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VECTORANDTENSORCALCULUSFORENGINEERS
EXAMPLE1.4
Demonstratethevalidityoftheformula
(a+b)2+(a–b)2=2(a2+b2)anddescribe
itsgeometricalmeaning.
Solution
L(
=
ab
+
)
2
+
(
ab
−
)
2
=
a
2
+
2
abb
+
2
+
a
2
−
2
abb
+
2
=
2
a
2
+
2
b
2
.
Usingthisformula(1.9)9letuscalculate:
a
2
=
a
2
=
a
2
and
b
2
=
b
2
=
b
2
9
therefore
2
a
2
+
2
b
2
=
2(
a
2
+
b
2
)P
=
.
Answer
Thesumofthesquarelengthsofthesidesoftheparallelogram
isequaltothesumofthesquaresofitsdiagonals.
EXAMPLE1.5
Findtheinterioranglesofthetrianglewiththefollowing
vertices
A
(
291
−9
)
B
(
193
)
9
C−
(
191
)
.
figure1.9
B
β
c
a
y
!
A
b
22
Solution
MethodI
ą
ą
ą
ą
ą
ą
AB
=−
BA
=−
[194]
9
BC
=−
CB
=−−
[292]
9
CA
=−
AC
=
[392]
−
9
ą
ą
ą
AB=
17
9
BC=
8
9
CA=
13
.
Letuscalculatetheanglesusingtheformula(1.9).
MaintainingtheappropriatesensesofthevectorsC
formingthetriangle(Fig.1.9)9weobtain:
C
