fisica1-youn-e-freedman-exercicios-resolvidos

1.91: A r and C r are perpendicular, so A
r
⋅C
r
= 0 . A C + A C = 0, , which gives
5.0 C
x
− 6.5C
y
= 0.
B
r ⋅C
r
= 15 .0, so − 3.5C x
+ 7.0C
y
= 15.0
We have two equations in two unknowns
C
8 .0 and C = 6.1.
x
=
y
x
x
C
x
and
y
y
C
y
. Solving gives
r
2
( − 5.00) + ( 2.00)
( 3.00)( 3.00)
1.92: A× B
r
= AB sin θ
A×
B
r
sin θ = = AB
r
( )
o
−
θ = sin 1 0.5984 = 36.8
2
= 0.5984
1.93: a) Using Equations (1.21) and (1.27), and recognizing that the vectors
r r r
A, B,and
C do not have the same meanings as they do in those equations,
r r r
( A B) ⋅C
= ( A B − A B ) iˆ
+ ( A B − A B ) ˆj
+ ( A B − A B ) kˆ
)
r
× ⋅ C
y
y
z
z
= A B C
x
z
z
y
− A B
y
C
A similar calculation shows that
r
A ⋅
x
z
z
x
+ A B C
x
y
x
z
− A B C
x
z
y
x
y
+ A B
x
y
y
C
z
x
− A B C .
r r
( B × C ) = Ax
ByCz
− Ax
BzCy
+ AyBzCx
− Ay
BxCz
+ Az
BxCy
− Az
ByCx
and a comparison of the expressions shows that they are the same.
b) Although the above expression could be used, the form given allows for ready
compuation of A
r × B
r
o
the magnitude is AB sin φ = ( 20.00) sin 37.0 and the direction
is, from the right-hand rule, in the +z-direction, and so
r r r
o
( A × B) ⋅C
= + ( 20.00 ) sin 37.0 ( 6.00) = + 72.2.
y
x
z