector triple-product identity and its componentwise proof: Wilkins, Notes on Vector Algebra and Spherical Trigometry, §5.6, Proposition 5.6, PDF pp. 10–11.

Cross-product determinant convention and related cross-product exercises: Taylor, Introduction to Analysis in Several Variables, §1.4, “Exercises on the cross product,” printed pp. 38–39 (PDF pp. 49–50).

The cross-product definition: The cross product in $R^{3}$ .

The Euclidean inner-product convention: The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ .

Facts & Assumptions

Given: Vectors $a, b, c \in R^{3}$ with the standard Euclidean coordinates.

[F1]

For $u = (u_{1}, u_{2}, u_{3})$ and $v = (v_{1}, v_{2}, v_{3})$ , $u \times v = (u_{2} v_{3} - u_{3} v_{2}, u_{3} v_{1} - u_{1} v_{3}, u_{1} v_{2} - u_{2} v_{1})$ (The cross product in $R^{3}$ ).

[F2]

$⟨ u, v ⟩ = u_{1} v_{1} + u_{2} v_{2} + u_{3} v_{3}$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

Proof

1.1givenF1algebra

Applying [F1] first to $b \times c$ and then to $a \times (b \times c)$ gives the three coordinates $(a_{2} (b_{1} c_{2} - b_{2} c_{1}) - a_{3} (b_{3} c_{1} - b_{1} c_{3}), a_{3} (b_{2} c_{3} - b_{3} c_{2}) - a_{1} (b_{1} c_{2} - b_{2} c_{1}), a_{1} (b_{3} c_{1} - b_{1} c_{3}) - a_{2} (b_{2} c_{3} - b_{3} c_{2})) .$

2.1givenF2step 1.1algebra∎

Expanding the first coordinate in step 1.1 gives $a_{2} b_{1} c_{2} - a_{2} b_{2} c_{1} - a_{3} b_{3} c_{1} + a_{3} b_{1} c_{3} = b_{1} (a_{2} c_{2} + a_{3} c_{3}) - c_{1} (a_{2} b_{2} + a_{3} b_{3})$ . Adding the zero term $b_{1} a_{1} c_{1} - c_{1} a_{1} b_{1} = 0$ gives $b_{1} (a_{1} c_{1} + a_{2} c_{2} + a_{3} c_{3}) - c_{1} (a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3})$ . Expanding the second coordinate gives $a_{3} b_{2} c_{3} - a_{3} b_{3} c_{2} - a_{1} b_{1} c_{2} + a_{1} b_{2} c_{1} = b_{2} (a_{1} c_{1} + a_{3} c_{3}) - c_{2} (a_{1} b_{1} + a_{3} b_{3})$ ; adding $b_{2} a_{2} c_{2} - c_{2} a_{2} b_{2} = 0$ gives $b_{2} (a_{1} c_{1} + a_{2} c_{2} + a_{3} c_{3}) - c_{2} (a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3})$ . Expanding the third coordinate gives $a_{1} b_{3} c_{1} - a_{1} b_{1} c_{3} - a_{2} b_{2} c_{3} + a_{2} b_{3} c_{2} = b_{3} (a_{1} c_{1} + a_{2} c_{2}) - c_{3} (a_{1} b_{1} + a_{2} b_{2})$ ; adding $b_{3} a_{3} c_{3} - c_{3} a_{3} b_{3} = 0$ gives $b_{3} (a_{1} c_{1} + a_{2} c_{2} + a_{3} c_{3}) - c_{3} (a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3})$ . By [F2] the three results are the coordinates of $b ⟨ a, c ⟩ - c ⟨ a, b ⟩$ .

The vector triple-product identity in $R^{3}$

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

The vector triple-product identity in R3R3

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

The vector triple-product identity in $R^{3}$