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# Vectors. Lecture 9

## 1.

VectorsKARASHBAYEVA ZH.O., SENIOR-LECTURER

## 2.

Contents:Scalar product of vectors and its properties

Vector product of vectors and its properties

Mixed product of vectors and its properties

Collinear and coplanar vectors

## 3.

THE DOT PRODUCTDefinition 1

• If a = ‹a1, a2, a3› and b = ‹b1, b2, b3›, then

the dot product of a and b is the number a • b given by:

a • b = a1b1 + a2b2 + a3b3

## 4.

SCALAR PRODUCTThe result is not a vector.

It is a real number, that is, a scalar.

◦ For this reason, the dot product is sometimes

called the scalar product (or inner product).

## 5.

DOT PRODUCTThough Definition 1 is given for three-dimensional (3-D) vectors, the dot product

of two-dimensional vectors is defined in

a similar fashion:

‹a1, a2› ∙ ‹b1, b2› = a1b1 + a2b2

## 6.

PROPERTIES OF DOT PRODUCTIf a, b, and c are vectors in V3 and c is a scalar, then

1. a a=|a|

2. a b = b a

3. a (b + c) = a b + a c

4. (ca) b = c(a b) = a (cb)

5. 0 a = 0

2

## 7.

DOT PRODUCT PROPERTY 1a∙a

2

2

2

= a1 + a2 + a3

2

= |a|

Proof

## 8.

DOT PRODUCT PROPERTY 3a • (b + c)

= ‹a1, a2, a3› ∙ ‹b1 + c1, b2 + c2, b3 + c3›

= a1(b1 + c1) + a2(b2 + c2) + a3(b3 + c3)

= a1b1 + a1c1 + a2b2 + a2c2 + a3b3 + a3c3

= (a1b1 + a2b2 + a3b3) + (a1c1 + a2c2 + a3c3)

=a∙b+a∙c

Proof

## 9.

GEOMETRIC INTERPRETATIONThe dot product a • b can be given a geometric interpretation in terms of the angle θ

between a and b.

◦ This is defined to be the angle between the representations of a and b that start

at the origin, where 0 ≤ θ ≤ π.

## 10.

GEOMETRIC INTERPRETATIONIn other words, θ is the angle between the line segments

OA and OB here.

◦ Note that if a and b

are parallel vectors,

then θ = 0 or θ = π.

## 11.

DOT PRODUCT—DEFINITIONIf θ is the angle between the vectors

a and b, then

a ∙ b = |a||b| cos θ

Theorem

## 12.

DOT PRODUCT—DEFINITIONIf we apply the Law of Cosines to triangle OAB here, we get:

|AB|2 = |OA|2 + |OB|2 – 2|OA||OB| cos θ

◦ Observe that

the Law of Cosines

still applies in

the limiting cases

when θ = 0 or π, or

a = 0 or b = 0

Proof—Equation 1

## 13.

DOT PRODUCT—DEFINITIONSo, Equation 1 becomes:

|a – b|2 = |a|2 + |b|2 – 2|a||b| cos θ

Proof—Equation 2

## 14.

DOT PRODUCT—DEFINITIONProof

Using Properties 1, 2, and 3 of the dot product, we can rewrite the left side of

the equation as follows:

|a – b|2 = (a – b) ∙ (a – b)

=a∙a–a∙b–b∙a+b∙b

= |a|2 – 2a ∙ b + |b|2

## 15.

DOT PRODUCT—DEFINITIONTherefore, Equation 2 gives:

|a|2 – 2a ∙ b + |b|2 = |a|2 + |b|2 – 2|a||b| cos θ

◦ Thus,

or

–2a ∙ b = –2|a||b| cos θ

a ∙ b = |a||b| cos θ

Proof

## 16.

THE CROSS PRODUCTThe cross product a x b of two vectors a and b, unlike the

dot product, is a vector.

◦ For this reason, it is also called the vector product.

◦ Note that a x b is defined only when a and b are three-dimensional (3-D)

vectors.

## 17.

THE CROSS PRODUCTIf a = ‹a1, a2, a3› and b = ‹b1, b2, b3›, then

the cross product of a and b is the vector

a x b = ‹a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1›

Definition 2

## 18.

CROSS PRODUCTEquation 3

We see that the cross product of the vectors

a = a1i +a2j + a3k and b = b1i + b2j + b3k

is:

a2

a b =

b2

a3

a1 a3

a1 a2

i−

j+

k

b3

b1 b3

b1 b2

i

j

a b = a1 a2

b1 b2

k

a3

b3

## 19.

CROSS PRODUCTExample 1

If a = <1, 3, 4> and b = <2, 7, –5>, then

i j k

a b = 1 3 4

2 7 −5

3 4

1 4

1 3

=

i−

j+

k

7 −5

2 −5

2 7

= (−15 − 28)i − (−5 − 8) j + (7 − 6)k

= −43i + 13 j + k

## 20.

CROSS PRODUCTExample 2

Show that a x a = 0 for any vector a in V3.

◦ If a = <a1, a2, a3>,

then

i

a a = a1

a1

j

a2

a2

k

a3

a3

= (a2 a3 − a3 a2 ) i − (a1a3 − a3a1 ) j

+ (a1a2 − a2 a1 ) k

= 0i − 0 j+ 0k = 0

## 21.

CROSS PRODUCTTheorem

The vector a x b is orthogonal to both a and b.

## 22.

CROSS PRODUCTProof

In order to show that a x b is orthogonal to

a, we compute their dot product as follows

(a b) a

a2

=

b2

a3

a1

a1 −

b3

b1

a3

a1

a2 +

b3

b1

a2

a3

b2

= a1 (a2b3 − a3b2 ) − a2 (a1b3 − a3b1 ) + a3 (a1b2 − a2b1 )

= a1a2b3 − a1b2 a3 − a1a2b3 + b1a2 a3 + a1b2 a3 − b1a2 a3

=0

## 23.

CROSS PRODUCTLet a and b be represented by directed

line segments with the same initial point,

as shown.

Then, the cross product a x b points in a direction

perpendicular to the plane through a and b.

## 24.

CROSS PRODUCTIt turns out that the direction of a x b is given by the right-hand rule, as follows.

If the fingers of your right hand curl in the direction of a rotation (through an angle

less than 180°) from a to b, then your thumb points in the direction of a x b.

## 25.

CROSS PRODUCTWe know the direction of the vector a x b.

The remaining thing we need to complete its geometric description is its

length |a x b|.

◦ This is given by the following theorem.

If θ is the angle between a and b (so 0 ≤ θ ≤ π), then

|a x b| = |a||b| sin θ

## 26.

CROSS PRODUCTProof

From the definitions of the cross product and length of a vector, we have:

|a x b|2 = (a2b3 – a3b2)2 + (a3b1 – a1b3)2 + (a1b2 – a2b1)2

= a22b32 – 2a2a3b2b3 + a32b22 + a32b12 – 2a1a3b1b3 + a12b32 + a12b22 – 2a1a2b1b2 + a22b12

= (a12 + a22 + a32)(b12 + b22 + b32) – (a1b1 + a2b2 + a3b3)2

= |a|2|b|2 – (a . b)2

= |a|2|b|2 – |a|2|b|2 cos2θ

= |a|2|b|2 (1 – cos2θ)

= |a|2|b|2 sin2θ

## 27.

CROSS PRODUCTTaking square roots and observing that

Proof

sin = sin

2

because sin θ ≥ 0 when

0 ≤ θ ≤ π, we have:

|a x b| = |a||b| sin θ

The length of the cross product a x b is equal to the area of the parallelogram

determined by a and b.

## 28.

CROSS PRODUCTExample 3

Find a vector perpendicular to the plane that passes through

the points

P(1, 4, 6), Q(-2, 5, -1), R(1, -1, 1)

## 29.

CROSS PRODUCTExample 3

The vector PQ PR is perpendicular to both PQ and PR .

◦ Therefore, it is perpendicular to the plane through P, Q, and R.

PQ = (−2 − 1) i + (5 − 4) j + (−1 − 6) k

= −3i + j − 7k

PR = (1 − 1) i + (−1 − 4) j + (1 − 6) k

= −5 j − 5k

## 30.

CROSS PRODUCTExample 3

We compute the cross product of these vectors:

i

j k

PQ PR = −3 1 −7

0 −5 −5

= (−5 − 35) i − (15 − 0) j + (15 − 0) k

= −40i − 15 j + 15k

Therefore, the vector ‹-40, -15, 15› is perpendicular to the given plane.

◦ Any nonzero scalar multiple of this vector,

such as ‹-8, -3, 3›, is also perpendicular

to the plane.

## 31.

CROSS PRODUCTExample 4

Find the area of the triangle with vertices

P(1, 4, 6), Q(-2, 5, -1), R(1, -1, 1)

## 32.

CROSS PRODUCTExample 4

In Example 3, we computed that

PQ PR = −40, −15,15

◦ The area of the parallelogram with adjacent sides PQ and

PR is the length of this cross product:

PQ PR = (−40) 2 + (−15) 2 + 152 = 5 82

The area A of the triangle PQR is half the area of this parallelogram, that is: 52

82

## 33.

CROSS PRODUCT PROPERTIESIf a, b, and c are vectors and c is a scalar, then

1. a x b = –b x a

2. (ca) x b = c(a x b) = a x (cb)

3. a x (b + c) = a x b + a x c

4. (a + b) x c = a x c + b x c

5. a · (b x c) = (a x b) · c

6. a x (b x c) = (a · c)b – (a · b)c

Theorem

## 34.

CROSS PRODUCT PROPERTY 5Let

a = <a1, a2, a3>,

b = <b1, b2, b3>, c = <c1, c2, c3>

Then,

a · (b x c) = a1(b2c3 – b3c2) + a2(b3c1 – b1c3) + a3(b1c2 – b2c1)

= a1b2c3 – a1b3c2 + a2b3c1 – a2b1c3 + a3b1c2 – a3b2c1

= (a2b3 – a3b2)c1 + (a3b1 – a1b3)c2+ (a1b2 – a2b1)c3

=(a x b) · c

Proof—Equation 4

## 35.

SCALAR TRIPLE PRODUCTThe product a . (b x c) that occurs

in Property 5 is called the scalar triple product

of the vectors a, b, and c.

a1 a2

a (b c) = b1 b2

c1 c2

a3

b3

c3

## 36.

SCALAR TRIPLE PRODUCTSThe geometric significance of the scalar triple product can be seen by considering

the parallelepiped determined by the vectors a, b, and c.

## 37.

SCALAR TRIPLE PRODUCTSThe area of the base parallelogram is:

A = |b x c|

If θ is the angle between a and b x c,

then the height h of the parallelepiped is:

h = |a||cos θ|

We must use |cos θ| instead of cos θ in

case θ > π/2.

## 38.

SCALAR TRIPLE PRODUCTSHence, the volume of the parallelepiped is:

V = Ah

= |b x c||a||cos θ|

= |a · (b x c)|

◦ Thus, we have proved the following formula.

The volume of the parallelepiped determined by the vectors a, b, and c is the magnitude of

their scalar triple product:

V = |a ·(b x c)|