I6-7. (a) A~·A~=|A~||A~|cos 0 =|A~|^2 =C^2
(b) d
dt(A~·A~) = d
dtC^2 =⇒ A~·d
A~
dt+d
A~
dt·A~= 0 =⇒ A~·d
A~
dt= 0which showsA~isperpendicular to d
A~
dtI6-8.
Equation of line is~r=~r 1 +tA~wheretis a
parameter. Use the property of right tri-
angles and write
d=|~r 0 −~r 1 |sinθ=|(~r 0 −~r 1 )×eˆA|where the absolute value sign insures that
dis positive and ˆeAis a unit vector in the
direction ofA~.I6-9. Area is 54 square units
I6-10.
A~+B~+C~+D~=~ 0
By construction
E~=^1
2
A~+^1
2
B~
F~=^1
2C~+^1
2D~G~=^1
2
B~+^1
2
C~H~ =^1
2
D~+^1
2
A~To showE~ is parallel toF~, showE~×F~ =~ 0 and to showG~ is parallel toH~, show
G~×H~ =~ 0E~×F~=(^1
2A~+^1
2B~)×(^1
2C~+^1
2D~) =^1
2(A~+B~)×(−^1
2)(A~+B~) =~ 0G~×H~ =(^1
2
B~+^1
2
C~)×(^1
2
D~+^1
2
A~) =^1
2 (
B~+C~)×(−^1
2 )(
B~+C~) =~ 0I6-11.
~r−~r 0 =(x−x 0 )ˆe 1 + (y−y 0 )ˆe 2 + (z−z 0 )ˆe 3
(~r−~r 0 )·(~r−~r 0 ) =ρ^2
(x−x 0 )^2 + (y−y 0 )^2 + (z−z 0 )^2 =ρ^2
Solutions Chapter 6