§ 7 Number product of two vectors

§ 7 Number product of two vectors

 

Definition 1For two vectorsAAndB, Modulo them |A|, |B| And their angleQThe product of the cosine is called the number product of the vector and recordedAB,That is

AB= |A|B| CosQ.

The relationship between the definition and projection is available.

AB= |B| Prjb
A = |A| PrjaB.

The nature of the number product:

(1)A.= |A| 2, noteA.=A2, thenA2 = |A| 2.

(2) For two non-zero vectorsA,B, IfA · B= 0, thenA^B

Otherwise, ifA^B, ThenA · B= 0.

If we assume that the zero vector is perpendicular to any vectorA^BBytesA·B= 0.

Theorem 1The number product satisfies the following calculation law:

(1) Exchange Law:A · B=
B.

(2) allocation law :(A+B) ×C=A×C+B×C
.

(3) (lA)·B=
A ·(LB) =
L (A · B),

(LA) · (MB) =
Lm (A · B), L, m number.

Certificate(1) clearly defined.

(2) proof:

Because whenC=0The above formula is obviously true;

WhenCBytes0Yes

(A+B) ×C= |C| PrjC(A+B)

= |C| (PrjCA+ PrjCB)

= |C| PrjCA+ |C| PrjCB

=A×C+B×C
.

(3) Similar proof.

Example 1Use vectors to prove the cosine theorem of triangles.

CertificateInABCMedium, largeBCA=, | =A, | =B, | =C, Certificate required

C2 =A2 +B2-2A
BCos.

Note =A, =B,=C, There areC=A-BTo

|C| 2 =C
×
C= (A-B)(A-B) =A2-2 ×AB+B2 = |A| 2 + |B| 2-2 |A|B| Cos (A, ^B),

That isC2 =A2 +B2-2A
BCos.

Coordinate representation of the number product:

Theorem 2SetA= {Ax,
Ay,
Az},B= {Bx,
By,
Bz}, Then

A · B=Axbx+Ayby+Azbz.

CertificateA · B= (AxI+
AyJ+
AzK)·(BxI+
ByJ+
BzK)

=Ax
BxI · I+
AxI · j+
Ax bzI · k

+Ay
BxJ·I+
AyJ·J+
Ay bzJ·K

+Az
BxK·I+
AzK·J+
Az bzK·K

=Ax
Bx
+Ay
+Az bz.

Theorem 3SetA= {}, Then the VectorAModule

|A| =.

CertificateKnown by Theorem 1.7.2

|A| 2 =A2 =,

So |A| =.

Coordinate representation of the cosine of the two vector angles:

Theorem 4

SetQ= (A, ^B), Then whenABytes0,BBytes0Yes

.

CertificateBecauseA · B= |A|B| CosQ
, So

.

 

Example 2Three known pointsM(1, 1 ),A(2, 2, 1) andB(2, 1, 2), locateAMB
.

SolutionSlaveMToAIs recordedA, FromMToBIs recordedB, Then beginAMBIs VectorAAndB.

A= {1, 1, 0 },B= {1, 0, 1 }.

Because

A×B= 1 '1 + 1' 0 + 0' 1 = 1,

,

.

So.

Thus.

Vector direction angle and direction cosine: the angle between the vector and the coordinate axis is called the vector direction angle, and the cosine of the direction angle is called the vector direction cosine..

Theorem 5SetA= {}, ThenAReturns the arc cosine

Cos =,

Cos,

Cos;

And,

These are vectors.AThe angle between X axis, Y axis, and Z axis.

CertificateBecauseAi=|A| COs andAi=,

So |A| Cos =,

Cos =.

Likewise, the CoS

 

Cos

Obviously