Cross product

For a quant, a useful small example is finding a three-asset position with zero exposure to two specified factors. The same construction also finds normals to planes. The operation is inherently tied to the usual oriented 3D geometry: it is not a general algorithm for hedging an arbitrary number of assets.

Compute the vector, then check its sign

In a right-handed orthonormal coordinate system, let $\mathbf{a}=(a_1,a_2,a_3)$ and $\mathbf{b}=(b_1,b_2,b_3)$. Then

The middle component is a frequent sign trap. Written as the negative of a minor, it is $-(a_1b_3-a_3b_1)$, which equals the second component above. A determinant with unit vectors in its first row is a useful mnemonic, but the component formula is the actual definition.

If $\mathbf{n}=\mathbf{a}\times\mathbf{b}$, the perpendicularity checks are

These checks alone cannot distinguish $\mathbf{n}$ from $-\mathbf{n}$. Orientation supplies the choice. For nonparallel inputs, curl the fingers of your right hand from the first input toward the second, through the smaller angle; your thumb points along their cross product. Reversing the order reverses the answer:

For example, the coordinate unit vectors satisfy $\mathbf{e}_1\times\mathbf{e}_2=\mathbf{e}_3$, rather than $-\mathbf{e}_3$.

A fictional two-factor hedge

Suppose three fictional assets have the following dimensionless return loadings. The columns are assets; the rows are two specified factors.

For this example, write each asset's return as $r_i=a_i f_A+b_i f_B+\varepsilon_i$, where $f_A,f_B$ are factor returns and $\varepsilon_i$ is the remaining asset-specific return. If $w_i$ is the signed dollar position divided by a common reference capital amount, the portfolio's factor coefficients are $\mathbf{a}\cdot\mathbf{w}$ and $\mathbf{b}\cdot\mathbf{w}$. The target is to set both to zero.

Both exposures vanish: $1-1+0=0$ for Factor A and $0-1+1=0$ for Factor B. There is also a direct way to verify the construction. The constraints say $w_1+w_2=0$ and $w_2+w_3=0$, so every solution is $\mathbf{w}=t(1,-1,1)$ for a scalar $t$.

Only the two specified common-factor terms cancel. The residual $\sum_i w_i\varepsilon_i$ remains, as do risks from omitted factors, changing loadings, trading costs and financing. This construction does not imply a riskless position, mean reversion or an arbitrage. Reversing the factor order gives the opposite trade with the same two zero exposures.

Length measures area, not portfolio risk

For nonzero inputs with angle $\theta\in[0,\pi]$,

This is base times perpendicular height: the parallelogram area. A triangle with the same two edge vectors has half that area. Here, $\lVert\mathbf{a}\rVert=\lVert\mathbf{b}\rVert=\sqrt2$ and $\mathbf{a}\cdot\mathbf{b}=1$, so $\cos\theta=1/2$ and $\theta=60^\circ$. Thus the cross-product length and parallelogram area are $\sqrt3$, while the triangle area is $\sqrt3/2$. These are geometric quantities in loading-coordinate space, not measures of portfolio variance.

An independent algebraic check is

If the inputs are parallel, or either is zero, the cross product is zero. For example, $(1,1,0)\times(2,2,0)=(0,0,0)$. There is then no cross-product direction to normalise. In the hedge interpretation, two nonzero proportional loading rows impose only one independent constraint, leaving a plane of possible three-asset positions rather than one line. If both rows are zero, neither constrains the position. Nearly proportional rows can also make the inferred direction sensitive to small estimation errors.

Continue with the underlying geometry

Sources

• Steven G. Krantz and Harold Parks, Vector Calculus, first edition, CRC Press, 2024, §1.4, especially Definition 1.4.1, Theorem 1.4.2 and §§1.4.2–1.4.3.
• MIT OpenCourseWare, “Cross Product,” 18.02SC Multivariable Calculus, Fall 2010, pp. 1–2: algebraic formula, orientation and geometric magnitude.