# Torsion of space curves, why dB/ds is perpendicular to tangent

• phucnguyen
In summary, the conversation discusses section 4.3 of George Cain & James Harod's multivariable calculus material, which focuses on torsion. It explains that the binormal, a unit vector perpendicular to both the unit tangent and principal unit normal, has a derivative that is also perpendicular to the tangent vector. This is deduced by differentiating the equation B ⋅ T = 0 with respect to s and taking into account that B is perpendicular to N. The scalar τ is then introduced as the torsion. The conversation concludes with the understanding that the derivative of B is perpendicular to both B and T, and thus has the direction of N.

#### phucnguyen

Hi,

I'm reading this piece from George Cain & James Harod's multivariable calculus material.

Section 4.3, which is about Torsion, says this:

Let R(t) be a vector description of a curve. If T is the unit tangent and N is the principal unit normal, the unit vector B = T × N is called the binormal. Note that the binormal is orthogonal to both T and N. Let’s see about its derivative dB/ds with respect to arclength s. First, note that B ⋅ B = 1, and so B ⋅ (dB/ds) = 0 , which means that being orthogonal to B, the derivative dB/ds is in the plane of T and N. Next, note that B is perpendicular to the tangent vector T, and so B ⋅ T = 0 . Thus (dB/ds) ⋅ T = 0 . So what have we here? The vector is perpendicular to both B and T, and so must have the direction of N (or, of course, - N). This means (dB/ds) = −τ N .
The scalar τ is called the torsion.

I don't understand how he deduces dB/ds is perpendicular to T? Where did I get lost?

Following the paragraph, it seems to me that T and N plays quite the same role to B, then suddenly dB/ds is perpendicular to T.