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Expand the vector triple product in Cartesian components. Its first component iswhere adding $a_x{b}_x{c}_x$ to both products introduces cancelling terms. Likewise the other two components areTogether these proveNow use the cyclic property of the scalar triple product and the just-proved identity:This proves Lagrange's identity in the requested form,

Let $\Delta =[\mathbf{a},\mathbf{b},\mathbf{c}]\ne 0$. These three vectors are linearly independent, so write $\mathbf{d}=\alpha \mathbf{a}+\beta \mathbf{b}+\gamma \mathbf{c}$. Taking the dot product with $\mathbf{b}\times \mathbf{c}$ gives $\alpha =[\mathbf{b},\mathbf{c},\mathbf{d}]/\Delta$. Likewise, the dot products with $\mathbf{c}\times \mathbf{a}$ and $\mathbf{a}\times \mathbf{b}$ giveHere the cyclic invariance and sign change under an interchange of the scalar triple product fix every sign. Substituting the three coefficients provesThis is precisely coordinate extraction using the reciprocal basis $(\mathbf{b}\times \mathbf{c},\mathbf{c}\times \mathbf{a},\mathbf{a}\times \mathbf{b})/\Delta$.

For $\mathbf{b}=0$, a nonzero coefficient vector $(x,y,z)$ with $x{\mathbf{c}}_1+y{\mathbf{c}}_2+z{\mathbf{c}}_3=0$ is precisely a linear dependence of the three columns. Three vectors in three dimensions are linearly dependent exactly when the determinant of their column matrix, or their scalar triple product, is zero. ThusFor this square homogeneous linear system, the condition is also sufficient.

At $\mu =-1$, ${\mathbf{c}}_2=-{\mathbf{c}}_1$, while ${\mathbf{c}}_1$ and ${\mathbf{c}}_3$ are independent. Their span is a plane. A solution requires $\mathbf{b}$ to lie in that plane, so every scalar triple product formed from two columns and $\mathbf{b}$ must vanish. The product with ${\mathbf{c}}_1,{\mathbf{c}}_2$ vanishes identically; the other two give the plane condition.

Explicitly, ${\mathbf{c}}_1\times {\mathbf{c}}_3=(-7,1,3)$, soThus $\lambda =7$ is necessary and, since the two columns span the plane, sufficient. The equations then give $x-y=1$ and $z=2$, soThe solution is an affine line parallel to the homogeneous null space from part (c), obtained by intersecting the compatible constraint planes. Every point on it is nonzero because $z=2$.

The position vectors end at the midpoints of $BC$, $CA$ and $AB$, respectively. The drawing places $E$, $F$ and $G$ on those edges and shows the arrows from the origin. The volume of their parallelepiped is the absolute scalar triple product:Equivalently, this is the absolute determinant of the matrix whose columns are the three vectors. Since it is nonzero, the vectors are linearly independent. Three linearly independent vectors in the three-dimensional vector space ${\mathbb{R}}^3$ form a basis.

The cross product of the direction vectors is ${\mathbf{n}}_1\times {\mathbf{n}}_2=(1,-1,0)$. It is nonzero, so a common perpendicular direction exists. Projecting ${\mathbf{a}}_2-{\mathbf{a}}_1=(-1,0,1)$ onto its unit vector gives the distance between skew lines:ThusThe numerator is a scalar triple product; movement along either line does not change this perpendicular projection.

Three vectors form a basis of three-dimensional space exactly when their scalar triple product is nonzero:It is the determinant of the matrix with these columns, so this condition is equivalent to linear independence and spanning the whole space.

Take the dot product of the expansion of $\mathbf{y}$ with $\mathbf{v}\times \mathbf{w}$. The terms containing $\mathbf{v}$ and $\mathbf{w}$ vanish, leaving $\mathbf{y}\cdot (\mathbf{v}\times \mathbf{w})=\alpha \Delta$. Cycling the same argument yieldsThe three cross-product numerators use the reciprocal basis.

If the three vectors lie in one plane through the origin and $\mathbf{u},\mathbf{v}$ are independent, their cross product is a normal vector to that plane, so its dot product with $\mathbf{w}$ is zero. If $\mathbf{u},\mathbf{v}$ are dependent, their cross product is already zero. Thus coplanarity always forces the scalar triple product to vanish.

Taking the contrapositive proves that a nonzero scalar triple product rules out coplanarity. Geometrically its absolute value is the nonzero volume of the spanned parallelepiped.

Use displacements from $P_1$, since coplanarity of points concerns an affine plane that need not contain the origin. The three displacement vectors areTheir cross product is $\mathbf{u}\times \mathbf{v}=(-1,1,-1{)}^T$ and its dot product with $\mathbf{w}$ is $-1$. Since the scalar triple product is nonzero,

The volume of the parallelepiped is the absolute value of the scalar triple product:

Equating vector components givesThe third component first forces $\gamma =0$, then the second forces $\beta =0$, and the first gives $\alpha =0$. ThusEquivalently, the nonzero scalar triple product proves that these three vectors are linearly independent.

Using Cartesian coordinates, the Levi-Civita symbol and the Einstein summation convention, the constant-vector assumption givesRecognizing the divergence and directional derivative provesHere the contractions use the Kronecker delta.

For an oriented piecewise smooth surface $S$ with boundary $C$, Stokes theorem stateswhere the orientations obey the right-hand rule and $\mathbf{Q}$ is continuously differentiable near $S$. Apply it to $\mathbf{Q}=\mathbf{F}\times \mathbf{a}$. The scalar triple product givesThe last integrand is $\mathbf{a}\cdot [(d\mathbf{S}\times \nabla )\times \mathbf{F}]$, as the component formulashows. Since $\mathbf{a}$ is arbitrary, the cross-product form of Stokes theorem follows:The derivatives in the surface expression act on $\mathbf{F}$, not on the surface element.

For the specified square, the path is clockwise when viewed from positive $z$, so the compatible unit normal is $-{\mathbf{e}}_z$. On the first and fourth edges, $d\mathbf{l}$ and $\mathbf{r}$ are parallel, giving zero cross product. On the second edge, $y=1$ and $x$ runs from $0$ to $1$, so $d\mathbf{l}\times \mathbf{r}={\mathbf{e}}_{z}dx$ contributes ${\mathbf{e}}_z$. On the third edge, $x=1$ and $y$ runs from $1$ to $0$, so the contribution is $-{\mathbf{e}}_z{\int }_1^{0}dy={\mathbf{e}}_z$. Thus the line integral is $2{\mathbf{e}}_z$.

For an independent surface integral, ${\partial }_i{F}_j={\delta }_{ij}$ and $\nabla \cdot \mathbf{F}=3$ for the position vector field. Its integrand is therefore $d\mathbf{S}-3d\mathbf{S}=-2d\mathbf{S}$. The square has unit area and downward orientation, giving

Take any constant vector $\mathbf{a}$. Its curl is zero, so part (a) and the divergence theorem givewhere $\mathbf{n}$ is the outward unit normal vector. The scalar triple product identity rewrites the last integrand as $\mathbf{a}\cdot (\mathbf{F}\times \mathbf{n})$. Since $\mathbf{a}$ is arbitrary, equality of these scalar products implies the vector identity

Dividing the coordinate tangents by their scale factors of orthogonal coordinates gives an orthonormal basisEvaluate the Cartesian cross products componentwise:ConsequentlyThe ordered cylindrical basis is right-handed: its scalar triple product ${\mathbf{e}}_{\rho }\cdot ({\mathbf{e}}_{\varphi }\times {\mathbf{e}}_z)$ equals $+1$.

The Jacobian determinant is the oriented scalar triple product of the three coordinate basis vectors:Here ${\mathbf{h}}_u\times {\mathbf{h}}_v=(u^2+v^2)(-\sin \varphi ,\cos \varphi ,0)$, soThe sign is nonnegative on the stated range, and the absolute Jacobian determinant also equals $h_u{h}_v{h}_{\varphi }$ as expected for orthogonal coordinates.

The divergence of a cross product and $\nabla \times \mathbf{a}=0$ give $\nabla \cdot (\mathbf{a}\times \mathbf{A})=-\mathbf{a}\cdot (\nabla \times \mathbf{A})$. On the boundary the scalar triple product gives $(\mathbf{a}\times \mathbf{A})\cdot \hat{\mathbf{n}}=-\mathbf{a}\cdot (\hat{\mathbf{n}}\times \mathbf{A})$. Thus the divergence theorem givesBecause $\mathbf{a}$ is arbitrary, the curl identity follows:

Use the magnetic-coupling convention of the supplied Hamiltonian; electromagnetic unit factors are absorbed into the stated $e$ and $B$. Let $\alpha =m_1/M$ and $\beta =m_2/M$, soThe kinetic energy separates into $M{\dot{\mathbf{R}}}^2/2+\mu {\dot{\mathbf{r}}}^2/2$. The symmetric gauge coupling ${\sum }_i{e}_i{\dot{\mathbf{r}}}_i\cdot (\mathbf{B}\times {\mathbf{r}}_i)/2$ becomesThe scalar triple product identity givesDiscarding $-e/2$ times that total derivative, which changes endpoint terms and canonical momenta but not the trajectories, leaves

Suppose $\mathbf{a}\ne 0$ and $\mathbf{a}\cdot \mathbf{b}=0$. The line $\mathbf{r}\times \mathbf{a}=\mathbf{b}$ has parametrization $\mathbf{r}={\mathbf{r}}_0+t\mathbf{a}$, where ${\mathbf{r}}_0=(\mathbf{a}\times \mathbf{b})/|\mathbf{a}{|}^2$. For a point $\mathbf{y}$ outside the line, a normal vector of the unique plane containing both is $\mathbf{N}=\mathbf{a}\times (\mathbf{y}-{\mathbf{r}}_0)=\mathbf{b}+\mathbf{a}\times \mathbf{y}$. Thus its equation is the one in the title. The vector triple product verifies ${\mathbf{r}}_0\times \mathbf{a}=\mathbf{b}$, and cyclic invariance of the scalar triple product verifies that every point of the line satisfies the equation. The excluded case $\mathbf{N}=0$ is exactly $\mathbf{y}$ lying on the line, when the plane is not unique.

A pseudoscalar is invariant under proper rotations but changes sign under a spatial reflection. In three dimensions the scalar triple product is a pseudoscalar.

For a basis $e_1,e_2,e_3$ with scalar triple product $\Delta$, the reciprocal vectors are cyclic cross products divided by $\Delta$ and satisfy $f_i\cdot e_j={\delta }_{ij}$. This construction is used for the basis of a reciprocal lattice.