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Review of vectors. The dot and cross products, Study Guides, Projects, Research of Calculus

If a and b are two vectors, their cross product is denoted by a × b. b × a = − a × b. | a × b| = | a|| b|sinθ. The cross product of a = 〈a1,a2,a3〉 and b = 〈b1, ...

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Calculus 3
Lia Vas
Review of vectors. The dot and cross products
Review of vectors in two and three dimensions. A two-dimensional vector is an ordered
pair ~a =ha1, a2iof real numbers. The coordinate representation of the vector ~a corresponds to the
arrow from the origin (0,0) to the point (a1, a2).Thus, the length of ~a is |~a|=qa2
1+a2
2.Analogously,
we have the following.
Athree-dimensional vector is an ordered
triple
~a =ha1, a2, a3i
of real numbers. The coordinate representation
of the vector ~a corresponds to the arrow from the
origin (0,0,0) to the point (a1, a2, a3).
The length of ~a is
|~a|=qa2
1+a2
2+a2
3.
Using the coordinate representation the vector addition and scalar multiplication can be realized
as follows.
Vector Addition - by coordinates ha1, a2, a3i+hb1, b2, b3i=ha1+b1, a2+b2, a3+b3i
Scalar multiplication - by coordinates kha1, a2, a3i=hka1, ka2, k a3i
This corresponds to the geometrical representation illustrated in the figure below.
Using its coordinates, a vector ~a =ha1, a2iin
xy-plane can be represented as a linear combina-
tion of vectors~
i=h1,0iand ~
j=h0,1ias follows.
~a =a1
~
i+a2~
j
The coordinates of a vector and geometrical
representation have analogous relation in three
dimensional space.
1
pf3
pf4
pf5

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Calculus 3

Lia Vas

Review of vectors. The dot and cross products

Review of vectors in two and three dimensions. A two-dimensional vector is an ordered

pair ~a = 〈a 1

, a 2

〉 of real numbers. The coordinate representation of the vector ~a corresponds to the

arrow from the origin (0, 0) to the point (a 1 , a 2 ). Thus, the length of ~a is |~a| =

a

2

1

  • a

2

2

. Analogously,

we have the following.

A three-dimensional vector is an ordered

triple

~a = 〈a 1

, a 2

, a 3

of real numbers. The coordinate representation

of the vector ~a corresponds to the arrow from the

origin (0, 0 , 0) to the point (a 1

, a 2

, a 3

The length of ~a is

|~a| =

a

2

1

  • a

2

2

  • a

2

3

Using the coordinate representation the vector addition and scalar multiplication can be realized

as follows.

Vector Addition - by coordinates 〈a 1

, a 2

, a 3

〉 + 〈b 1

, b 2

, b 3

〉 = 〈a 1

  • b 1

, a 2

  • b 2

, a 3

  • b 3

Scalar multiplication - by coordinates k〈a 1

, a 2

, a 3

〉 = 〈ka 1

, ka 2

, ka 3

This corresponds to the geometrical representation illustrated in the figure below.

Using its coordinates, a vector ~a = 〈a 1

, a 2

〉 in

xy-plane can be represented as a linear combina-

tion of vectors

i = 〈 1 , 0 〉 and

j = 〈 0 , 1 〉 as follows.

~a = a 1

i + a 2

j

The coordinates of a vector and geometrical

representation have analogous relation in three

dimensional space.

If

i = 〈 1 , 0 , 0 〉, ~j = 〈 0 , 1 , 0 〉, and

k = 〈 0 , 0 , 1 〉 and a vector ~a can be represented as ~a = 〈a 1

, a 2

, a 3

then

~a = a 1

i + a 2

j + a 3

k.

In the next section, it will be relevant to determine the coordinates of the vector from one point

to the other. Let P = (a 1

, a 2

, a 3

) and Q = (b 1

, b 2

, b 3

), be two points in space. If O denotes the origin

(0, 0 , 0), then the vector

OP can be represented as 〈a 1 , a 2 , a 3 〉, and the vector

OQ as 〈b 1 , b 2 , b 3 〉.

Since

OP +

P Q =

OQ we have that

P Q =

OQ −

OP = 〈b 1 , b 2 , b 3 〉 − 〈a 1 , a 2 , a 3 〉 = 〈b 1 − a 1 , b 2 − a 2 , b 3 − a 3 〉.

In some cases, we may need to find the vector

with same direction and sense as a nonzero vec-

tor ~a but of length 1. Such vector is called the

normalization of ~a.

The normalization of ~a is

the vector of length 1 in the direction of ~a,

ˆa =

~a

|~a|

Practice problems.

  1. Let P be the point (2, −1) and Q be the point (1, 3). Determine and sketch the vector

P Q.

  1. Let ~a = 〈 2 , − 1 〉 and

b = 〈 1 , 3 〉. Sketch ~a +

b, ~a −

b, 2 ~a, 2 ~a − 3

b.

  1. Let ~a = 〈 3 , 4 , 0 〉 and

b = 〈− 1 , 4 , 2 〉. Determine |~a|, 2 ~a + 3

b, 3 ~a − 2

b.

  1. Let ~a =

i + 4

j − 8

k and

b = − 2

i +

j + 2

k. Determine |~a|, ~a +

b, 2 ~a − 3

b.

  1. Find the normalization of the vector ~a =

i + 4

j + 8

k.

  1. Find the normalization of the vector ~a = 〈 3 , 4 , 0 〉.

Solutions. 1.

P Q = 〈− 1 , 4 〉 3. |~a| = 5, 2 ~a + 3

b = 〈 3 , 20 , 6 〉, 3 ~a − 2

b = 〈 11 , 4 , − 4 〉

  1. |~a| = 9, ~a +

b = 〈− 1 , 5 , − 6 〉, 2 ~a − 3

b = 〈 8 , 5 , − 22 〉 5. The length of ~a is |~a| =

2

  • 8

2

81 = 9 so ˆa =

1

9

i +

4

9

j +

8

9

k 6. The length of ~a is |~a| =

2

  • 4

2

  • 0

2

25 = 5 so ˆa = 〈

3

5

4

5

The Dot Product

If one is to define a meaningful product of two vectors, ~a = 〈a 1

, a 2

, a 3

〉 and

b = 〈b 1

, b 2

, b 3

the first idea that comes to mind would probably be to consider coordinate-wise multiplication

〈a 1 b 1 , a 2 b 2 , a 3 b 3 〉. However, since this type of product is geometrically not very meaningful nor appli-

cable, one consider two other types of multiplication, the dot and the cross product.

  1. Find the angle between the vectors ~a = 〈 3 , − 1 , 2 〉 and

b = 〈 2 , 4 , − 1 〉.

  1. Find the projection of ~a onto

b if ~a = 〈 1 , − 1 , 0 〉 and

b = 〈 1 , 0 , 1 〉.

Solutions. 1. ~a ·

b = −2 + 9 − 4 = 3. 2. If θ denotes the angle between the vectors, then

cos θ =

~a·

~ b

|~a||

~ b|

15+

(5)(13)

63

65

= 969. θ = cos

− 1 (.969) = .249 radians or 14.25 degrees.

  1. cos θ =

~a·

~ b

|~a||

~ b|

0

|~a||

~ b|

= 0. So, θ = 90 degrees and the vectors are perpendicular.

  1. proj ~ b

~a =

~a·

~ b

|

~ b|

2

b =

1

(

2

1

2

1

2

1

2

The Cross Product

As opposed to the dot product which results in a scalar, the cross product of two vectors is again

a vector. If ~a and

b are two vectors, their cross product is denoted by ~a ×

b.

The vector ~a ×

b is perpendicular to the plane determined by ~a and

b. This determines the

direction of ~a ×

b. The sense of ~a ×

b is determined by the right hand rule: if ~a and is the thumb

and

b the middle finger, the index finger has the same sense as ~a ×

b. Using the right hand rule, you

can see that the cross product is not not commutative, ~a ×

b 6 =

b × ~a in general, and that

b × ~a = −~a ×

b.

The length of ~a ×

b is the same as the area of the parallelogram determined by ~a and

b. If

θ is the angle between ~a and

b, then

|~a ×

b| = |~a||

b| sin θ.

The cross product of ~a = 〈a 1 , a 2 , a 3 〉 and

b = 〈b 1 , b 2 , b 3 〉 can be computed using the coordinates

as follows.

~a ×

b = 〈a 2

b 3

− a 3

b 2

, a 3

b 1

− a 1

b 3

, a 1

b 2

− a 2

b 1

i

j

k

a 1

a 2

a 3

b 1 b 2 b 3

Since |~a ×

b| = |~a||

b| sin θ, ~a and

b are parallel exactly when sin θ = 0 which happens exactly when

~a ×

b =

  1. Thus,

~a and

b are parallel if and only if ~a ×

b =

Another way to check if the two vectors are parallel is to check if one is a scalar multiple of the other

(i.e. if ~a = k

b for some k). In this case, for

b 6 =

0 , the coordinates are such that

a 1

b 1

a 2

b 2

a 3

b 3

Practice Problems.

  1. Let ~a = 〈 1 , 2 , 0 〉 and

b = 〈 0 , 3 , 1 〉. Find ~a ×

b.

  1. Do the same for ~a = 2

i +

j −

k and

b =

j + 2

k.

  1. Let ~a = 〈− 5 , 3 , 7 〉 and

b = 〈 6 , − 8 , 2 〉. Determine if the vectors are parallel, perpendicular or

neither.

  1. Do the same for ~a = −

i + 2

j + 5

k and

b = 3

i + 4

j −

k.

  1. Find a vector perpendicular to the plane through the points P (1, 0 , 0), Q(0, 2 , 0) and R(0, 0 , 3)

and find the area of the triangle P QR.

  1. Do the same for P (0, 0 , 0), Q(1, − 1 , 1) and R(4, 3 , 7).

Solutions. 1. 〈 2 , − 1 , 3 〉 2. 3

i − 4

j + 2

k

  1. ~a·

b = − 40 6 = 0 so the vectors are not perpendicular. Also, the coordinates are not proportional

− 5

6

3

− 8

7

2

) so the vectors are not parallel either. Alternatively, find that the cross product is

a ×

b = 〈 62 , 52 , 22 〉 6 = 〈 0 , 0 , 0 〉 so the vectors are not parallel.

  1. ~a ·

b = 0, thus the vectors are perpendicular.

  1. Since vectors

P Q and

P R are in the plane, their cross product

P Q ×

P R is perpendicular to

the plane. Calculate

P Q = 〈− 1 , 2 , 0 〉 and

P R = 〈− 1 , 0 , 3 〉,

P Q ×

P R = 〈 6 , 3 , 2 〉. The area of the

triangle determined by P, Q, and R is half of the area of the parallelogram determined by the vectors

P Q and

P R which is the magnitude of

P Q ×

P R. Thus the triangle area is

1

2

7

2

  1. Similarly to previous problem, find a vector perpendicular to the plane to be 〈− 10 , − 3 , 7 〉 and

the area of the triangle to be