The Right-Hand Rule for Finding the Direction of the Standard Unit Normal

Let

(v,

^w

)

be an ordered pair of two nonparallel vectors. Represent

v

and ^w by directed line segments with a common initial point. Point the fingers of your right hand along the first vector

v

and then curl them toward the second vector ^w, as in Figure 9. The direction that the thumb points during this process is the direction of the standard unit normal vector for the vectors

v,

^w. Look again at Figure

8.

The direction of vector ⁿ is the standard unit normal for the ordered pair

(v,

^w

)

^.

Standard unit normal

_. Figure 9

� EXAMPLE 5 Find the standard unit normal for the pairs

(i,j)

and

(j,k)

and

(k, i).

Find also the standard unit normal for the pairs

(j, i)

and

(k,j)

and

(i,k).

Solution If you curl the fingers of your right hand from

i

toward

j,

then your thumb will point in the direction of the positive z-axis. Thus the standard unit normal for the pair

(i,j)

is

k.

Similar reasoning shows that the standard unit normal for

(j,i)

is

-k.

We leave it as an exercise to check that the standard unit normal for the pair

(j, k)

is

i,

and the standard unit normal for the pair

(k, i)

is

j.

In addition, the standard unit normal for

(k,j)

is

-i,

and the standard unit normal for

(i, k)

is

-j.

^<Ill

INSIGHT

According to the right-hand rule, the standard unit normal vector for the ordered pair (v, w) points in the opposite direction to the standard unit normal vector for the ordered pair (w, v). Example 4 provides particular cases of this general fact.

If v ^x w i- ^0, then we may form the unit vector dir(v ^x w) =

ll

v ^x w

r

1 (v ^x w).

Because v

X

w is perpendicular to both v and w, the vector dir(v

X

w) must be the standard unit normal for either the ordered pair (v, w) or the ordered pair (w, v).

Our next theorem tells us which. It also specifies the length of v

X

w and tells us precisely when the equation v

X

w ^{= 0} holds.

THEOREM 3 Let v and w be vectors. Then:

a.

ll

v

X

w

ll

^{2 =}

ll

^v

ll

2

ll

^w

ll

^{2 - (v · w)2.}

b. If v and w are nonzero, then

ll

^v

^X

^w

ll

=

ll

v

ll ll

^w

ll

sin(O) where 0 ^E

[O, rr]

denotes the angle between v and w.

c. v and w are parallel if and only if v

X

w = ^0.

d. If v and w are not parallel, then v ^x w points in the direction of the standard unit normal for the pair (v, w). In particular, dir(v

X

w) =

ll

v

X

w

r

1 (v

X

w) is the standard unit normal vector for the pair (v, w).

Proot Part a is an identity among the entries of v = (v1, v2, v3) and w = (w1, w2, w3).

We have

ll

v

X

^w

ll

^{2 =} (v2w3 - w2v3)2

+

(v1w3 - w1v3)2

+

(v1w2 - w1v2)2

=

(Vt+ V� + V�)(wr + W� + W�) - (V1W1 + VzWz + V3W3)2

=

ll

^v

ll

²

ll

^w

ll

^{2 - (v · w)2.}

The second equality in this chain is not obvious, but it may be verified by multiplying everything out.

If neither v nor w is the zero vector, then the scalars

ll

^v

ll

^and

ll

^w

ll

are nonzero.

In this case, the identity of part a becomes

ll

^v

^X

^w

ll

2 =

ll

v

ll

2

ll

w

f

- (v · w)2 =

ll

v

ll

2

ll

w

ll

2

(i

^{- (v}

_ll

_v

_{ll ll}

_w

_ll

⁼

^ll

^v

^ll

²

^ll

^w

^ll

²

⁽

^{1- cos2(0)}

⁾

^,

where the last equality is obtained by using formula (9.3.6). It follows that

ll

^v

^X

^w

ll

2 =

ll

v

ll

2

ll

w

ll

2sin2(0) and, on taking the square root,

ll

v

X

w

ll

=

ll

v

ll ll

w

ll

lsin(O)I. Because ^{0 :5} 0 ^:5 rr, it follows that ^{0 :5} sin(O). Therefore lsin(O)I = sin(O) and

ll

v

X

w

ll

=

ll

v

ll ll

w

ll

sin(O), as asserted in part b.

Theorem 3 of Section 3 tells us that vectors v and w are parallel ^if and only ^if Iv· wl =

ll

v

ll ll

w

ll

^· Using the identity of part a, we conclude that v and w are parallel if and only ^if

ll

v

X

w

ll

= ^0. Because ⁰ is the only zero length vector, assertion c follows.

For part d, we will limit our verification to a few special cases. Notice that

i X j

=

k, j X k

=

i,

and

k X i

=

j.

Based on our observations in Example 4, we conclude that,

for the basic vectors

i,j,

and

k,

the operation of cross product produces the standard unit normal. Geometric reasoning can be used to show that the cross product v

X

w always points in the direction of the standard unit normal for the vectors v, w. •

� EXAMPLE 6 Let v=(l,-3,2) and w=(l,-1,4). What is the standard unit normal vector for (v, w)?

Cross Products and the Calculation of Area

9.4 The Cross Product and Triple Product 759

Solution We calculate the vector

v x w = ( ( -3)

^{. 4}

- ( -1) . 2) i - ( 1 .

⁴

- 1 . 2 )j + ( 1 . ( -1) - 1 . ( -3)) k

=

-10i - 2j + 2k

and its length

llv x wll = + (-2)2 + 22 = v'T08 = = 6v'3.

Part d of Theorem

2

tells us that

is the vector we seek. <11111

Now we learn a connection between cross products and areas of triangles and parallelograms. To be specific, if

v

^and

w

are two nonparallel vectors that are

-

-represented by directed line segments

OP

and

OQ

with common initial point 0, then we speak of triangle

OPQ

as "the triangle determined by the vectors

v

and

w."

See Figure

10

for this triangle. The parallelogram determined by

v

and

w

is shown in Figure

11.

Notice that, when we declare

OPQ

to be the triangle deter­

mined by

v

^and

w,

the third side

QP

represents

v-w.

Had we positioned

v

^{so that}

its initial point coincided with the terminal point

Q

of

w,

then we would have obtained triangle

OQR

with third side

v + w.

Notice that the areas of triangles

OPQ

and

OQR

are the same because each is half the area of the parallelogram.

Thus had we declared

OQR

as the triangle determined by the vectors

v

^and

w

instead of 0

PQ,

we would have been speaking of a triangle with the same area.

_. Figure 11

THEOREM ₄ Suppose that

v

^and

w

are nonparallel vectors. The area of the triangle determined by

v

^and

w

^is

� II v

X

w II

^· The area of the parallelogram determined by

v

^and

w

^is

II ^v

^X

^w II·

Proof. As we see from Figure

10,

the altitude from vertex

P

to base

OQ

has length

llvll

^sin(

O

). The area of triangle

OPQ,

namely half the product of its base and height, is therefore

� llwll llvllsin(O).

This quantity is

� llv x wll

by Theorem 3b. As

A Physical Application

In document Calculus Multivariable 2nd Edition Blank & Krantz - Vector Calculus PDF (Page 55-58)