Reciprocal System of Vectors

In short

Given three non-coplanar vectors \vec{a}, \vec{b}, \vec{c}, their reciprocal system is a new triple \vec{a}', \vec{b}', \vec{c}' defined so that each reciprocal vector is perpendicular to two of the original vectors and has unit dot product with the third. The formulas are \vec{a}' = \frac{\vec{b} \times \vec{c}}{[\vec{a}\;\vec{b}\;\vec{c}]} and cyclically. The reciprocal system lets you decompose any vector into components along non-orthogonal directions.

Take the standard basis vectors \hat{i}, \hat{j}, \hat{k}. They are perpendicular to each other, and each has magnitude 1. If you want to find the component of a vector \vec{v} along, say, \hat{i}, you simply compute \vec{v} \cdot \hat{i}. The dot product does all the work because the basis vectors are orthonormal — perpendicular and unit-length.

But what if your three directions are not perpendicular? Suppose you have three vectors \vec{a}, \vec{b}, \vec{c} that are linearly independent (non-coplanar), but tilted at odd angles to each other. You want to write some vector \vec{v} as \vec{v} = x\vec{a} + y\vec{b} + z\vec{c}. How do you find x, y, z?

If you try x = \vec{v} \cdot \vec{a}, the answer is wrong — because \vec{a}, \vec{b}, \vec{c} are not perpendicular, the dot product picks up unwanted cross-contributions from the other two directions. The trick of "just dot with the basis vector" only works for orthonormal bases.

The fix is to build a second set of three vectors — call them \vec{a}', \vec{b}', \vec{c}' — designed specifically so that \vec{a}' \cdot \vec{a} = 1, \vec{a}' \cdot \vec{b} = 0, \vec{a}' \cdot \vec{c} = 0. Then x = \vec{v} \cdot \vec{a}', and the unwanted cross-terms vanish. This second set is called the reciprocal system of \vec{a}, \vec{b}, \vec{c}.

Building the reciprocal system

You need \vec{a}' to be perpendicular to \vec{b} and \vec{c}, but to have dot product 1 with \vec{a}. The vector perpendicular to both \vec{b} and \vec{c} is their cross product \vec{b} \times \vec{c}. So \vec{a}' must be a scalar multiple of \vec{b} \times \vec{c}.

What scalar? The condition \vec{a}' \cdot \vec{a} = 1 determines it:

\vec{a}' = \frac{\vec{b} \times \vec{c}}{\vec{a} \cdot (\vec{b} \times \vec{c})} = \frac{\vec{b} \times \vec{c}}{[\vec{a}\;\vec{b}\;\vec{c}]}

Check: \vec{a}' \cdot \vec{a} = \frac{(\vec{b} \times \vec{c}) \cdot \vec{a}}{[\vec{a}\;\vec{b}\;\vec{c}]} = \frac{[\vec{a}\;\vec{b}\;\vec{c}]}{[\vec{a}\;\vec{b}\;\vec{c}]} = 1. And \vec{a}' \cdot \vec{b} = \frac{(\vec{b} \times \vec{c}) \cdot \vec{b}}{[\vec{a}\;\vec{b}\;\vec{c}]} = \frac{[\vec{b}\;\vec{b}\;\vec{c}]}{[\vec{a}\;\vec{b}\;\vec{c}]} = 0, because a scalar triple product with two equal vectors is zero. Similarly \vec{a}' \cdot \vec{c} = 0.

The same logic applies to \vec{b}' and \vec{c}'.

Reciprocal system of vectors

Given three non-coplanar vectors \vec{a}, \vec{b}, \vec{c} (i.e., [\vec{a}\;\vec{b}\;\vec{c}] \neq 0), the reciprocal system is:

\vec{a}' = \frac{\vec{b} \times \vec{c}}{[\vec{a}\;\vec{b}\;\vec{c}]}, \qquad \vec{b}' = \frac{\vec{c} \times \vec{a}}{[\vec{a}\;\vec{b}\;\vec{c}]}, \qquad \vec{c}' = \frac{\vec{a} \times \vec{b}}{[\vec{a}\;\vec{b}\;\vec{c}]}

These satisfy the orthogonality conditions:

\vec{a}' \cdot \vec{a} = 1, \quad \vec{b}' \cdot \vec{b} = 1, \quad \vec{c}' \cdot \vec{c} = 1

\vec{a}' \cdot \vec{b} = 0, \quad \vec{a}' \cdot \vec{c} = 0, \quad \vec{b}' \cdot \vec{a} = 0, \quad \vec{b}' \cdot \vec{c} = 0, \quad \vec{c}' \cdot \vec{a} = 0, \quad \vec{c}' \cdot \vec{b} = 0

In compact notation, \vec{e}_i' \cdot \vec{e}_j = \delta_{ij} (the Kronecker delta: 1 if i = j, 0 otherwise).

The original system $(\vec{a}, \vec{b}, \vec{c})$ in dark ink and the reciprocal system $(\vec{a}', \vec{b}', \vec{c}')$ in red. Each reciprocal vector is perpendicular to the plane of the other two original vectors, and its dot product with the matching original vector is exactly $1$.

The nine dot products between the original and reciprocal systems form the $3 \times 3$ identity matrix. Diagonal entries (matching pairs) are $1$; off-diagonal entries (mismatched pairs) are $0$.

Why "reciprocal"?

The name comes from the fact that the two systems undo each other. If you start with \vec{a}, \vec{b}, \vec{c}, build the reciprocal system \vec{a}', \vec{b}', \vec{c}', and then build the reciprocal of that, you get back the original \vec{a}, \vec{b}, \vec{c}.

The reciprocal operation is its own inverse: applying it twice returns you to the original system.

This is like numbers: the reciprocal of 5 is \frac{1}{5}, and the reciprocal of \frac{1}{5} is 5 again. The operation is its own inverse. For vectors, "reciprocal" means the same thing — applying it twice returns you to where you started.

Properties

Each property below follows directly from the definition and the properties of the scalar triple product.

1. The reciprocal of the reciprocal is the original system.

(\vec{a}')' = \vec{a}, \qquad (\vec{b}')' = \vec{b}, \qquad (\vec{c}')' = \vec{c}

To see this for \vec{a}: the reciprocal of \vec{a}' in the system (\vec{a}', \vec{b}', \vec{c}') is \frac{\vec{b}' \times \vec{c}'}{[\vec{a}'\;\vec{b}'\;\vec{c}']}. The numerator and denominator both involve the reciprocal system, and when you work through the algebra using the original definitions, everything simplifies back to \vec{a}. The key fact is in Property 2 below.

2. The scalar triple product of the reciprocal system is the reciprocal of the original.

[\vec{a}'\;\vec{b}'\;\vec{c}'] = \frac{1}{[\vec{a}\;\vec{b}\;\vec{c}]}

Here is the derivation. Write V = [\vec{a}\;\vec{b}\;\vec{c}] for the original scalar triple product. Then:

[\vec{a}'\;\vec{b}'\;\vec{c}'] = \vec{a}' \cdot (\vec{b}' \times \vec{c}')

Substitute the definitions:

= \frac{\vec{b} \times \vec{c}}{V} \cdot \left(\frac{\vec{c} \times \vec{a}}{V} \times \frac{\vec{a} \times \vec{b}}{V}\right)

The scalars \frac{1}{V} pull out — one from the dot product, two from the cross product:

= \frac{1}{V^3} \; (\vec{b} \times \vec{c}) \cdot [(\vec{c} \times \vec{a}) \times (\vec{a} \times \vec{b})]

Use the BAC-CAB formula on (\vec{c} \times \vec{a}) \times (\vec{a} \times \vec{b}). Treat \vec{p} = \vec{c} \times \vec{a} and \vec{q} = \vec{a} \times \vec{b} as the two vectors:

\vec{p} \times \vec{q} = (\vec{p} \cdot (\vec{a} \times \vec{b})) \cdot \vec{a} - \text{...}

A cleaner route: note that (\vec{c} \times \vec{a}) \times (\vec{a} \times \vec{b}) = [(\vec{c} \times \vec{a}) \cdot \vec{b}]\,\vec{a} - [(\vec{c} \times \vec{a}) \cdot \vec{a}]\,\vec{b}. The second term vanishes because (\vec{c} \times \vec{a}) \cdot \vec{a} = [\vec{a}\;\vec{c}\;\vec{a}] = 0 (two equal vectors in the scalar triple product). And (\vec{c} \times \vec{a}) \cdot \vec{b} = [\vec{b}\;\vec{c}\;\vec{a}] = V. So:

(\vec{c} \times \vec{a}) \times (\vec{a} \times \vec{b}) = V\vec{a}

Substituting back:

[\vec{a}'\;\vec{b}'\;\vec{c}'] = \frac{1}{V^3}(\vec{b} \times \vec{c}) \cdot (V\vec{a}) = \frac{V}{V^3} \cdot (\vec{b} \times \vec{c}) \cdot \vec{a} = \frac{V}{V^3} \cdot V = \frac{V^2}{V^3} = \frac{1}{V}

So [\vec{a}'\;\vec{b}'\;\vec{c}'] = \frac{1}{[\vec{a}\;\vec{b}\;\vec{c}]}.

3. The dot products form the identity matrix.

Arrange all nine dot products between the original and reciprocal systems in a matrix:

\begin{pmatrix} \vec{a}' \cdot \vec{a} & \vec{a}' \cdot \vec{b} & \vec{a}' \cdot \vec{c} \\ \vec{b}' \cdot \vec{a} & \vec{b}' \cdot \vec{b} & \vec{b}' \cdot \vec{c} \\ \vec{c}' \cdot \vec{a} & \vec{c}' \cdot \vec{b} & \vec{c}' \cdot \vec{c} \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = I

This is the defining property of the reciprocal system, compactly stated.

4. Decomposition of a vector.

If \vec{v} = x\vec{a} + y\vec{b} + z\vec{c}, then the coefficients can be read off using the reciprocal system:

x = \vec{v} \cdot \vec{a}', \qquad y = \vec{v} \cdot \vec{b}', \qquad z = \vec{v} \cdot \vec{c}'

The proof is direct: dot both sides of \vec{v} = x\vec{a} + y\vec{b} + z\vec{c} with \vec{a}':

\vec{v} \cdot \vec{a}' = x(\vec{a} \cdot \vec{a}') + y(\vec{b} \cdot \vec{a}') + z(\vec{c} \cdot \vec{a}') = x(1) + y(0) + z(0) = x

This is the whole point. The reciprocal system is a tool for extracting components along non-orthogonal axes.

5. For an orthonormal basis, the reciprocal system is the basis itself.

If \vec{a} = \hat{i}, \vec{b} = \hat{j}, \vec{c} = \hat{k}, then [\hat{i}\;\hat{j}\;\hat{k}] = 1 and:

\vec{a}' = \frac{\hat{j} \times \hat{k}}{1} = \hat{i}, \qquad \vec{b}' = \frac{\hat{k} \times \hat{i}}{1} = \hat{j}, \qquad \vec{c}' = \frac{\hat{i} \times \hat{j}}{1} = \hat{k}

The reciprocal of an orthonormal basis is itself. This is why dot products with \hat{i}, \hat{j}, \hat{k} directly give components — the standard basis is self-reciprocal.

6. Cross products between original and reciprocal vectors.

You can also express cross products between the two systems cleanly:

\vec{a}' \times \vec{b}' = \frac{(\vec{b} \times \vec{c}) \times (\vec{c} \times \vec{a})}{V^2}

Using the vector triple product (BAC-CAB), this simplifies. The result is that:

\vec{a}' \times \vec{b}' = \frac{\vec{c}}{[\vec{a}\;\vec{b}\;\vec{c}]}

and similarly for cyclic permutations. This means the cross product of two reciprocal vectors gives (up to a scalar) the third original vector — another expression of the duality between the two systems.

7. A useful identity for JEE problems.

If \vec{r} is any vector, then:

\vec{r} = (\vec{r} \cdot \vec{a}')\vec{a} + (\vec{r} \cdot \vec{b}')\vec{b} + (\vec{r} \cdot \vec{c}')\vec{c}

and equally:

\vec{r} = (\vec{r} \cdot \vec{a})\vec{a}' + (\vec{r} \cdot \vec{b})\vec{b}' + (\vec{r} \cdot \vec{c})\vec{c}'

Both formulas decompose \vec{r} — the first along the original basis using reciprocal vectors for the coefficients, and the second along the reciprocal basis using original vectors for the coefficients. These dual decomposition formulas appear frequently in JEE Advanced vector problems.

Computing one from start to finish

Example 1: Finding the reciprocal system

Find the reciprocal system of \vec{a} = \hat{i} + \hat{j}, \vec{b} = \hat{j} + \hat{k}, \vec{c} = \hat{k} + \hat{i}.

Step 1. Compute the scalar triple product [\vec{a}\;\vec{b}\;\vec{c}].

[\vec{a}\;\vec{b}\;\vec{c}] = \begin{vmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{vmatrix} = 1(1-0) - 1(0-1) + 0(0-1) = 1 + 1 + 0 = 2

Why: the scalar triple product appears in the denominator of all three reciprocal vectors. It must be nonzero, otherwise the system has no reciprocal (the vectors would be coplanar).

Step 2. Compute \vec{b} \times \vec{c}.

\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{vmatrix} = (1-0)\hat{i} - (0-1)\hat{j} + (0-1)\hat{k} = \hat{i} + \hat{j} - \hat{k}

Why: this cross product, divided by [\vec{a}\;\vec{b}\;\vec{c}], gives \vec{a}'.

Step 3. Compute \vec{c} \times \vec{a} and \vec{a} \times \vec{b}.

\vec{c} \times \vec{a} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{vmatrix} = (0-1)\hat{i} - (0-1)\hat{j} + (1-0)\hat{k} = -\hat{i} + \hat{j} + \hat{k}

\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix} = (1-0)\hat{i} - (1-0)\hat{j} + (1-0)\hat{k} = \hat{i} - \hat{j} + \hat{k}

Why: these cross products divided by 2 give \vec{b}' and \vec{c}' respectively.

Step 4. Divide each by [\vec{a}\;\vec{b}\;\vec{c}] = 2.

\vec{a}' = \frac{\hat{i} + \hat{j} - \hat{k}}{2}, \qquad \vec{b}' = \frac{-\hat{i} + \hat{j} + \hat{k}}{2}, \qquad \vec{c}' = \frac{\hat{i} - \hat{j} + \hat{k}}{2}

Result: \vec{a}' = \frac{1}{2}(\hat{i} + \hat{j} - \hat{k}), \vec{b}' = \frac{1}{2}(-\hat{i} + \hat{j} + \hat{k}), \vec{c}' = \frac{1}{2}(\hat{i} - \hat{j} + \hat{k}).

The original vectors $\vec{a} = \hat{i}+\hat{j}$, $\vec{b} = \hat{j}+\hat{k}$, $\vec{c} = \hat{k}+\hat{i}$ (solid) and their reciprocal system (dashed, in red). The dot product checks confirm that each reciprocal vector is orthogonal to the two non-matching original vectors and has unit dot product with the matching one.

You can verify: \vec{a}' \cdot \vec{a} = \frac{1}{2}(1 \cdot 1 + 1 \cdot 1 + (-1) \cdot 0) = \frac{1}{2}(1 + 1) = 1. And \vec{a}' \cdot \vec{b} = \frac{1}{2}(1 \cdot 0 + 1 \cdot 1 + (-1) \cdot 1) = \frac{1}{2}(0 + 1 - 1) = 0. The reciprocal system works as advertised.

Example 2: Decomposing a vector along non-orthogonal axes

Using the reciprocal system from Example 1, express \vec{v} = 3\hat{i} + 5\hat{j} + 7\hat{k} as a linear combination of \vec{a} = \hat{i} + \hat{j}, \vec{b} = \hat{j} + \hat{k}, \vec{c} = \hat{k} + \hat{i}.

Step 1. Compute x = \vec{v} \cdot \vec{a}'.

x = (3\hat{i} + 5\hat{j} + 7\hat{k}) \cdot \frac{\hat{i} + \hat{j} - \hat{k}}{2} = \frac{3 + 5 - 7}{2} = \frac{1}{2}

Why: \vec{a}' is designed so that \vec{v} \cdot \vec{a}' directly extracts the \vec{a}-coefficient in \vec{v} = x\vec{a} + y\vec{b} + z\vec{c}.

Step 2. Compute y = \vec{v} \cdot \vec{b}'.

y = (3\hat{i} + 5\hat{j} + 7\hat{k}) \cdot \frac{-\hat{i} + \hat{j} + \hat{k}}{2} = \frac{-3 + 5 + 7}{2} = \frac{9}{2}

Why: same idea — the reciprocal vector \vec{b}' filters out all contributions except the \vec{b}-component.

Step 3. Compute z = \vec{v} \cdot \vec{c}'.

z = (3\hat{i} + 5\hat{j} + 7\hat{k}) \cdot \frac{\hat{i} - \hat{j} + \hat{k}}{2} = \frac{3 - 5 + 7}{2} = \frac{5}{2}

Why: completing the decomposition.

Step 4. Verify: x\vec{a} + y\vec{b} + z\vec{c}.

= \frac{1}{2}(\hat{i} + \hat{j}) + \frac{9}{2}(\hat{j} + \hat{k}) + \frac{5}{2}(\hat{k} + \hat{i})

= \frac{1}{2}\hat{i} + \frac{1}{2}\hat{j} + \frac{9}{2}\hat{j} + \frac{9}{2}\hat{k} + \frac{5}{2}\hat{k} + \frac{5}{2}\hat{i}

= \left(\frac{1}{2} + \frac{5}{2}\right)\hat{i} + \left(\frac{1}{2} + \frac{9}{2}\right)\hat{j} + \left(\frac{9}{2} + \frac{5}{2}\right)\hat{k}

= 3\hat{i} + 5\hat{j} + 7\hat{k} = \vec{v} \quad \checkmark

Result: \vec{v} = \frac{1}{2}\vec{a} + \frac{9}{2}\vec{b} + \frac{5}{2}\vec{c}.

The vector $\vec{v} = 3\hat{i} + 5\hat{j} + 7\hat{k}$ decomposed as $\frac{1}{2}\vec{a} + \frac{9}{2}\vec{b} + \frac{5}{2}\vec{c}$, where the coefficients are found by dotting $\vec{v}$ with the reciprocal vectors. Without the reciprocal system, finding these coefficients would require solving a system of three simultaneous equations.

Without the reciprocal system, you would need to solve the system x + z = 3, x + y = 5, y + z = 7 — three equations in three unknowns. That is doable, but the reciprocal-system approach gives each coefficient independently, with a single dot product per coefficient.

Applications

Left: with orthogonal axes, dotting with the basis vectors directly gives the components. Right: with non-orthogonal axes, dotting with the basis vectors gives the *wrong* components — you need the reciprocal vectors to filter out cross-contributions.

1. Decomposing vectors along non-orthogonal axes. This is the primary application, shown in Example 2. In crystallography, crystal axes are often non-orthogonal (oblique lattices), and the reciprocal system is the standard tool for computing components.

2. The reciprocal lattice in physics. In solid-state physics, the atoms of a crystal sit at positions described by three lattice vectors \vec{a}, \vec{b}, \vec{c}. The reciprocal lattice — built from \vec{a}', \vec{b}', \vec{c}' — is fundamental to understanding X-ray diffraction and Bragg's law. Every diffraction pattern you see in a physics lab is a picture of the reciprocal lattice.

3. Solving simultaneous equations. Writing a system of three linear equations in vector form, the reciprocal system effectively inverts the coefficient matrix. The formulas for \vec{a}', \vec{b}', \vec{c}' are equivalent to Cramer's rule for a 3 \times 3 system.

Common confusions

"The reciprocal of \vec{a} is 1/\vec{a}." There is no such thing as dividing by a vector. The reciprocal system is not a single-vector operation — you need all three vectors to build the reciprocal of any one of them. \vec{a}' depends on \vec{b} and \vec{c}, not just on \vec{a}.
"Reciprocal vectors are unit vectors in the same direction." No. A unit vector \hat{a} points in the same direction as \vec{a} with magnitude 1. The reciprocal vector \vec{a}' points in a completely different direction — perpendicular to \vec{b} and \vec{c}, not along \vec{a}.
"The reciprocal system exists for any three vectors." Only if the three vectors are non-coplanar, meaning [\vec{a}\;\vec{b}\;\vec{c}] \neq 0. If the three vectors are coplanar, the scalar triple product is zero, and the definition involves division by zero.
"The scalar triple product of the reciprocal system is the negative of the original." It is the reciprocal (the multiplicative inverse), not the negative. [\vec{a}'\;\vec{b}'\;\vec{c}'] = \frac{1}{[\vec{a}\;\vec{b}\;\vec{c}]}, not -[\vec{a}\;\vec{b}\;\vec{c}].

Going deeper

If you came here to learn what the reciprocal system is, how to compute it, and how to use it for decompositions, you have it. The rest of this article explores the connection to dual bases, the metric tensor, and crystallography.

The metric tensor and dual bases

In the language of linear algebra, the reciprocal system is the dual basis (or contravariant basis) associated with the original covariant basis \vec{a}, \vec{b}, \vec{c}.

The matrix of dot products between the original vectors,

G = \begin{pmatrix} \vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} & \vec{a} \cdot \vec{c} \\ \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{b} & \vec{b} \cdot \vec{c} \\ \vec{c} \cdot \vec{a} & \vec{c} \cdot \vec{b} & \vec{c} \cdot \vec{c} \end{pmatrix}

is called the metric tensor (or Gram matrix). It encodes all the geometric information — lengths and angles — about the original basis.

The reciprocal system has a beautiful relationship to this matrix: the matrix of dot products between original and reciprocal vectors is the identity (Property 3 above), which means the reciprocal basis vectors are the rows of G^{-1} expressed in the original basis. Computing the reciprocal system is equivalent to inverting the metric tensor.

This becomes the standard language in general relativity and Riemannian geometry, where the distinction between covariant and contravariant components is central to the entire theory.

Crystallographic conventions

In crystallography, the reciprocal lattice vectors are defined with an extra factor of 2\pi:

\vec{a}^* = \frac{2\pi(\vec{b} \times \vec{c})}{[\vec{a}\;\vec{b}\;\vec{c}]}

This factor makes the Fourier-transform relationship between the real lattice and reciprocal lattice come out cleanly. The mathematical structure is the same; only the normalisation differs.

Indian physicist C. V. Raman's work on crystal optics and lattice dynamics relied heavily on reciprocal-space analysis. The reciprocal lattice is not an abstract construction — it is the framework that explains why crystals diffract light and X-rays into specific patterns.

A proof that the reciprocal of the reciprocal is the original

Start from the reciprocal system \vec{a}' = \frac{\vec{b} \times \vec{c}}{V}, \vec{b}' = \frac{\vec{c} \times \vec{a}}{V}, \vec{c}' = \frac{\vec{a} \times \vec{b}}{V}, where V = [\vec{a}\;\vec{b}\;\vec{c}].

The reciprocal of \vec{a}' in the system (\vec{a}', \vec{b}', \vec{c}') is \frac{\vec{b}' \times \vec{c}'}{[\vec{a}'\;\vec{b}'\;\vec{c}']}.

From Property 2, [\vec{a}'\;\vec{b}'\;\vec{c}'] = 1/V.

Now compute \vec{b}' \times \vec{c}':

\vec{b}' \times \vec{c}' = \frac{\vec{c} \times \vec{a}}{V} \times \frac{\vec{a} \times \vec{b}}{V} = \frac{1}{V^2}[(\vec{c} \times \vec{a}) \times (\vec{a} \times \vec{b})]

From the derivation in Property 2, (\vec{c} \times \vec{a}) \times (\vec{a} \times \vec{b}) = V\vec{a}. So:

\vec{b}' \times \vec{c}' = \frac{V\vec{a}}{V^2} = \frac{\vec{a}}{V}

Therefore:

(\vec{a}')' = \frac{\vec{b}' \times \vec{c}'}{[\vec{a}'\;\vec{b}'\;\vec{c}']} = \frac{\vec{a}/V}{1/V} = \vec{a}

The reciprocal of the reciprocal is the original. The same proof works for \vec{b} and \vec{c} by cyclic symmetry.

Where this leads next

The reciprocal system sits at the junction of vectors and linear algebra.

Scalar Triple Product — the denominator in every reciprocal vector formula, and the key to the coplanarity condition that determines whether the reciprocal system exists.
Vector Triple Product — the BAC-CAB formula used in proving that the reciprocal of the reciprocal is the original.
Cross Product — the operation that builds each reciprocal vector from the other two original vectors.
Dot Product — the operation that makes the reciprocal system useful for decomposition.
Collinearity and Coplanarity — the condition [\vec{a}\;\vec{b}\;\vec{c}] \neq 0 that the reciprocal system requires is exactly the non-coplanarity condition.