Begin2.DVI

I8-52. IfA~=αE~ 1 +βE~ 2 +γE~ 3 andE~ 1 ·E~ 2 = 0,E~ 1 ·E~ 3 = 0,E~ 2 ·E~ 3 = 0, then show

A~·E~ 1 =αE~ 1 ·E~ 1 or α= A~·E~^1

E~ 1 ·E~ 1

A~·E~ 2 =βE~ 2 ·E~ 2 or β=

A~·E~ 2

E~ 2 ·E~ 2

A~·E~ 3 =γE~ 3 ·E~ 3 or γ=

A~·E~ 3

E~ 3 ·E~ 3

I8-53. IfE~^1 had the same direction asE~ 2 ×E~ 3 , thenE~^1 =αE~ 2 ×E~ 3. IfE~^1 ·E~ 1 = 1,

then1 =αE~ 1 ·(E~ 2 ×E~ 3 )orα=^1

V

forV =E~ 1 ·(E~ 2 ×E~ 3 )

Do the same type of arguments forE~^2 andE~^3.

Reverse the roles ofE~ 1 ,E~ 2 ,E~ 3 withE~^1 ,E~^2 ,E~^3 to show

E~^1 ·(E~^2 ×E~^3 ) =^1

V

(E~ 2 ×E~ 3 )·

[

1

V

(E~ 3 ×E~ 1 )×^1

V

(E~ 1 ×E~ 2 )

]

then use the triple scalar product relation to show

E~^1 ·(E~^2 ×E~^3 ) =^1

V^3 (

E~ 1 ×E~ 2 )

[

(E~ 2 ×E~ 3 )×(E~ 3 ×E~ 1 )

]

Use another vector identity to show

E~^1 ·(E~^2 ×E~^3 ) =^1

V^3

[E~ 1 ·(E~ 2 ×E~ 3 )]^2 =^1

V^3

V^2 =^1

V

Solutions Chapter 8