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Points are coplanar when they belong to a common plane. For three vectors based at the origin of three-dimensional Euclidean space, this means their span has dimension at most two. Equivalently their scalar triple product, and hence the determinant with those vectors as columns, is zero. This includes dependent or zero vectors.

The cross product of two cross products sharing their first vector is parallel to that vector, with scalar factor equal to the scalar triple product. In suffix notation, use ${ϵ}_{ijk}{ϵ}_{kpq}={\delta }_{ip}{\delta }_{jq}-{\delta }_{iq}{\delta }_{jp}$. The component of the double cross product becomes $A_i{ϵ}_{q\ell m}{C}_q{A}_{\ell }{B}_m-C_i{ϵ}_{p\ell m}{A}_p{A}_{\ell }{B}_m$. The second term vanishes by antisymmetry of the Levi-Civita symbol; the first equals $A_i(A\cdot (B\times C))$, proving the identity even when the vectors are dependent.

The determinant assigns a scalar to a square matrix. For a linear map on the real plane, it measures signed area scaling: its absolute value tells how much area grows, and its sign records whether the orientation of a vector space is preserved or reversed. In three dimensions, the same idea measures signed volume. Over a field, a square matrix is an invertible matrix exactly when its determinant is nonzero. These are two views of one phenomenon: losing an independent direction collapses full-dimensional volume and makes the original input impossible to recover uniquely.^[ref]

Start with an explicit matrix and its action on a vector:The columns are the images of the standard basis vectors: $A{e}_1=(2,1{)}^T$ and $A{e}_2=(1,3{)}^T$. Consequently the unit square with corners $(0,0),(1,0),(1,1),(0,1)$ becomes the parallelogram with corners $(0,0),(2,1),(3,4),(1,3)$. Its area is five. Every small patch of this square undergoes the same linear map, so the factor describes the whole deformation, rather than just one specially chosen boundary. The matrix mixes horizontal and vertical components, but its determinant compresses that combined effect into one area factor.^[ref]

Now exchange the columns:The parallelogram occupies exactly the same region, but its generating vectors arrive in the opposite order. Moving from the first column to the second now turns clockwise rather than counterclockwise. Ordinary area remains five; the oriented volume in two dimensions changes sign. A negative determinant therefore signals reversed orientation of a vector space, not a negative amount of material. Keeping the sign makes successive reversals calculable: two reversals restore the original orientation of a vector space.

Compare a singular matrix:Its columns are linearly dependent: the second is twice the first. The unit square collapses onto the segment from $(0,0)$ to $(3,6)$, which has zero area despite nonzero length. Since $C(-2,1{)}^T=0$, the kernel of a linear map contains a nonzero vector; inputs differing by that vector have identical outputs. The collapsed picture explains why an inverse matrix cannot exist.

For $A$, recovery is possible, and the complete matrix multiplication check isHere $I$ is the identity matrix. The inverse matrix restores the original square, shrinking its area by $1/5$. Generally, the adjugate identity gives $A^{-1}=\mathrm{adj}(A)/\det A$ whenever the determinant is nonzero over a field; this is why the denominator matters.^[ref]

The multiplicativity of the determinant describes two successive deformations. Matrix multiplication means that $AB$ applies $B$ first, then $A$. TakeThe first linear map doubles area; the second multiplies the resulting area by five. Thus the unit square passes through areas $1,2,10$, with the last parallelogram bounded by $(0,0),(2,1),(4,7),(2,6)$. Direct calculation agrees:This explains physically why area factors multiply instead of adding. The general identity holds for square matrices of the same size, including singular matrices. If either stage loses a direction, later processing cannot recover the discarded information.^[ref]

The same interpretation works in three dimensions without drawing a new picture. The diagonal matrixsends the unit cube to a rectangular box with side lengths two, three, and one. Its ordinary volume is six. The minus sign comes from reversing the third direction, so the oriented volume is minus six. Replacing the last diagonal entry by zero flattens the cube into a planar region: its area may be positive, but its three-dimensional volume is zero. This distinction matters whenever a linear map changes shape: the determinant measures size in the full input dimension, rather than the size of whichever lower-dimensional object remains.

The determinant measures neither trace nor perimeter. For $A$, the trace happens also to be five, but the identity matrix in two dimensions has trace two and determinant one. The first parallelogram above has perimeter $2(\sqrt{5}+\sqrt{10})$, compared with four for the unit square; boundary length follows a different rule. Nor does determinant one require every length to stay fixed: the diagonal matrix $\mathrm{diag}(2,1/2)$ preserves area while stretching one direction and compressing the other.

These properties make the determinant useful when coordinates obscure what a deformation does. A nonzero determinant certifies that linearly independent input directions remain distinguishable; its magnitude counts the area or volume represented by a transformed cell. That second role reappears in the change of variables formula, where the Jacobian determinant corrects the sizes of small cells. Signed volume also separates a deformation that reverses the orientation of a vector space from one that merely stretches it, even when both give identical ordinary volume factors.^[ref]

For general $n$, regard the columns $v_1,\dots ,v_n$ as ordered arguments. The determinant is the unique alternating multilinear map to the scalar field normalized by $\det I=1$. The multilinearity of the determinant means linearity in each column separately, with all others held fixed:Alternation means equal columns give zero; exchanging columns negates the result. Normalization sets the oriented volume of the standard basis to one. These properties also hold in rows. For real three-dimensional columns, their determinant is the scalar triple product, and its absolute value is the enclosed volume. Over other fields, the algebraic definition still works even without a geometric notion of positive volume.^[ref]

Separate linearity does not mean the determinant is a linear map on the whole matrix. Scaling every column of an $n$-by-$n$ matrix by $t$ gives $\det (tA)=t^n\det A$. For our two-dimensional example, $\det (2A)=4\det A=20$: doubling both directions quadruples area. The power records how many independent directions are being scaled together.

The properties suggest practical computations. Adding a multiple of one column to another preserves the determinant: the extra term has repeated columns and vanishes. This is the algebra behind an area-preserving shear mapping. Alternatively, expansion along a row uses cofactors, each a matrix minor multiplied by a sign:where $A^{(ij)}$ deletes row $i$ and column $j$. The Leibniz formula for determinants supplies the general entry formula. Both calculations encode the same geometric invariant; the defining properties explain their structure without requiring memorization of a separate recipe for every size.^[ref]

For three non-collinear points with position vectors $\mathbf{a},\mathbf{b},\mathbf{c}$, a nonzero normal vector isThe plane is the set of $\mathbf{r}$ satisfying $(\mathbf{r}-\mathbf{a})\cdot \mathbf{N}=0$. Expanding the cross product and using the scalar triple product givesNon-collinearity is essential: otherwise $\mathbf{N}=0$ and the equation fails to select a unique plane.

The determinant of a Gram matrix is nonnegative and vanishes precisely when its vectors are linearly dependent. In Euclidean three-dimensional space, three column vectors form a square matrix $V$, whose Gram matrix is $V^{T}V$. Its determinant is $(\det V{)}^2=[v_1\cdot (v_2\times v_3){]}^2$, the square of the scalar triple product and of the parallelepiped volume.

Consider lines $r=a+sm$ and $r=b+tn$ with nonzero direction vectors. If they intersect, then $a-b=tn-sm$, so the necessary coplanarity condition is $[a-b,m,n]=0$. For nonparallel directions it is also sufficient, since the orthogonal complement of $m\times n$ is the linear span of $m,n$. For parallel directions the condition alone is insufficient: distinct parallel lines also satisfy it.

Assume intersection and $D=[a,m,n]\ne 0$. Take the scalar triple products $[a,b,n]$ and $[a,b,m]$ of $b=a+sm-tn$. The scalar triple product is an alternating multilinear map, giving respectively $sD$ and $tD$, proving the title formulas. The nonzero denominator fixes a convenient determinant chart; intersections with $D=0$ can still exist and require a different auxiliary vector or another method.

On the smooth boundary of an orientable surface, let $n$ be the chosen unit normal vector and $t$ the positively oriented unit tangent vector used by Stokes theorem. The outward conormal is $u=t\times n$. It is a unit vector, tangent to the surface and orthogonal to its boundary. This convention turns $(n\times v)\cdot d\mathbf{r}$ into $u\cdot vds$ by cyclic invariance of the scalar triple product.

Put $v={\mathbf{a}}_1\cdot ({\mathbf{a}}_2\times {\mathbf{a}}_3)\ne 0$. DefineThe scalar triple product gives ${\mathbf{a}}_i\cdot {\mathbf{b}}_j=2\pi {\delta }_{ij}$. Consequently every integer combination $\mathbf{G}={\sum }_j{m}_j{\mathbf{b}}_j$ satisfiesThis proves that the proposed vectors generate points in the reciprocal lattice, but primitiveness also requires the converse. Let $\mathbf{G}$ satisfy the exponential condition for every lattice vector, and in particular for $\mathbf{R}={\mathbf{a}}_i$. ThenSince the ${\mathbf{b}}_i$ form a basis of three-dimensional space, the vector ${\sum }_i{m}_i{\mathbf{b}}_i$ has the same dot products with each ${\mathbf{a}}_j$ as $\mathbf{G}$. Their difference is orthogonal to three independent vectors and is zero. Henceand no additional reciprocal lattice vectors lie outside that integer span. Thus the ${\mathbf{b}}_i$ are primitive, not merely three possible reciprocal translations.

Given independent primitive lattice translation vectors ${\mathbf{a}}_1,{\mathbf{a}}_2,{\mathbf{a}}_3$, defineThen ${\mathbf{a}}_i\cdot {\mathbf{b}}_j=2\pi {\delta }_{ij}$. The reciprocal lattice consists ofequivalently all wavevectors for which $e^{i\mathbf{G}\cdot \mathbf{R}}=1$ for every direct lattice vector $\mathbf{R}$.

For the face-centered cubic lattice, chooseTheir scalar triple product is $a^3/4$, and direct evaluation of the cross products givesEvery integer combination has coordinates $(2\pi /a)(H,K,L)$ with $H,K,L$ all even or all odd. Conversely, for such coordinates the coefficients $(K+L)/2,(H+L)/2,(H+K)/2$ are integers, so every such point occurs. The even-coordinate points are the corners of cubes of edge $4\pi /a$, and the odd-coordinate points are their body centres. ThereforeThis is a property of the Bravais lattice; the two-atom atomic basis changes structure factors rather than these reciprocal lattice translation vectors.

Use the unit normal from part (c), and orient ${\gamma }_a$ by increasing $v$. Write $T={\gamma }_a^{\prime }/|{\gamma }_a^{\prime }|$ for its unit tangent vector. The signed geodesic curvature convention used below iswhere $s$ is arc length. The second equality follows by differentiating $T$: the term parallel to ${\gamma }_a^{\prime }$ disappears in the scalar triple product. It makes the correction for a parameter which is not arc-length parametrization explicit.

Put $C=\cos a$, $S=\sin a$, $D=1+S^2$, and $G=3+2C$. The speed is the constant $\sqrt{G}$. At $v=0$,soTaking its dot product with $N=(C,S,-S)/\sqrt{D}$ yields $-S(3+2C)/\sqrt{D}=-SG/\sqrt{D}$. ThereforeThe screw motion of part (b) makes this independent of $v$. Thus ${\gamma }_a$ is a geodesic precisely when $a$ is an integer multiple of $\pi$. Reversing the direction of traversal or the unit normal changes the sign of signed geodesic curvature; its magnitude is the absolute value of the displayed expression. The signed convention is needed in part (e).

Let $D=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})\ne 0$ be the scalar triple product. The reciprocal basis satisfies $\mathbf{a}\cdot {\mathbf{a}}^{\prime }=1$ and all mismatched dot products vanish, with cyclic analogues. Apply the given vector triple product identity to obtainDotting with ${\mathbf{a}}^{\prime }$ givesThe sign is retained: a negatively oriented original basis has a negatively oriented reciprocal basis. Equivalently, if the original vectors are columns of $B$, their reciprocals are columns of $B^{-T}$, whose determinant is $1/\det B$.

Three vectors form a basis of ${\mathbb{R}}^3$ exactly when they are linearly independent. The determinant with those vectors as columns is their scalar triple product, so the condition isGeometrically their parallelepiped must have nonzero volume.

In components the scalar triple product is $A_i{ϵ}_{ijk}{B}_j{C}_k$. A cyclic permutation has positive sign of a permutation, so ${ϵ}_{ijk}={ϵ}_{kij}$. Relabelling the summed indices therefore giveswhich proves $A\cdot (B\times C)=(A\times B)\cdot C$. Equivalently both expressions are the same three-column determinant.

Subtract the first vertex from the remaining three to obtain edge vectors $a=(1,2,-2)$, $b=(0,1,3)$, $c=(-1,-1,0)$. Their scalar triple product isA tetrahedron occupies one sixth of the corresponding parallelepiped, so its volume isThe absolute value removes the orientation sign of the determinant.

Let $P$ have columns $a,b,c$ and let $G$ have columns $A,B,C$. The cross-product formulas defining the reciprocal basis giveprovided the original scalar triple product $T=\det P$ is nonzero. Taking determinants yields the reciprocal basis involution identityWhen $T=0$, the displayed reciprocal vectors are undefined, so there is no additional zero-denominator case to evaluate.

The scalar triple product is the determinant with the three vectors as columns. A cyclic permutation of three columns is even, while exchanging two columns reverses the determinant's sign. ConsequentlyEquivalently, expand the first expression as ${ϵ}_{ijk}{a}_i{b}_j{c}_k$ and interchange the last two indices, using the antisymmetry of the Levi-Civita symbol. This argument includes linearly dependent vectors, for which both sides vanish.

Noncoplanarity makes the scalar triple product $\Delta$ nonzero, so the three displayed vectors form a reciprocal basis. Their scalar products obeywith all six cross-pairings zero. Therefore, assuming the nonzero $\alpha ,\beta ,\gamma$ needed to define the three given points,Every difference of two of these points is orthogonal to $\mathbf{X}$. The two independent differences span their plane, so $\mathbf{X}$ is a normal to the plane, whose equation is $\mathbf{X}\cdot \mathbf{r}=1$. Also $\mathbf{X}\ne 0$, because its scalar product with each specified point is $1$.

A plane through the three points also contains the origin exactly when their position vectors are linearly dependent. The scalar triple product isThus $\boxed{\gamma =9}$. The resulting plane is $\boxed{x-2y+z=0}$, which contains all three points and the origin. The points are not collinear: the differences $(3,3,3)$ and $(\gamma -1,6,4)$ cannot be proportional, since their last two components would require $6=4$. Therefore the plane used in the argument is well defined.

Let $\Delta =v_1\cdot (v_2\times v_3)\ne 0$, since the original vectors form a basis. The reciprocal basis isTaking the dot product with each $v_j$ gives $u^i\cdot v_j={\delta }_{ij}$, using orthogonality of a cross product and cyclic symmetry of the scalar triple product. If ${\sum }_i{a}_i{u}^i=0$, taking its dot product with $v_j$ yields $a_j=0$. The three reciprocal vectors are therefore linearly independent and form a basis of ${\mathbb{R}}^3$.

Equivalently, if $V$ has columns $v_j$ and $U$ columns $u^i$, then $U^{T}V=I$, so $U=V^{-T}$. In particular any vector has coordinates $r={\sum }_i(r\cdot u^i)v_i$, even when the original basis is not orthonormal.

For nonparallel lines, $\mathbf{n}={\mathbf{m}}_1\times {\mathbf{m}}_2$ is perpendicular to both direction vectors. Every displacement between the lines differs from ${\mathbf{a}}_2-{\mathbf{a}}_1$ by a vector in their span, so its component along $\mathbf{n}$ is fixed. Varying the two parameters cancels the component in that span. The shortest distance between skew lines is thereforeThis also gives zero for intersecting nonparallel lines. For parallel lines the denominator vanishes and the correct separate formula is $|({\mathbf{a}}_2-{\mathbf{a}}_1)\times {\mathbf{m}}_1|/|{\mathbf{m}}_1|$.

The given lines have vector formsHere ${\mathbf{m}}_1\times {\mathbf{m}}_2=(-3,0,3)$ and ${\mathbf{a}}_2-{\mathbf{a}}_1=(-4,-4,2)$. Their scalar triple product is $18$, givingAs a constructive check, the closest parameters are $t=2/3$ and $u=7/3$, where the connecting vector is $(3,0,-3)$ and is perpendicular to both lines.