Vector identities

Dot product symmetry.#rvi‑ed

$\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$

Derivation

Using the coordinate expression #rvv-es gives: $\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 = b_1 a_1 + b_2 a_2 + b_3 a_3 = \vec{b} \cdot \vec{a}.$

Dot product vector length.#rvi‑eg

$\vec{a} \cdot \vec{a} = \|a\|^2$

Derivation

Using the coordinate expression #rvv-es gives: $\vec{a} \cdot \vec{a} = a_1 a_1 + a_2 a_2 + a_3 a_3 = \|a\|^2.$

Dot product bi-linearity.#rvi‑ei

$\begin{aligned} \vec{a} \cdot (\vec{b} + \vec{c}) &= \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} \\ (\vec{a} + \vec{b}) \cdot \vec{c} &= \vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{c} \\ \vec{a} \cdot (\beta \vec{b}) &= \beta (\vec{a} \cdot \vec{b}) = (\beta \vec{a}) \cdot \vec{b}\end{aligned}$

Derivation

Using the coordinate expression #rvv-es gives: $\begin{aligned} \vec{a} \cdot (\vec{b} + \vec{c}) &= a_1 (b_1 + c_1) + a_2 (b_2 + c_2) + a_3 (b_3 + c_3) \\ &= (a_1 b_1 + a_2 b_2 + a_3 b_3) + (a_1 c_1 + a_2 c_2 + a_3 c_3) \\ &= \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} \\ (\vec{a} + \vec{b}) \cdot \vec{c} &= (a_1 + b_1) c_1 + (a_2 + b_2) c_2 + (a_3 + b_3) c_3 \\ &= (a_1 c_1 + a_2 c_2 + a_3 c_3) + (b_1 c_1 + a_2 c_2 + a_3 c_3) \\ &= \vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{c} \\ \vec{a} \cdot (\beta \vec{b}) &= a_1 (\beta b_1) + a_2 (\beta b_2) + a_3 (\beta b_3) \\ &= \beta (a_1 b_1 + a_2 b_2 + a_3 b_3) \\ &= \beta (\vec{a} \cdot \vec{b}) \\ &= (\beta a_1) b_1 + (\beta a_2) b_2 + (\beta a_3) b_3 \\ &= (\beta \vec{a}) \cdot \vec{b}. \end{aligned}$

Cross product anti-symmetry.#rvi‑ea

$\begin{aligned} \vec{a} \times \vec{b} = - \vec{b} \times \vec{a}\end{aligned}$

Derivation

Writing the component expression #rvv-ex gives: $\begin{aligned} \vec{a} \times \vec{b} &= (a_2 b_3 - a_3 b_2) \,\hat{\imath} + (a_3 b_1 - a_1 b_3) \,\hat{\jmath} + (a_1 b_2 - a_2 b_1) \,\hat{k} \\ &= -(a_3 b_2 - a_2 b_3) \,\hat{\imath} - (a_1 b_3 - a_3 b_1) \,\hat{\jmath} - (a_2 b_1 - a_1 b_2) \,\hat{k} \\ &= -\vec{b} \times \vec{a}. \end{aligned}$

Cross product self-annihilation.#rvi‑ez

$\begin{aligned} \vec{a} \times \vec{a} = 0 \end{aligned}$

Derivation

From anti-symmetry #rvi-ea we have: $\begin{aligned} \vec{a} \times \vec{a} &= - \vec{a} \times \vec{a} \\ 2 \vec{a} \times \vec{a} &= 0 \\ \vec{a} \times \vec{a} &= 0. \end{aligned}$

Cross product bi-linearity.#rvi‑eb

$\begin{aligned} \vec{a} \times (\vec{b} + \vec{c}) &= \vec{a} \times \vec{b} + \vec{a} \times \vec{c} \\ (\vec{a} + \vec{b}) \times \vec{c} &= \vec{a} \times \vec{c} + \vec{b} \times \vec{c} \\ \vec{a} \times (\beta \vec{b}) &= \beta (\vec{a} \times \vec{b}) = (\beta \vec{a}) \times \vec{b} \end{aligned}$

Derivation

Writing the component expression #rvv-ex for the first equation gives: $\begin{aligned} \vec{a} \times (\vec{b} + \vec{c}) &= (a_2 (b_3 + c_3) - a_3 (b_2 + c_2)) \,\hat{\imath} \\ &\quad + (a_3 (b_1 + c_1) - a_1 (b_3 + c_3)) \,\hat{\jmath} \\ &\quad + (a_1 (b_2 + c_2) - a_2 (b_1 + c_1)) \,\hat{k} \\ &= \Big((a_2 b_3 - a_3 b_2) \,\hat{\imath} + (a_3 b_1 - a_1 b_3) \,\hat{\jmath} + (a_1 b_2 - a_2 b_1) \,\hat{k} \Big) \\ &\quad + \Big((a_2 c_3 - a_3 c_2) \,\hat{\imath} + (a_3 c_1 - a_1 c_3) \,\hat{\jmath} + (a_1 c_2 - a_2 c_1) \,\hat{k} \Big) \\ &= \vec{a} \times \vec{b} + \vec{a} \times \vec{c}. \\ \end{aligned}$ The second equation follows similarly, and for the third equation we have: $\begin{aligned} \vec{a} \times (\beta \vec{b}) &= (a_2 (\beta b_3) - a_3 (\beta b_2)) \,\hat{\imath} + (a_3 (\beta b_1) - a_1 (\beta b_3)) \,\hat{\jmath} + (a_1 (\beta b_2) - a_2 (\beta b_1)) \,\hat{k} \\ &= \beta \Big( (a_2 b_3 - a_3 b_2) \,\hat{\imath} + (a_3 b_1 - a_1 b_3) \,\hat{\jmath} + (a_1 b_2 - a_2 b_1) \,\hat{k} \Big) \\ &= \beta (\vec{a} \times \vec{b}). \end{aligned}$ The last part of the third equation can be seen with a similar derivation.

Scalar triple product formula.#rvi‑es

$\begin{aligned} \vec{a} \cdot (\vec{b} \times \vec{c}) = \vec{b} \cdot (\vec{c} \times \vec{a}) = \vec{c} \cdot (\vec{a} \times \vec{b})\end{aligned}$

Derivation

Writing the first two expressions out in components, we can check that they give the same result:

$\begin{aligned} \vec{a} \cdot (\vec{b} \times \vec{c}) &= (a_1 \,\hat{\imath} + a_2 \,\hat{\jmath} + a_3 \,\hat{k}) \\ &\qquad \cdot \big( (b_1 \,\hat{\imath} + b_2 \,\hat{\jmath} + b_3 \,\hat{k}) \times (c_1 \,\hat{\imath} + c_2 \,\hat{\jmath} + c_3 \,\hat{k}) \big) \\ &= (a_1 \,\hat{\imath} + a_2 \,\hat{\jmath} + a_3 \,\hat{k}) \\ &\qquad \cdot \big( (b_2 c_3 - b_3 c_2) \,\hat{\imath} + (b_3 c_1 - b_1 c_3) \,\hat{\jmath} + (b_1 c_2 - b_2 c_1) \,\hat{k} \big) \\ &= a_1 b_2 c_3 - a_1 b_3 c_2 + a_2 b_3 c_1 - a_2 b_1 c_3 + a_3 b_1 c_2 - a_3 b_2 c_1 \\ \vec{b} \cdot (\vec{c} \times \vec{a}) &= (b_1 \,\hat{\imath} + b_2 \,\hat{\jmath} + b_3 \,\hat{k}) \\ &\qquad \cdot \big( (c_1 \,\hat{\imath} + c_2 \,\hat{\jmath} + c_3 \,\hat{k}) \times (a_1 \,\hat{\imath} + a_2 \,\hat{\jmath} + a_3 \,\hat{k}) \big) \\ &= (b_1 \,\hat{\imath} + b_2 \,\hat{\jmath} + b_3 \,\hat{k}) \\ &\qquad \cdot \big( (c_2 a_3 - c_3 a_2) \,\hat{\imath} + (c_3 a_1 - c_1 a_3) \,\hat{\jmath} + (c_1 a_2 - c_2 a_1) \,\hat{k} \big) \\ &= b_1 c_2 a_3 - b_1 c_3 a_2 + b_2 c_3 a_1 - b_2 c_1 a_3 + b_3 c_1 a_2 - b_3 c_2 a_1.\end{aligned}$

The third expression is also the same.

Vector triple product expansion.#rvi‑ev

$\begin{aligned} \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}\end{aligned}$

Derivation

In components we have:

$\begin{aligned} \vec{a} \times (\vec{b} \times \vec{c}) &= (a_1 \,\hat{\imath} + a_2 \,\hat{\jmath} + a_3 \,\hat{k}) \\ &\qquad \times \Big( (b_1 \,\hat{\imath} + b_2 \,\hat{\jmath} + b_3 \,\hat{k}) \times (c_1 \,\hat{\imath} + c_2 \,\hat{\jmath} + c_3 \,\hat{k}) \Big) \\ &= (a_1 \,\hat{\imath} + a_2 \,\hat{\jmath} + a_3 \,\hat{k}) \\ &\qquad \times \Big( (b_2 c_3 - b_3 c_2) \,\hat{\imath} + (b_3 c_1 - b_1 c_3) \,\hat{\jmath} + (b_1 c_2 - b_2 c_1) \,\hat{k} \Big) \\ &= \Big(a_2 (b_1 c_2 - b_2 c_1) - a_3 (b_3 c_1 - b_1 c_3)\Big) \,\hat{\imath} \\ &\quad+ \Big(a_3 (b_2 c_3 - b_3 c_2) - a_1 (b_1 c_2 - b_2 c_1)\Big) \,\hat{\jmath} \\ &\quad+ \Big(a_1 (b_3 c_1 - b_1 c_3) - a_2 (b_2 c_3 - b_3 c_2)\Big) \,\hat{k} \\ &= (a_2 b_1 c_2 - a_2 b_2 c_1 - a_3 b_3 c_1 + a_3 b_1 c_3) \,\hat{\imath} \\ &\quad+ (a_3 b_2 c_3 - a_3 b_3 c_2 - a_1 b_1 c_2 + a_1 b_2 c_1) \,\hat{\jmath} \\ &\quad+ (a_1 b_3 c_1 - a_1 b_1 c_3 - a_2 b_2 c_3 + a_2 b_3 c_2) \,\hat{k} \\ &= (a_1 b_1 c_1 + a_2 b_1 c_2 + a_3 b_1 c_3 - a_1 b_1 c_1 - a_2 b_2 c_1 - a_3 b_3 c_1) \,\hat{\imath} \\ &\quad+ (a_1 b_2 c_1 + a_2 b_2 c_2 + a_3 b_2 c_3 - a_1 b_1 c_2 - a_2 b_2 c_2 - a_3 b_3 c_2) \,\hat{\jmath} \\ &\quad+ (a_1 b_3 c_1 + a_2 b_3 c_2 + a_3 b_3 c_3 - a_1 b_1 c_3 - a_2 b_2 c_3 - a_3 b_3 c_3) \,\hat{k} \\ &= (a_1 b_1 + a_2 c_2 + a_3 c_3) b_1 \,\hat{\imath} + (a_1 c_1 + a_2 c_2 + a_3 c_3) b_2 \,\hat{\jmath} \\ &\qquad + (a_1 c_1 + a_2 c_2 + a_3 c_3) b_3 \,\hat{k} \\ &\quad- (a_1 b_1 + a_2 b_2 + a_3 b_3) c_1 \,\hat{\imath} - (a_1 b_1 + a_2 b_2 + a_3 b_3) c_2 \,\hat{\jmath} \\ &\qquad - (a_1 b_1 + a_2 b_2 + a_3 b_3) c_3 \,\hat{k} \\ &= (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}.\end{aligned}$

Cross product orthogonality.#rvi‑eo

$\begin{aligned} \vec{a} \times \vec{b} \text{ is orthogonal to both } \vec{a} \text{ and } \vec{b}\end{aligned}$

Derivation

This follows immediately from the scalar triple product formula #rvi-es:

$\begin{aligned} \vec{a} \cdot (\vec{a} \times \vec{b}) = \vec{b} \cdot (\vec{a} \times \vec{a}) = 0,\end{aligned}$

and similarly for $\vec{b}$ .

Binet-Cauchy identity.#rvi‑eb

$\begin{aligned} (\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{d}) - (\vec{a} \cdot \vec{d})(\vec{b} \cdot \vec{c})\end{aligned}$

Derivation

From the scalar triple product formula #rvi-es and the vector triple product expansion #rvi-ev:

$\begin{aligned} (\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) &= \vec{c} \cdot \big( \vec{d} \times (\vec{a} \times \vec{b}) \big) \\ &= \vec{c} \cdot \big( (\vec{d} \cdot \vec{b}) \vec{a} - (\vec{d} \cdot \vec{a}) \vec{b} \big) \\ &= (\vec{d} \cdot \vec{b}) (\vec{c} \cdot \vec{a}) - (\vec{d} \cdot \vec{a}) (\vec{c} \cdot \vec{b}).\end{aligned}$

Lagrange's identity.#rvi‑el

$\begin{aligned} \| \vec{a} \times \vec{b} \|^2 = \|\vec{a}\|^2 \|\vec{b}\|^2 - (\vec{a} \cdot \vec{b})^2\end{aligned}$

Derivation

Cross product length.#rvi‑ee

$\begin{aligned} \| \vec{a} \times \vec{b} \| = \|\vec{a}\| \|\vec{b}\| \sin\theta\end{aligned}$

Derivation

From Lagrange’s identity #rvi-el:

$\begin{aligned} \| \vec{a} \times \vec{b} \|^2 &= \|\vec{a}\|^2 \|\vec{b}\|^2 - (\vec{a} \cdot \vec{b})^2 \\ &= \|\vec{a}\|^2 \|\vec{b}\|^2 - (\|\vec{a}\| \|\vec{b}\| \cos\theta)^2 \\ &= \|\vec{a}\|^2 \|\vec{b}\|^2 (1 - \cos^2\theta) \\ &= \|\vec{a}\|^2 \|\vec{b}\|^2 \sin^2\theta.\end{aligned}$

Jacobi's identity.#rvi‑ej

$\begin{aligned} \vec{a} \times (\vec{b} \times \vec{c}) + \vec{b} \times (\vec{c} \times \vec{a}) + \vec{c} \times (\vec{a} \times \vec{b}) = 0\end{aligned}$

Derivation

Using the vector triple product expansion #rvi-ev:

$\begin{aligned} &\vec{a} \times (\vec{b} \times \vec{c}) + \vec{b} \times (\vec{c} \times \vec{a}) + \vec{c} \times (\vec{a} \times \vec{b}) \\ &\qquad= (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} + (\vec{b} \cdot \vec{a}) \vec{c} - (\vec{b} \cdot \vec{c}) \vec{a} + (\vec{c} \cdot \vec{b}) \vec{a} - (\vec{c} \cdot \vec{a}) \vec{b} \\ &\qquad= 0.\end{aligned}$

Vector quadruple product expansion.#rvi‑eq

$\begin{aligned} (\vec{a} \times \vec{b}) \times (\vec{a} \times \vec{c}) = \big( \vec{a} \cdot (\vec{b} \times \vec{c}) \big) \vec{a}\end{aligned}$

Derivation

Take $\vec{d} = (\vec{a} \times \vec{b}) \times (\vec{a} \times \vec{c})$ . Then $\vec{d}$ is in the $\vec{a},\vec{b}$ plane and in the $\vec{a},\vec{c}$ plane, so it is a scalar multiple of $\vec{a}$ . We use the scalar triple product formula #rvi-es and the vector triple product expansion #rvi-ev to compute:

$\begin{aligned} \vec{d} \cdot \vec{a} &= \vec{a} \cdot \big( (\vec{a} \times \vec{b}) \times (\vec{a} \times \vec{c}) \big) \\ &= (\vec{a} \times \vec{c}) \cdot \big( \vec{a} \times (\vec{a} \times \vec{b}) \big) \\ &= (\vec{a} \times \vec{c}) \cdot \big( (\vec{a} \cdot \vec{b}) \vec{a} - (\vec{a} \cdot \vec{a}) \vec{b} \big) \\ &= - (\vec{a} \cdot \vec{a}) \vec{b} \cdot (\vec{a} \times \vec{c}) \\ &= (\vec{a} \cdot \vec{a}) \big( \vec{a} \cdot (\vec{b} \times \vec{c}) \big).\end{aligned}$

Then

$\begin{aligned} \vec{d} = \operatorname{Proj}(\vec{d}, \vec{a}) = \left(\frac{\vec{d} \cdot \vec{a}}{\vec{a} \cdot \vec{a}}\right) \vec{a} = \big( \vec{a} \cdot (\vec{b} \times \vec{c}) \big) \vec{a}.\end{aligned}$

TAM 212: Introductory Dynamics

Vector identities #rvi