76 CHAPTER 6. VECTOR PRODUCTS

This is√(2,1,4,2) · (2,1,4,2) = 5.

The dot product satisfies a fundamental inequality known as the Cauchy Schwarz in-equality. It has already been proved but here is another proof. This proof will be basedonly on the above axioms for the dot product.

Theorem 6.1.5 The dot product satisfies the inequality

|a ·b| ≤ |a| |b| . (6.6)

Furthermore equality is obtained if and only if one of a or b is a scalar multiple of theother.

Proof: First note that if b= 0, both sides of 6.6 equal zero and so the inequality holdsin this case. Indeed,

a ·0 = a·(0+0) = a ·0+a ·0

so a ·0= 0. Therefore, it will be assumed in what follows that b ̸= 0.Define a function of t ∈ R

f (t) = (a+ tb) · (a+ tb) .

Then by 6.2, f (t)≥ 0 for all t ∈ R. Also from 6.3,6.4,6.1, and 6.5

f (t) = a · (a+ tb)+ tb · (a+ tb)

= a ·a+ t (a ·b)+ tb ·a+ t2b ·b

= |a|2 +2t (a ·b)+ |b|2 t2.

Now

f (t) = |b|2(

t2 +2ta ·b|b|2

+|a|2

|b|2

)

= |b|2t2 +2t

a ·b|b|2

+

(a ·b|b|2

)2

−

(a ·b|b|2

)2

+|a|2

|b|2

= |b|2

(t +a ·b|b|2

)2

+

 |a|2|b|2−

(a ·b|b|2

)2≥ 0

for all t ∈ R. In particular f (t)≥ 0 when t = −(a ·b/ |b|2

)which implies

|a|2

|b|2−

(a ·b|b|2

)2

≥ 0. (6.7)

Multiplying both sides by |b|4,|a|2 |b|2 ≥ (a ·b)2

which yields 6.6.From Theorem 6.1.5, equality holds in 6.6 whenever one of the vectors is a scalar

multiple of the other. It only remains to verify this is the only way equality can occur.