Resolução Marion[Mec.Clássica] - Capítulo 1

Resolução Marion[Mec.Clássica] - Capítulo 1

(Parte 1 de 5)

Matrices, Vectors, and Vector Calculus

1-1.

So the transformation matrix is:

1-2. a) βγ O E x

From this diagram, we have cosOEOAα= cosOEODγ= Taking the square of each equation in (1) and adding, we find

But and

Therefore,

Thus, b) x

First, we have the following trigonometric relation:

MATRICES, VECTORS, AND VECTOR CALCULUS 3

But,

2 cos cos cos cos

cos cos or, cos cos cos cos cos cos 2 cos cos cos cos cos cos

2 cos cos cos cos cos cos cos cos cos cos cos cos cosθ αα β β γ γ=+ +′ ′ ′ (10)

1-3.

e e e A e ′

Denote the original axes by , , , and the corresponding unit vectors by e,, . Denote the new axes by , , and the corresponding unit vectors by 1x 2x 3x 1 2e 3e

e e e e e e e e e e e e

1-4. a) Let C = AB where A, B, and C are matrices. Then,

( )t ji jkk i ki jkij

Identifying ()t ki ik

B=B and ()t jk kj

1-5. Take λ to be a two-dimensional matrix:

Then,

2 2 λ λλ λλ λ λ λ λ λλ λ λ λλ λ λ λ λ λ λ λ λ λλ λλ λ λ λ λ

But since λ is an orthogonal transformation matrix, ijkjikj λλδ=∑.

Thus,

1-6. The lengths of line segments in the jx and jx′ systems are

MATRICES, VECTORS, AND VECTOR CALCULUS 5

If , then L=′

The transformation is

Then, ji k k ji k

ki k i ki x x λλ λλ i x

But this can be true only if

which is the desired result.

1-7.

There are 4 diagonals:

1D, from (0,0,0) to (1,1,1), so (1,1,1) – (0,0,0) = (1,1,1) = D; 1 2D, from (1,0,0) to (0,1,1), so (0,1,1) – (1,0,0) = (–1,1,1) = ; 2D 3D, from (0,0,1) to (1,1,0), so (1,1,0) – (0,0,1) = (1,1,–1) = ; and 3D

4D, from (0,1,0) to (1,0,1), so (1,0,1) – (0,1,0) = (1,–1,1) = D. 4 The magnitudes of the diagonal vectors are

so that

Similarly,

1-8. Let θ be the angle between A and r. Then, 2A⋅=Ar can be written as 2cosArAθ=

This implies

Therefore, the end point of r must be on a plane perpendicular to A and passing through P.

1-92=+−Aijk23=−++Bijk

b) component of B along A

The length of the component of B along A is B cos θ. cosABθ⋅=AB

The direction is, of course, along A. A unit vector in the A direction is

MATRICES, VECTORS, AND VECTOR CALCULUS 7

So the component of B along A is

+−ijk

(Parte 1 de 5)

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