fisica1-youn-e-freedman-exercicios-resolvidos

1.88: a) This is a statement of the law of cosines, and there are many ways to
derive it. The most straightforward way, using vector algebra, is to assume the linearity
of the dot product (a point used, but not explicitly mentioned in the text) to show that
r r
the square of the magnitude of the sum A + B is
r r r r
( A + B) ⋅ ( A + B)
r r r r r r r r
= A ⋅ A + A ⋅ B + B ⋅ A + B ⋅ B
r r r r r r
= A ⋅ A + B ⋅ B + 2 A ⋅ B
r r
2 2
= A + B + 2A
⋅ B
= A
2
+
B
2
+ 2AB
cosφ
Using components, if the vectors make angles θ A and θ B with the x-axis, the components
of the vector sum are A cos θ A + B cos θ B and A sin θ A + B sin θ B , and the square of the
magnitude is
( A cos θ + Bcosθ
) 2 + ( A sin θ + Bsinθ
) 2
A
B
= A
2
= A
2
= A
2
B
2
θA
2 2
2
( cos θ + sin ) + B ( cos θ + sin θ )
+ B
+ B
A
2
A
2
2
+ 2AB
cos
+ 2AB
cos φ
where φ = θ A – θ B is the angle between the vectors.
+ 2AB( cosθ
A
cos θB
+ sin θA
sin θB
)
( θ −θ
)
A
B
b) A geometric consideration shows that the vectors A
r , B
r
r r
and the sum A + B
r r
must be the sides of an equilateral triangle. The angle between A , and B is
120 o , since one vector must shift to add head-to-tail. Using the result of part
(a), with A = B,
the condition is that A 2 = A
2 + A
2 + 2 A 2 cosφ
, which solves
1
for 1 = 2 + 2 cos φ, cos φ = − , and φ = 120 o .
2
c) Either method of derivation will have the angle φ replaced by 180 o – φ, so
2 2
the cosine will change sign, and the result is A + B − 2AB
cosφ.
d) Similar to what is done in part (b), when the vector difference has the same
magnitude, the angle between the vectors is 60 o . Algebraically,φ is obtained
1
from 1 = 2 – 2 cos φ, so cos φ = 2
and φ = 60 o .
B
B