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Binormal Vector — Definition, Formula & Examples

The binormal vector is a unit vector perpendicular to both the unit tangent vector and the principal normal vector of a space curve. It completes the right-handed orthonormal frame (called the Frenet-Serret frame) that moves along the curve.

Given a smooth space curve r(t)\mathbf{r}(t) with unit tangent vector T\mathbf{T} and principal unit normal vector N\mathbf{N}, the binormal vector is defined as B=T×N\mathbf{B} = \mathbf{T} \times \mathbf{N}. Since T\mathbf{T} and N\mathbf{N} are orthogonal unit vectors, B\mathbf{B} is also a unit vector, and the triple {T,N,B}\{\mathbf{T}, \mathbf{N}, \mathbf{B}\} forms an orthonormal basis at each point of the curve.

Key Formula

B=T×N\mathbf{B} = \mathbf{T} \times \mathbf{N}
Where:
  • B\mathbf{B} = Binormal vector (unit vector)
  • T\mathbf{T} = Unit tangent vector to the curve
  • N\mathbf{N} = Principal unit normal vector to the curve

How It Works

First, compute the unit tangent vector T=r′(t)/∥r′(t)∥\mathbf{T} = \mathbf{r}'(t)/\|\mathbf{r}'(t)\|. Next, find the principal normal vector N=T′(t)/∥T′(t)∥\mathbf{N} = \mathbf{T}'(t)/\|\mathbf{T}'(t)\|. Then take their cross product: B=T×N\mathbf{B} = \mathbf{T} \times \mathbf{N}. The binormal vector points in the direction perpendicular to the osculating plane (the plane that best fits the curve locally). Its rate of change dB/dsd\mathbf{B}/ds is related to the torsion τ\tau of the curve, which measures how the curve twists out of its osculating plane.

Worked Example

Problem: Find the binormal vector for the helix r(t)=⟨cos⁡t, sin⁡t, t⟩\mathbf{r}(t) = \langle \cos t,\, \sin t,\, t \rangle at any point tt.
Find T: Compute the derivative and normalize it.
\mathbf{r}'(t) = \langle -\sin t,\, \cos t,\, 1 \rangle, \quad \|\mathbf{r}'\| = \sqrt{\sin^2 t + \cos^2 t + 1} = \sqrt{2}$$$$\mathbf{T} = \frac{1}{\sqrt{2}}\langle -\sin t,\, \cos t,\, 1 \rangle
Find N: Differentiate T with respect to t and normalize.
\mathbf{T}'(t) = \frac{1}{\sqrt{2}}\langle -\cos t,\, -\sin t,\, 0 \rangle, \quad \|\mathbf{T}'\| = \frac{1}{\sqrt{2}}$$$$\mathbf{N} = \langle -\cos t,\, -\sin t,\, 0 \rangle
Compute B = T × N: Take the cross product of T and N.
B=12∣ijk−sin⁡tcos⁡t1−cos⁡t−sin⁡t0∣=12⟨sin⁡t, −cos⁡t, 1⟩\mathbf{B} = \frac{1}{\sqrt{2}}\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -\sin t & \cos t & 1 \\ -\cos t & -\sin t & 0 \end{vmatrix} = \frac{1}{\sqrt{2}}\langle \sin t,\, -\cos t,\, 1 \rangle
Answer: B(t)=12⟨sin⁡t, −cos⁡t, 1⟩\mathbf{B}(t) = \frac{1}{\sqrt{2}}\langle \sin t,\, -\cos t,\, 1 \rangle, which is indeed a unit vector at every point along the helix.

Why It Matters

The binormal vector is essential in differential geometry and physics for describing how curves twist through three-dimensional space. Engineers use the Frenet-Serret frame (built from T, N, and B) to analyze the motion of particles along curved paths, design roller coasters, and study the geometry of DNA helices.

Common Mistakes

Mistake: Computing N × T instead of T × N, which reverses the direction of B.
Correction: The binormal vector is defined as B = T × N (in that order). Reversing the cross product gives −B, which breaks the right-handed orientation of the Frenet-Serret frame.