8.3 Dot product

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DEFINITION 8.3.1 If X =

2

4

a1

b1

c1

3

5 and Y =

2

4

a2

b2

c2

3

5, then X ¢ Y , the

dot product of X and Y , is de¯ned by

X ¢ Y = a1a2 + b1b2 + c1c2:

8.3. DOT PRODUCT 157

¢

¢

¢

¢®

¢

¢

¢

¢®

- -

6 6

½

½

½

½> ½

½

½

½=

A

B

A

B

v =

-

AB ¡v =

-

BA

Figure 8.5: The negative of a vector.

¢

¢

¢

¢®

¢

¢

¢

¢®

- -

6 6

½

½

½

½>

½

½

½

½>

½

½

½

½>

»»»»»»»»:

XXXXz

A

B

C

D

A

B

C

(a) (b)

-

AB=

-

CD

-

AC=

-

AB +

-

BC

-

BC=

-

AC ¡

-

AB

Figure 8.6: (a) Equality of vectors; (b) Addition and subtraction of vectors.

The dot product has the following properties:

(i) X ¢ (Y + Z) = X ¢ Y + X ¢ Z;

(ii) X ¢ Y = Y ¢ X;

(iii) (tX) ¢ Y = t(X ¢ Y );

(iv) X ¢ X = a2 + b2 + c2 if X =

2

4

a

b

c

3

5;

(v) X ¢ Y = XtY ;

(vi) X ¢ X = 0 if and only if X = 0.

The length of X is de¯ned by

jjXjj =

p

a2 + b2 + c2 = (X ¢ X)1=2:

We see that jjPjj = OP and more generally jj

-

P1P2 jj = P1P2, the

distance between P1 and P2.

158 CHAPTER 8. THREE{DIMENSIONAL GEOMETRY

¢

¢

¢®

-

6

y

z

x

O

Q

Q

Q

Q

Q

Q

,

,

,

,

,

,

¢

¢

¢

®

®

i

j -

6

-

6

ck

k

Q

Q

Q

Q

Q

Qs

ai ai + bj ¢

¢

¢

¢

bj

P = ai + bj + ck

Figure 8.7: Position vector as a linear combination of i; j and k.

Vectors having unit length are called unit vectors.

The vectors

i =

2

4

1

0

0

3

5 ; j =

2

4

0

1

0

3

5 ; k =

2

4

0

0

1

3

5

are unit vectors. Every vector is a linear combination of i; j and k:

2

4

a

b

c

3

5 = ai + bj + ck:

(See Figure 8.7.)

It is easy to prove that

jjtXjj = jtj ¢ jjXjj;

if t is a real number. Hence if X is a non{zero vector, the vectors

§

1

jjXjj

X

are unit vectors.

A useful property of the length of a vector is

jjX § Y jj2 = jjXjj2 § 2X ¢ Y + jjY jj2: (8.2)

8.3. DOT PRODUCT 159

The following important property of the dot product is widely used in

mathematics:

THEOREM 8.3.1 (The Cauchy{Schwarz inequality)

If X and Y are vectors in R3, then

jX ¢ Y j · jjXjj ¢ jjY jj: (8.3)

Moreover if X 6= 0 and Y 6= 0, then

X ¢ Y = jjXjj ¢ jjY jj , Y = tX; t > 0;

X ¢ Y = ¡jjXjj ¢ jjY jj , Y = tX; t < 0:

Proof. If X = 0, then inequality 8.3 is trivially true. So assume X 6= 0.

Now if t is any real number, by equation 8.2,

0 · jjtX ¡ Y jj2 = jjtXjj2 ¡ 2(tX) ¢ Y + jjY jj2

= t2jjXjj2 ¡ 2(X ¢ Y )t + jjY jj2

= at2 ¡ 2bt + c;

where a = jjXjj2 > 0; b = X ¢ Y; c = jjY jj2.

Hence

a(t2 ¡

2b

a

t +

c

a

) ¸ 0

µ

t ¡

b

a

¶2

+

ca ¡ b2

a2 ¸ 0 :

Substituting t = b=a in the last inequality then gives

ac ¡ b2

a2 ¸ 0;

so

jbj · pac = papc

and hence inequality 8.3 follows.

To discuss equality in the Cauchy{Schwarz inequality, assume X 6= 0

and Y 6= 0.

Then if X ¢ Y = jjXjj ¢ jjY jj, we have for all t

jjtX ¡ Y jj2 = t2jjXjj2 ¡ 2tX ¢ Y + jjY jj2

= t2jjXjj2 ¡ 2tjjXjj ¢ jjY jj + jjY jj2

= jjtX ¡ Y jj2:

Taking t = jjXjj=jjY jj then gives jjtX ¡ Y jj2 = 0 and hence tX ¡ Y = 0.

Hence Y = tX, where t > 0. The case X¢Y = ¡jjXjj¢jjY is proved similarly.

160 CHAPTER 8. THREE{DIMENSIONAL GEOMETRY

COROLLARY 8.3.1 (The triangle inequality for vectors)

If X and Y are vectors, then

jjX + Y jj · jjXjj + jjY jj: (8.4)

Moreover if X 6= 0 and Y 6= 0, then equality occurs in inequality 8.4 if and

only if Y = tX, where t > 0.

Proof.

jjX + Y jj2 = jjXjj2 + 2X ¢ Y + jjY jj2

· jjXjj2 + 2jjXjj ¢ jjY jj + jjY jj2

= (jjXjj + jjY jj)2

and inequality 8.4 follows.

If jjX + Y jj = jjXjj + jjY jj, then the above proof shows that

X ¢ Y = jjXjj ¢ jjY jj:

Hence if X 6= 0 and Y 6= 0, the ¯rst case of equality in the Cauchy{Schwarz

inequality shows that Y = tX with t > 0.

The triangle inequality for vectors gives rise to a corresponding inequality

for the distance function:

THEOREM 8.3.2 (The triangle inequality for distance)

If A; B; C are points, then

AC · AB + BC: (8.5)

Moreover if B 6= A and B 6= C, then equality occurs in inequality 8.5 if and

only if

-

AB= r

-

AC, where 0 < r < 1.

Proof.

AC = jj

-

AC jj = jj

-

AB +

-

BC jj

· jj

-

AB jj + jj

-

BC jj

= AB + BC:

Moreover if equality occurs in inequality 8.5 and B 6= A and B 6= C, then

X =

-

AB6= 0 and Y =

-

BC6= 0 and the equation AC = AB + BC becomes

8.4. LINES 161

jjX + Y jj = jjXjj + jjY jj. Hence the case of equality in the vector triangle

inequality gives

Y =

-

BC= tX = t

-

AB; where t > 0:

Then

-

BC =

-

AC ¡

-

AB= t

-

AB

-

AC = (1 + t)

-

AB

-

AB = r

-

AC;

where r = 1=(t + 1) satis¯es 0 < r < 1.

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