- The binormal vector B(t) is defined as the cross product of the unit tangent vector T(t) and unit normal vector N(t). - It is proven that B(t) is a unit vector, meaning it has constant length. Its derivative dB/ds is therefore orthogonal to B(t). - The torsion τ of a space curve is defined as the rate of change of the binormal vector with respect to arc length s, or τ = -dB/ds·N. Torsion measures how much a curve twists as one moves along it. - For a plane curve, the torsion is always zero since the cross product that defines torsion is equal to the

1. Unit Binormal Vector B and Torsion
If r(t) is a smooth space curve, |T(t)| = 1 for all t, we have T′
and T are
orthogonal.
Proof: Since the tangent vector, T, is of constant length (recall that |T(t)| = 1
so it is of constant length) we have
T · T = c2
d
dt
[T · T] = 0
T′
· T + T · T′
= 0
2T′
· T = 0
and since the dot product between two vectors is zero implies the vectors are
orthogonal, we have our result.
Note that T′
(t) is not necessarily a unit vector, but if r′
(t) is also smooth
then the unit tangent and unit normal vectors can be defined as:
T =
v
|v|
N =
T′
|T′
|
.
The last vector in the TNB frame is called the binormal vector. It is perpendic-
ular to both the T(t) and N(t) vectors and is also a unit vector. It is computed
via the cross product of the unit tangent and the unit normal vectors. Thus:
B(t) = T(t) × N(t)
Below we prove the binormal vector is a unit vector.
Proof that B is a unit vector: We have
|B(t)| = |T(t) × N(t)|
= |T||N| sin(θ)
= (1)(1) sin(π/2)
= 1
Thus, the binormal vector is also a unit vector. This also establishes that the
binormal has constant length. Hence its derivative will also be orthogonal to it.
Because we are evaluating several cross products and we typically use deter-
minants to do these evaluations, let us review how to interpret the determinant
notation for 2 × 2 and 3 × 3 matrices:
a b
c d
= ad − bc
a b c
d e f
g h i
= aei + bfg + cdh − ceg − fha − idb
1

2. Examples
Example 1: Compute the binormal vector for r(t) = t2
, t3
, t .
Answer: To compute the binormal vector we need to perform the cross product
of T and N. So we need to find these vectors,
Component First Derivative Second Derivative
f(t) = t2
2t 2
g(t) = t3
3t2
6t
h(t) = t 1 0
Now that we have our derivatives, we need to find the magnitude of r′
(t) in order
to compute the unit tangent vector. So (2t)2 + (3t2)2 + 12 =
√
9t4 + 4t2 + 1.
Hence
T(t) =
2ti + 3t2
j + 1k
√
9t4 + 4t2 + 1
is the unit tangent vector. Now we need the unit normal vector. This is com-
puted by first taking the derivative of the unit tangent vector and then dividing
by its magnitude. So we have
N(t) =
dT/dt
|dT/dt|
=
(1 − 9t4
)i + (3t + 6t3
)j + (2t + 9t3
)k
1 + 4t2 + 9t4
Now we need to cross these vectors. So we use the determinant and we have
T × N =
i j k
2t
√
9t4 + 4t2 + 1
3t2
√
9t4 + 4t2 + 1
1
√
9t4 + 4t2 + 1
1 − 9t4
1 + 4t2 + 9t4
3t + 6t3
1 + 4t2 + 9t4
2t + 9t3
1 + 4t2 + 9t4
=
3ti + j + 3t2
k
√
1 + 4t2 + 9t4
Torsion
Let us examine how dB/ds behaves in relation to T, N, and B. From our
differentiation rules we have
dB
ds
=
dT
ds
× N + T ×
dN
ds
Since N is in the direction of dT/ds,
dT
ds
× N = 0. So our result reduces to
dB
ds
= T ×
dN
ds
So dB/ds is orthogonal to both T and dN/ds, but N is also orthogonal to T
and dN/ds, so dB/ds must be parallel to N. Since dB/ds is parallel to N, it
must be a scalar multiple of N. So let’s write
dB
ds
= −τN
2

3. If we consider the dot product of dB/ds and N we have
dB
ds
· N = −τN · N = −τ
since N is a unit vector. Thus we may conclude this scalar multiple can be
determined via
−
dB
ds
· N = τ
and this quantity is called the torsion. This is a measure of how a space curve
twists while particles move along it.
Summary
• T is the unit tangent vector. T =
v
|v|
.
• N is the unit normal vector. N =
dT/dt
|dT/dt|
.
• B is the unit binormal vector. B = T × N.
• κ is the curvature. κ =
dT
ds
=
|v × a|
|v|3
.
• τ is the torsion. τ = −
dB
ds
· N =
˙x ˙y ˙z
¨x ¨y ¨z
...
x
...
y ...
z
|v × a|2
.
• The osculating circle (x−a)2
+(y−b)2
= r2
is the circle that is tangent to
a curve at a point and provides the best circle to approximate the curve
at the point in question. The center may be determined by the vector
equation: C = r(t0) +
1
κ(t0)
N(t0). The radius is given be
1
κ
.
• The radius of torsion is the reciprocal of the torsion, so ρ =
1
τ
.
Example
Example 2: Find the torsion of r(t) = cosh(t)i − sinh(t)j + tk.
Answer: In order to answer this question we need to find the first three deriva-
tives of the components and then we need to consider the magnitude of the
cross-product of the velocity and acceleration vectors. We begin with the com-
putation of the derivatives.
Component First Derivative Second Derivative Third Derivative
f(t) = cosh(t) sinh(t) cosh(t) sinh(t)
g(t) = sinh(t) cosh(t) sinh(t) cosh(t)
h(t) = t 1 0 0
3

4. Now we need the cross product between the velocity and acceleration vectors:
v × a =
i j k
sinh(t) − cosh(t) 1
cosh(t) − sinh(t) 0
= sinh(t)i + cosh(t)j + (cosh2
(t) − sinh2
(t))k
= sinh(t)i + cosh(t)j + 1k
Now we may fill in our torsion formula using the determinant:
τ =
sinh(t) − cosh(t) 1
cosh(t) − sinh(t) 0
sinh(t) − cosh(t) 0
sinh2
(t) + cosh2
(t) + 1
=
− cosh2
(t) + sinh2
(t)
sinh2
(t) + cosh2
(t) + 1
=
−1
2 cosh2
(t)
Alternate Approach Using B, T, and N
If we choose to use the formula −
dB
ds
· N we may also do so. We first find the
unit tangent and unit normal vectors. The unit tangent vector is computed
below.
v = sinh(t)i + cosh(t)j + 0k
|v| = (sinh2
(t))2 + (− cosh(t))2 + 1
=
√
2 cosh(t)
T =
v
|v|
=
sinh(t)i
√
2 cosh(t)
+
− cosh(t)j
√
2 cosh(t)
+
k
√
2 cosh(t)
=
1
√
2
tanh(t)i −
j
√
2
+
sech(t)
√
2
k
Now we may find the unit normal vector by differentiating the unit tangent
4

5. vector with respect to t. We have
dT
dt
=
sech2
(t)
√
2
i −
sech(t) tanh(t)
√
2
k
dT
dt
=
sech4
(t)
2
+
sech2
(t) tanh2
(t)
2
=
sech(t)
√
2
N =
dT/dt
|dT/dt|
= sech(t)i − tanh(t)k
So we may now consider the binormal vector as the cross product between T
and N. Hence
B = T × N =
i j k
tanh(t)
√
2
−
1
√
2
sech(t)
√
2
sech(t) 0 − tanh(t)
=
tanh(t)
√
2
i +
1
√
2
j +
sech(t)
√
2
k
Now we may take the derivative of B with respect to s. Since rewritting the
parameter t in terms of s will lead to inverse hyperbolic trigonometric functions
which have restrictions on their domains, not to mention that the formulas
themselves are cumbersome computationally to work with when considering
derivatives we shall impose the chain rule to have differentiation of B in terms
of t instead. Thus we have
dB
ds
=
dB
dt
dt
ds
=
dB
dt
1
|v|
We already know |v| from our calculation of the unit tangent vector. Differen-
tiating B with respect to t is done below.
dB
dt
=
d
dt
tanh(t)
√
2
i +
sech(t)
√
2
k
=
sech2
(t)
√
2
i −
sech(t) tanh(t)
√
2
k
dB
dt
1
|v|
=
sech2
(t)
√
2
i −
sech(t) tanh(t)
√
2
k
1
√
2 cosh(t)
=
1
2
sech3
(t)i + sech2
(t) tanh(t)k
5

6. Now perform the dot product between this vector and N to obtain the
−
dB
dt
1
|v|
· N = −
1
2
sech3
(t)i + sech2
(t) tanh(t)k · (sech(t)i − tanh(t)k)
=
−1
2
sech4
(t) − tanh2
(t) sech2
(t)
=
−1
2
sech2
(t) sech2
(t) − tanh2
(t)
=
−1
2
sech2
(t)
=
−1
2 cosh2
(t)
and we can see the results are identical.
Theorem: Plane curves do not exhibit torsion.
Proof: Consider a curve r(t) = f(t)i+g(t)j. Now our formula for torsion would
reduce to
˙x ˙y ˙z
¨x ¨y ¨z
...
x
...
y ...
z
|v × a|2
=
˙x ˙y 0
¨x ¨y 0
...
x
...
y 0
|v × a|2
= 0.
since a matrix with an entire row or an entire column of zeros has a zero deter-
minant.
Tangential and Normal Components of
Acceleration
When a body experiences a form of acceleration, it is common to ask how much
of the acceleration is in the direction of motion. This implies computing the
tangential component of the acceleration. If we consider the velocity vector, v,
we have
v =
dr
dt
=
dr
ds
ds
dt
= T
ds
dt
Differentiating both sides of this equation with respect to t yields
a =
d2
s
dt2
T + κ
ds
dt
2
N
What this implies is that the tangential component of acceleration is the second
derivative of the arc length with respect to t while the normal component of the
acceleration is the curvature times the square of the first derivative of the arc
length with respect to t. So we have
a = aT T + aN N where
aT =
d2
s
dt2
=
d
dt
|v| and aN = κ
ds
dt
2
= κ|v|2
6

7. are the tangential and normal scalar components of acceleration.
Things to note:
• B does not appear in the components, so the acceleration always lies in
the plane formed by T and N, orthogonal to B.
• The tangential component of acceleration measures the rate of change of
the length of v. The normal component of acceleration measures the rate
of change of the direction of v.
• The normal scalar component involves curvature and the square of the
speed.
We note that the formula for the normal component for acceleration is usually
very cumbersome to compute. If we examine the formula for the acceleration
vector we have
a =
d2
s
dt2
T + κ
ds
dt
2
N
a · a =
d2
s
dt2
T + κ
ds
dt
2
N ·
d2
s
dt2
T + κ
ds
dt
2
N
|a|2
=
d2
s
dt
2
+ κ2 ds
dt
4
|a|2
= a2
T + a2
N
a2
N = |a|2
− a2
T
aN = |a|2 − a2
T
and this formula is often used to compute the normal component, rather than
perform the computations previously mentioned.
We may also express the normal and tangential components of the accelera-
tion in terms of r(t). The formulas are:
aT =
r′
(t) · r′′
(t)
|r′(t)|
aN =
|r′
(t) × r′′
(t)|
|r′(t)|
These formulas are convenient when performing computations via technology.
Examples
Example 3: Compute the tangential and normal scalar components to the
acceleration at t = 0 if the position vector is given as r(t) = cos(t)i − et
j.
Answer: The acceleration vector is given by a(t) = − cos(t)i − et
j If we
7

8. consider the magnitude of the velocity vector v(t) = − sin(t)i − et
j we have
− sin2
(t) + e2t. The derivative of this with respect to t is
d
dt
sin2
(t) + e2t =
2e2x
+ 2 cos(x) sin(x)
2 e2x + sin2
(x)
Now if we evaluate these vectors at t = 0 we have a(0) = −1i − 1j and the
tangential scalar component of the acceleration is aT (0) = 1. So the normal
scalar component may be computed via |a|2 − a2
T =
√
2 − 1 = 1. Thus the
acceleration vector may be written as a(0) = T + N.
Example 4: Compute the tangential and normal components to the accelera-
tion vector if the position vector is r(t) = cos(t)i + sin(t)j + tk.
Answer: If we find the acceleration vector a(t) = − cos(t)i − sin(t)j. Now we
may find the tangential scalar component by taking the derivative of the speed.
Thus
d
dt
|v| =
d
dt
(− sin(t))2 + (cos(t))2 + 12 =
d
dt
√
2 = 0
So the tangential scalar component is 0. The normal scalar component for this
problem is found by |a|2 = |a| so we have
|a| = (− cos(t))2 + (− sin(t))2 = 1
So the normal component is 1 and the tangential component is 0.
Planes of the TNB Frame
Now that we have the vectors for the TNB frame we may now discuss the planes
that are formed when using the TNB frame. I shall refer to these collectively
as the TNB Planes. They are the
• Rectifying plane: the plane formed by the vectors T and B.
• Normal plane: the plane formed by the vectors N and B.
• Osculating plane: the plane formed by the vectors T and N.
The diagram below demonstrates this:
8

9. x y
z
T N
B
Osculating Plane
Rectifying Plane Normal Plane
If we recall that the equation of a plane may be determined by knowing a point
P(x, y, z) and a perpendicular vector n = Ai + Bj + Cj. I exclude the use of
the word “normal” as that has been taken by the N vector in the TNB frame.
Hopefully this will cut down on confusion. Since we are using the TNB frame,
we may exploit the following facts:
• The vector T (or any vector parallel to it, so we do not need the unit
vector itself) will act as a perpendicular vector to the Normal Plane.
• The vector N (or any vector parallel to it, so we do not need the unit
vector itself) will act as a perpendicular vector to the Rectifying Plane.
• The vector B (or any vector parallel to it, so we do not need the unit
vector itself) will act as a perpendicular vector to the Osculating Plane.
With this in mind, we may derive equations for these planes. Suppose r(t) =
f(t)i + g(t)j + h(t)k is 2nd-order smooth (so neither r′
(t) = 0 nor r′′
(t) = 0 at
any t value in the domain). If we wish to find the TNB planes at a point t = t0
9

10. along r(t) then the TNB Planes may be determined via the following formulas
TNB Plane Equation
Normal Plane f′
(t0)(x − f(t0)) + g′
(t0)(y − g(t0)) + h′
(t0)(z − h(t0)) = 0
Osculating Plane
x − f(t0) y − g(t0) z − h(t0)
f′
(t0) g′
(t) h′
(t0)
f′′
(t0) g′′
(t0) h′′
(t0)
= 0
Rectifying Plane
x − f(t0) y − g(t0) z − h(t0)
f′
(t0) g′
(t0) h′
(t0)
g′
(t0) h′
(t0)
g′′
(t0) h′′
(t0)
h′
(t0) f′
(t0)
h′′
(t0) f′′
(t0)
f′
(t0) g′
(t0)
f′′
(t0) g′′
(t0)
= 0
Note that each of the evaluations for these determinants occur at the point in
question, t = t0. Also note that x, y and z are variables for the equation and
are not replaced by a value.
Example
Example 5: Find the rectifying plane, osculating plane and normal plane for
the following vector function r(t) = cos(t)i + sin(t)j + tk at the value t = 0.
Answer: To answer this question we need the derivatives of the components.
So we have
Component First Derivative Second Derivative
f(t) = cos(t) − sin(t) − cos(t)
g(t) = sin(t) cos(t) − sin(t)
h(t) = t 1 0
Now we need to have these evaluated at t = 0.
Component First Derivative Second Derivative
f(0) = 1 0 −1
g(0) = 0 1 0
h(0) = 0 1 0
Now we substitute these values into our equations provided above and compute.
10

11. TNB Plane Equation Simplified Result
Normal Plane 0(x − 1) + 1(y − 0) + 1(z − 0) = 0 y + z = 0
Osculating Plane
x − 1 y − 0 z − 0
0 1 1
−1 0 0
= 0 y − z = 0
Rectifying Plane
x − 1 y − 0 z − 0
0 1 1
1 1
0 0
1 0
0 −1
0 1
−1 0
= 0 x = 1
11