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Contract the Levi-Civita symbol identity ${\epsilon }_{ijk}{\epsilon }_{i\ell m}={\delta }_{j\ell }{\delta }_{km}-{\delta }_{jm}{\delta }_{k\ell }$. Using Einstein summation convention,This Lagrange identity for the cross product includes zero and parallel vectors without a separate division by their norms.

Use the arc length coordinate oriented in the direction of increasing $t$: $s(t)={\int }_0^{t}2e^{\tau }d\tau =2(e^t-1)$, with $s>-2$. Also ${\mathbf{r}}^{\prime \prime }=e^t(\sqrt{2},-2\cos t,-2\sin t)$, giving $|{\mathbf{r}}^{\prime \prime }{|}^2=6e^{2t}$ and ${\mathbf{r}}^{\prime }\cdot {\mathbf{r}}^{\prime \prime }=4e^{2t}$. The Lagrange identity for the cross product yieldsThe curvature of a space curve is thereforeFor the forward portion $t\ge 0$, this has ordinary nonnegative arc length $s\ge 0$. If instead nonnegative distance is measured backwards from $t=0$, put $s_b=2(1-e^t)$ for $t\le 0$; that branch has $\kappa =1/[\sqrt{2}(2-s_b)]$, $0\le s_b<2$. Choosing orientation avoids confusing the two points at the same unoriented distance.

Expanding a sum of squares gives the $n$-dimensional Lagrange identity for the cross product:The left-hand side is nonnegative, so

Let $\mathbf{r}(u,v)$ parametrize $S$. Then $A\mathbf{r}$ parametrizes its image, with tangent vectors $A{\mathbf{r}}_u$ and $A{\mathbf{r}}_v$. An orthogonal matrix preserves dot products, so the Lagrange identity for the cross product gives, for any $\mathbf{p},\mathbf{q}$,The area element of a surface is therefore the same pointwise before and after applying $A$. Integrating provesThis includes orthogonal transformations with determinant $-1$: reversal of orientation does not affect the Euclidean norm in the surface area formula.

Use the convention $[\mathbf{a},\mathbf{b},\mathbf{c}]=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})$ for the scalar triple product. With $\mathbf{q},\mathbf{w}$ as above, $\mathbf{q}=\mathbf{n}\times \mathbf{w}$. The vector triple product identity givesThe Lagrange identity for the cross product, $|\mathbf{a}\times \mathbf{b}{|}^2=|\mathbf{a}{|}^2|\mathbf{b}{|}^2-(\mathbf{a}\cdot \mathbf{b}{)}^2$, also givesTaking the dot product with the unit vector $\mathbf{n}$ proves