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For the numerical vectors, $\mathbf{b}\times \mathbf{c}=(-2,1,-2)$ and $D=(-1,1,0)\cdot (-2,1,-2)=3\ne 0$, proving noncoplanarity. The reciprocal basis isReciprocal dot products extract the coefficients of any vector in the original basis. Here $\mathbf{A}\cdot \mathbf{d}=-1/3$, $\mathbf{B}\cdot \mathbf{d}=2/3$, and $\mathbf{C}\cdot \mathbf{d}=5/3$, soSubstitution in Cartesian components recovers $(2,1,-1)$.

Set $D=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})>0$. Since these vectors form a basis, write $\mathbf{r}=\alpha \mathbf{a}+\beta \mathbf{b}+\gamma \mathbf{c}$. The reciprocal basis givesA point is in the closed tetrahedron precisely when it is a convex combination of the four vertices, with weights $1-\alpha -\beta -\gamma ,\alpha ,\beta ,\gamma$. Thus the necessary and sufficient conditions areThese four half-spaces are the four inward sides of its faces. This is tetrahedron membership via scalar triple products; making all inequalities strict describes the interior.

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 $(\mathbf{b}\times \mathbf{c},\mathbf{c}\times \mathbf{a},\mathbf{a}\times \mathbf{b})/\Delta$.

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.

In a real finite-dimensional inner product space, let two bases satisfy $⟨u_i,w_j⟩={\delta }_{ij}$. Taking inner products with $u_i$ in $v={\sum }_j{b}_j{w}_j$ gives $b_i=⟨u_i,v⟩$. If also $v={\sum }_j{a}_j{u}_j$, then $b_i={\sum }_j⟨u_i,u_j⟩a_j$. Thus the Gram matrix $G_U$ converts the original coordinates into reciprocal basis coordinates, and $G_W=G_U^{-1}$ converts back. With Euclidean Cartesian column basis matrices $U,W$, reciprocity is $U^{T}W=I$, hence $W=U^{-T}$ and $W^{-1}U=U^{T}U=G_U$. This is why dot products with nonorthogonal basis vectors are coordinates in the reciprocal basis, rather than coordinates in the original basis.

For a three-dimensional basis with nonzero scalar triple product $\Delta$, its reciprocal basis satisfies ${\hat{e}}_2\times {\hat{e}}_3=e_1/\Delta$ and the cyclic analogues. Consequently its scalar triple product is $1/\Delta$, and taking the reciprocal basis again recovers the original basis. The coordinates of $V$ in the original basis are $V\cdot {\hat{e}}_i$; in the reciprocal basis they are $V\cdot e_i$.

Let $D=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})>0$. The reciprocal basis gives the coordinates of any position vector asThe point lies in the closed tetrahedron with vertices $0,\mathbf{a},\mathbf{b},\mathbf{c}$ exactly when $\alpha ,\beta ,\gamma \ge 0$ and $\alpha +\beta +\gamma \le 1$. Indeed, these conditions express it as a convex combination of the four vertices, with fourth weight $1-\alpha -\beta -\gamma$. They also describe the intersection of the four inward half-spaces bounded by its faces. Strict inequalities characterize its interior.