fig.3

(1.3 )

With these elements we can introduce the notion of scalar product on the vector space of free vectors.

2. The Scalar Product

Let V₃ be real vector space of free vectors

2.1 Theorem. The function V V R, defined with

(2.1)

defines a scalar product on vector space of free vectors.

Demonstration. We verify the four conditions that define a scalar product.

1. V₃

From the definition of the scalar product and the property (1.3) we have

=| | | | = | | | | , I V₃ so =| | | | =

= = = + .

a ) = a( ), , I V a I R.

a ) = | | | |=a | | | |=a ( ), , IV aI R.

= , , I V .The comutativity result from the product's comutativity in the set of the real numbers.

4. 0, = 0 = , I V

< , > = | |² 0, | |² = 0 = .

For the case where at least one factor of the scalar product is the null vector the properties results immediately.

2.2 Consequence. The vector space of free vectors V endowed with the scalar product (2.1) is a real Euclidean vector space.

2.3 Consequence. The affine space A = ( E₃, V j ) having like vector associate the Euclidean space V becomes an Euclidean punctual space which we will note with _3.

Remarks.

1 In the last paragraph there were showed the natural one-to-one mapping between the spaces E₃, V₃ and R³. In this way, having fixed a Cartesian mark R (O; ) in the affine space A₃, the function of coordinates f: V₃ R³, defined by f ( ) = ( x₁, x₂, x₃) I R³, I V₃, realize a one-to-one mapping between the two spaces. This one-to-one mapping represents an isomorphism of vector spaces which permits the transport of the canonical Euclidean structure defined on R³ on the vector pace of free vectors V₃.

We can easily verify the application :

< ,> :V₃ V₃ R, ( , ) < , > =: <f( ), f( )>_R (2.2)

it's a scalar product on V₃, where < , >_R is the scalar product defined on R³.

With the help of this scalar product we can define in a natural way the norm || ||= = _{R .}

If we consider two arbitrary points A, B I E₃ and the position vectors and characterized by the triple ( x₁, x₂, x₃) I R³ , and ( y₁, y₂, y₃) I R³, then the vector will be characterized by the triple (y₁ - x₁, y₂ - x₂, y₃ - x₃) and will have the norm given by

= = _R =

= = = | |.

This result shows that the norm || || defined by the scalar product (2.2) coincides with the geometrical length | | , of the vector .

The angle between two vectors that aren't zero and I V₃ defined by the scalar product < , > coincides with the (geometric) angle defined by the directions of the semilines |OA and |OB . Really,

= = = = .

In consequence, the scalar product (2.2), induce by the one-to-one mapping f, on the vector space V₃ of free vectors, coincides with the scalar product (2.1).

2 Knowing the scalar product on the vector space of free vectors permits the calculus of the vector's length and the angle between two vectors :

|| || = , , (2.3)

3 Two vectors that are not zero are orthogonal their scalar product is zero.

Let be B = a basis in the vector space V₃.

If and , then we obtain:

(2.4)

So, the scalar product of two vectors is perfect determined if we know the scalar of the multiplication of the vectors of basis B.

A basis in V₃ formed by vectors orthogonal two by two in named orthonormal basis and the coordinates of a vector in an orthonormal basis are called Euclidean coordinates.

In Euclidean geometry we demonstrate that into a point there exists three perpendicular lines two by two from where results the existing of a orthonormal Cartesian mark in the Euclidean punctual space ₃.

If B = is an orthonormal basis in V₃ then , , that is the scalar product of the vectors of the basis B is given by the table

The scalar product of two vectors and will have the canonical expression

(2.5)

The orthogonal projection of the vector on the direction of the vector is given by

, analogous and .

In this way the Euclidean coordinates of vector represents the sizes f the orthogonal projections of on the three axes of the orthonormal Cartesian mark . The analytic expression of the vector's norm and the angle between two vectors will be given by

|| ||= (2.6)

= , j I [0, p] (2.7)

In private the vectors and are orthogonal if and only if

a₁b₁ + a₂b₂ + a₃b₃ = 0 (2.8)

3. The vector product

Let be the vectors and I V₃. For and we note with j I [0, p] the angle between and .

3.1 Definition. We call vector product the binary internal operation " ":V₃ V₃ V₃ , which to the ordinate pair ( , ) associates the vector noted with , characterized by

1 || || = || || || || sin j

2 = is orthogonal on and

3 The sense of the vector = is given by the law of the right hand when we rotate over under a pointed angle(the law of the right auger)(fig.4)

fig.4

If we noted with the versor of the orthogonal direction on and then

= || || || || sin j .

3.2Propozitie. The vector product has the following properties:

Demonstration.

1.From the definition of the vector product results that the vectors and have the same norm, the same direction, but different senses.

2.Let's consider the representants , and of the vectors and and respective in point O IE and the plane p through O perpendicular on the direction of the vector .(fig.5)

fig.5

If we note with , we obtain

= || || || || sin j = || || || || = =

analogous = =

If + = and the parallelogram OB'D'C' is the projection of the parallelogram OBCD on the plane p, then = .

The vector is obtained rotating with an angle of p/2 the vector in the plane p, and multiplying the vector obtained with || ||. Analogous we obtain from . By rotateing the parallelogram OB'D'C' with an angle of p/2 and multiplying with || || we obtain a parallelogram, too (like the first one) from where results that and || || = || || || ||.

So

_{Similarly we can demonstrate the distributivity of the vector product to respect to the first factor, that is}

3. For a > 0, the vector a has the direction and sense of and (a ) and a( ) have the same direction and sense. In plus,

For a < 0, the vector a has the direction and sense of - and (a ) and a( ) have the same direction and sense. In plus,

Analogous we can demonstrate that a a ( ).

For a = 0, from 3 results that = .

4. If , , from = 0 results that || || = 0 sinj = 0, that is the vectors and are collinear, = l .

From the definition of the vector product we have that || || = 0, from where we obtain = 0 .

If = l then = l ) = l ( ) = 0.

5. For and that are not collinear we construct the parallelogram OACB (fig.6)

C

B

h

j        A

0

fig.6

that is the area of the parallelogram determined by and .

If B ( , , ) is an orthonormal basis in V ₃ then using the definition of the vector space and its properties, we obtain the table

				(3.1)
			-
	-
		-

Thus, the vector product of two vectors and , and , will have the canonical expression

_{The canonical expression (3.2) can be obtained by developing by the first line the formal determinant.}

(3.2)

Two vectors are collinear ( = ) if and only if

(3.3)

3.3 Propozitie. For any two vectors and is satisfy the Lagrange's identity :

( )² + ( )² = || (3.4)

Demonstration. From the definition of the scalar product we have

( )² = , and ( )² =

Summing these two vectors we obtain the equality (3.4).

4. The double vector product

4.1 Definition. Let be the vectors , , I V . We call double vector product of vectors , , the vector

   

From the definition of the vector product results that the vector is coplanar with the vectors and (the vectors from the brackets).If we construct the vector ( ) = , this will be a coplanar vector with and from where results that

We can demonstrate easy, using the analytic expressions of the vectors , , , the developing formula of the double vector product

(4.2)

or under the shape of the symbolic determinant

(4.3)

5. The mixed product

5.1 Definition.    Let be the vectors , , I V . We call mixed product of vectors , , the real number ( , , ) given by

( , , ) = : ( ) (5.1)

5.2 Theorem. The mixed product has the following properties:

) ( ) = ( , , ) + ( , , )

a , , ) = a ( , , )

3) ( , , ) = e_s ( ) , s I S₃, e_s

4) ( , , ) = 0 , , are linear dependent (coplanar)

) |( , , )| = , for , , I V

The properties 1) and 2), additivity and homogeneity, results from the definition of the mixed product and it spreads for any factor.

The property 3) can be equivalent expressed through the properties:

( , , ) = ( , , ) = ( , , )

which express the unvariation of the mixed product at the circular permutation, that is e_s

s I S₃ - even permutation ) and

( , , ) = - ( , , ) ,

and the other suitable relations for the uneven permutations which express the anticomutativity property for any two enclosed factors.

The equivalence 4) immediately results for almost a factor equal with the null vector, and for , , I V ₃ , the cancellation of the mixed product is equivalent with the vector's orthogonality and , that is the coplanarity of the vectors , and .

If we note with the volume of the parallelepiped formed by the vector's representants , , into a point O I E₃ (fig.7 ) and note with q = < ( , ) , with j = < ( , ), we obtain

A

B

O

C

j

q

h

fig.7

If B = ( , , ) is a orthonormal basis in vector space of free vectors V , and = + + , and are analytic expressions of vectors , and respective , then the mixed product has the canonical expression given by

Counting the properties of the determinants and the analytic canonical expression of

the mixed product con be easily checked the properties 1-5.

We say that a basis B V ₃ is positive (negative) orientated if the mixed product ( , , ) is positive (negative).

6.Problems

Demonstrate that in any triangle: a) medians; b) highs; c) bisections; d) mediations, are concurrent.

Being given two cords AMB and CMD perpendicular between them in a circle with the center O, demonstrate that:

H is the orthocenter of the triangle ABC if and only if takes place the equalities:

In a thethraedru ABCD, the opposite edges are perpendicular two by two if and only if : .

The lines that join the middles of the opposite edges of a thethraedru are concurrent.

Demonstrate that, if the vectors and are collinear, then the vectors and are collinear.

7. Let be , where ||

( . Calculate :

a)      the length of the parallelogram's diagonals constructed on and  

b)      the angle between the diagonals;

c)      the area of the parallelogram determined by and .

  Let be and  two perpendicular vectors, with ||

Calculate ||

Let be , , vectors that are not coplanar. Study the liniar independence

of the vectors

_{Demonstrate the Jacobi's identity:}

Demonstrate the relation . If , , are coplanar, then they are also collinear.

Prove the Lagrange's identity:

+ || ||² = .

Let be , , that are not coplanar. Show that exists vectors , , with the properties:

If , , are the mutual of the vectors , , , we're asking:

a)      to express the mixed product ( , , ) function to the mixed product

, , ). In which conditions , , are collinear?

b)      demonstrate that the volume of the thethraedru constructed on the vectors , , is equal with the volume of the thethraedru constructed on , , .

c)      deduce the equality .

Let be . Determine a and b such that is perpendicular on the vectors and . Calculate the angle between and .

The vectors , , are given. Calculate: ( , , ), ( ) and + + .

Determine l such that the vectors = , = , = = are coplanar and find the relation of linear dependence.

Calculate the area and the height from A in the triangle ABC given by A (0, 1, 0), B ( 2, 0, 1), C ( -1, 0, -4).

Calculate the volume of the thethraedru ABCD an the height from A, where

A ( 3, 2, -1), B ( 4, 3, -1), C ( 5, 3, -1), D ( 4, 2, 1).

Consider the points A (a, 0, 0), B (0, b, 0), C ( 0, 0, c). Show that the area of the triangle ABC is at least equal with a⁴ + b⁴ + c⁴. In which conditions the equality takes place?

21. Are given the vectors that are not coplanar and

+3 a

a)      determine a I R such that V₍ = 5V₍

b)      in the case when a = 2 and the vectors form a regulate thethraedru with the side 1, determine the angle between and the plane determined by and .

Solve the equation - given).

23. If ( 0, solve the system:

_{Consider the system}

where

a)    show that the system has solutions if and only if || || = || || and

solve it in this case

b) if , then .Is the reciprocal true?