binormal ˆeband unit normal ˆenis called the normal plane. The plane which is per-
pendicular to the principal normal ˆenis called the rectifying plane. Let r (t∗)denote
the position vector to a fixed point on the given curve and let r =xˆe 1 +yˆe 2 +zˆe 3
denote the position vector to a variable point in one of these planes. One can then
show
The osculating plane can be written (r −r (t∗)) ·eˆb= 0
The normal plane can be written (r −r(t∗)) ·ˆet= 0
The rectifying plane can be written (r −r (t∗)) ·ˆen= 0
(7 .15)
The equations of the straight lines through the fixed point r (t∗) and having the
directions of ˆet,eˆn or ˆebare given by
(r −r (t∗)) ׈et= 0 Tangent line
(r −r (t∗)) ׈en= 0 Line normal to curve
(r −r (t∗)) ×eˆb= 0 Line in binormal direction
(7 .16)
Let us examine the three unit vectors ˆet,ˆeband ˆen and their derivatives with
respect to the arc length parameter s. One can calculate the derivatives
dˆet
ds ,
dˆeb
ds
and
dˆen
ds with the aid of the triple scalar product relations
A·(B×C) = B·(C×A) = C·(A×B)
It has been demonstrated that dˆet
ds
=κˆen and ˆeb=ˆet׈en, consequently one finds
that
dˆeb
ds
=ˆet×dˆen
ds
+dˆet
ds
׈en=ˆet×dˆen
ds
+κˆen×eˆn=ˆet×dˆen
ds
(7 .17)
Take the dot product of both sides of equation (7.17) with the vector ˆetand use the
above triple scalar product result to show
ˆet·d
ˆeb
ds
=ˆet·eˆt×d
ˆen
ds
= 0
This result shows that the vector eˆt is perpendicular to the vector d
ˆeb
ds
. By differ-
entiating the relation ˆeb·ˆeb= 1 one finds that
ˆeb·dˆeb
ds
+dˆeb
ds
·ˆeb= 2 ˆeb·dˆeb
ds
= 0
