2. Vector Product of two vectors

2. Vector Product of two vectors

 

Definition 1Set VectorCIs composed of two vectorsAAndBDetermined by the following means:CModule
|C| = |A|B| SinQ
, WhereQIsAAndBThe angle between them;CIs perpendicularAAndBDetermined plane,CFromARedirectionBTo determine, we put this vectorCIt is called vector.AAndBIs recordedA'B, That is

C=A'B.

The property of the zhiyserx product is defined as follows:

(1)A'A=0

(2) For two non-zero vectorsA,B, IfA'B=0, ThenA//BOtherwise, ifA//B, ThenA'B
=0.

Theorem 1Two VectorsAAndBThe necessary and sufficient condition for the collinearity isA'B=0.

CertificateWhenAAndBBecause sin (A,B) = 0, so |A'B| = |A|B| Sin (A,B) = 0, thusA'B=0;Otherwise, whenA'B=0By definition,A=0, OrB=0, OrA//BBecause the zero vector can be viewed as having the same line with any vector, there is alwaysA//B, That isAAndB.

Theorem 2The yarn product satisfies the following calculation law:

(1) Exchange LawA'B=-B'A,

(2) allocation Law
(A+B)'C=A'C+B'C,

(3) combination law of number factors (lA)'B=A'(LB) = L (A'B) (L number ).

Certificate (omitted ).

Theorem 3

SetA
=AxI+
AyJ+
AzK,B
=BxI+
ByJ+
BzK, Then
A'B= (Aybz
-Azby)I+ (Azbx
-Axbz)J+ (Axby
-Aybx)K.

CertificateObtained by the calculation Law of the Vector Product

A'B= (AxI+AyJ+AzK)'(BxI+ByJ+BzK)

=AxbxI'I+AxbyI'J+AxbzI'K

+AybxJ'I+AybyJ'J+AybzJ'K

+AzbxK'I+AzbyK'
+AzbzK'K.

BecauseI'I=J'J=K'K=0,I'J=K,J'K=I,K'I=J,

SoA'B= (Aybz
-Azby)I+ (Azbx
-Axbz)J+ (Axby
-Aybx)K.

 

In order to help the memory, the upper form can be written using the third-order Determinant symbol.

=AybzI+AzbxJ+AxbyK-AybxK-AxbzJ-AzbyI

= (Ay bz
-Az)I+ (Az
Bx
-Ax bz)J+ (Ax
-Ay
Bx)K..

Example 1SetA= (2, 1,
-1 ),
B= (1,
-1, 2), computingA'B
.

Solution
2I-J-2K-K-4J-I
=I-5J-3K.

Example 2Known triangleABCThe vertices areA(1, 2, 3 ),B(3, 4, 5 ),C(2, 4, 7), triangleABCArea.

SolutionAccording to the vector product definition, we can see the triangleABCArea

.

Because = (2, 2, 2), = (1, 2, 4 ),

= 4I-6J+ 2K.

So.

 

Example 3Set the isoangular velocity of a rigid bodyW
BypassLRotate the axis to calculate the point on the rigid bodyMLine speed.

SolutionRigid Body WrapLWhen the axis is rotated, we can useLA vector on the AxisNIndicates the angular velocity. Its size is equal to the angular velocity. Its direction is determined by the right-hand rule:LAxis. When the orientation of the four fingers on the right hand is the same as that of the rigid bodyNDirection.

SettingsMTo axis of rotationLDistance isA
, And thenLTake any point in the axisOVectorR
=, AndQIndicatesNAndR, Then

A
=
|R| SinQ
.

Set the line speedV, We can see from the relationship between the physical online speed and the angular velocity,VIs

|V| = |
N|A
= |N|R| SinQ

VIs perpendicularMPoints andLThe plane of the axis, that isVPerpendicularNAndR, AndVIsN,R,VIt complies with the right-hand rule. Therefore

V
=
N'R.