2.4 Tangent, Normal and Binormal Vectors

90

CHAPTER 2. VECTOR FUNCTIONS

2.4

Tangent, Normal and Binormal Vectors

Three vectors play an important role when studying the motion of an object

along a space curve. These vectors are the unit tangent vector, the principal normal vector and the binormal vector. We have already de¡­ned the unit tangent

vector. In this section, we de¡­ne the other two vectors.

Let us start by reviewing the de¡­nition of the unit tangent vector.

De¡­nition 147 (Unit Tangent Vector) Let C be a smooth curve with posi!

tion vector !

r (t). The unit tangent vector, denoted T (t) is de¡­ned to be

!

!

r 0 (t)

T (t) = !0

k r (t)k

2.4.1

Normal and Binormal Vectors

De¡­nition 148 (Normal Vector) Let C be a smooth curve with position vector !

r (t). The principal normal vector or simply the normal vector, denoted

!

N (t) is de¡­ned to be:

!0

!

T (t)

N (t) = !

(2.13)

T 0 (t)

The name of this vector suggests that it is normal to something, the question

!

!

is to what? By de¡­nition, T is a unit vector, that is T (t) = 1. From

!

!

!

!

proposition 120, it follows that T 0 (t) ? T (t). Thus, N (t) ? T (t). In fact,

!

!

N (t) is a unit vector, perpendicular to T pointing in the direction where the

curve is bending.

De¡­nition 149 (Binormal Vector) Let C be a smooth curve with position

!

vector !

r (t). The binormal vector, denoted B (t), is de¡­ned to be

!

!

B (t) = T (t)

!

N (t)

!

!

Since both T (t) and N (t) are unit vectors and perpendicular, it follows that

!

!

!

B (t) is also a unit vector. It is perpendicular to both T (t) and N (t).

Example 150 Consider the circular helix !

r (t) = hcos t; sin t; ti. Find the unit

tangent, normal and binormal vectors.

!

!

r 0 (t)

Unit Tangent: Since T (t) = !0

, we need to compute !

r 0 (t) and

k r (t)k

k!

r 0 (t)k.

!

r 0 (t) = h sin t; cos t; 1i

2.4. TANGENT, NORMAL AND BINORMAL VECTORS

91

and

!

r 0 (t)

=

=

Thus

p

p

!

T (t) =

sin2 t + cos2 t + 1

2

sin t cos t 1

p ; p ;p

2

2

2

!0

!

!

!

T (t)

, we need to compute T 0 (t) and T 0 (t) .

Normal: Since N (t) = !

T 0 (t)

!0

T (t) =

sin t

cos t

p ; p ;0

2

2

and

!0

T (t)

=

=

Thus

s

cos2 t sin2 t

+

2

2

1

p

2

!

N (t) = h cos t;

sin t; 0i

Binormal:

!

B (t)

!

!

T (t) N (t)

sin t cos t 1

p ; p ;p

=

2

2

2

!

sin t

cos t 1

p ; p ;p

B (t) =

2

2

2

=

h cos t;

sin t; 0i

The pictures below (¡­gures 2.5, 2.6 and 2.7) show the helix for t 2 [0; 2 ]

!

!

!

as well as the three vectors T (t), N (t) and B (t) plotted for various

values of t. If the three vectors do not appear to be exactly orthogonal, it

is because the scale is not the same in the x; y and z directions.

2.4.2

Osculating and Normal Planes

De¡­nition 151 (Osculating and Normal Planes) Let C be a smooth curve

with position vector !

r (t). Let P be a point on the curve corresponding to !

r (t0 )

for some value of t.

!

!

1. The plane through P determined by N (t0 ) and B (t0 ) is called the normal

!

plane of C at P . Note that its normal will be T (t).

92

CHAPTER 2. VECTOR FUNCTIONS

!

!

!

Figure 2.5: Helix and the vectors T (0), N (0) and B (0)

!

!

!

Figure 2.6: Helix and the vectors T (1), N (1) and B (1)

2.4. TANGENT, NORMAL AND BINORMAL VECTORS

93

!

!

!

Figure 2.7: Helix and the vectors T (4), N (4) and B (4)

!

!

2. The plane through P determined by T (t0 ) and N (t0 ) is called the oscu!

lating plane of C at P . Note that its normal will be B (t).

Example 152 Find the normal and osculating planes to the helix given by

!

.

r (t) = hcos t; sin t; ti at the point 0; 1;

2

Earlier, we found that

!

T (t) =

sin t cos t 1

p ; p ;p

2

2

2

!

N (t) = h cos t;

sin t; 0i

and

!

B (t) =

At the point 0; 1;

2

, that is when t =

!

T

=

2

!

N

cos t 1

p ;p

2

2

sin t

p ;

2

2

2

, we have

1

1

p ; 0; p

2

2

= h0; 1; 0i

94

CHAPTER 2. VECTOR FUNCTIONS

and

!

B

2

1

1

p ; 0; p

2

2

=

Normal Plane: It is the plane through

0; 1;

1

1

p ; 0; p . Thus its equation is

2

2

1

1

p (x 0) + p z

2

2

p

Multiplying each side by 2 gives

(x

0) + z

2

=

=0

2

2

!

with normal T

2

=0

or

z

x=

2

Osculating Plane: It is the plane through 0; 1;

1

1

p ; 0; p

2

2

!

with normal B

2

2

=

Thus its equation is

1

1

p (x 0) + p z

2

2

p

Multiplying each side by 2 gives

(x

0) + z

2

2

=0

=0

or

x+z =

2

Make sure you have read, studied and understood what was done above

before attempting the problems.

2.4.3

Problems

Do # 11, 13, 37, 39, 43 at the end of 10.3 in your book

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