Demostración de las propiedades de las operaciones vectoriales Maestría en Enseñanza de la Física

1. FACULTAD DE CIENCIAS BÁSICAS
MAESTRÍA EN ENSEÑANZA DE LA FÍSICA
ELECTROMAGNETISM
Professor: Jimmy A. Cortes, José A. Chaves
VECTOR ANALYSIS AN INTRODUCTION
Verification of the properties of the scalar and vector product in R3
using general vector
expressions
Produced by:
Jhon F. González, Gelver Osorio, Julián F. Villada
April 8, 2021

2. Part I
Scalar product
1 Properties of dot product (Proof)
Carry out the verification of the properties of the scalar and vector product in R3
using general vector expres-
sions such as:
~
u = (ux,uy,uz); ~
v = (vx,vy,vz); ~
w = (wx,wy,wz).
1. The dot product is distributive over vector addition:
~
u·(~
v+ ~
w) = ~
u·~
v+~
u· ~
w
Proof:
Assuming that the vectors ~
u, ~
v and ~
w are represented by ~
u = (ux,uy,uz), ~
v = (vx,vy,vz) and ~
w =
(wx,wy,wz), then we have:
~
u·(~
v+ ~
w) = (ux,uy,uz)· (vx,vy,vz)+(wx,wy,wz)
~
u·(~
v+ ~
w) = (ux,uy,uz)·(vx,vy,vz)
+ (ux,uy,uz)·(wx,wy,wz)
~
u·(~
v+ ~
w) = ux ·vx +uy ·vy +uz ·vz +ux ·wx +uy ·wy +uz ·wz
~
u·(~
v+ ~
w) = ux ·vx +uy ·vy +uz ·vz
+ ux ·wx +uy ·wy +uz ·wz
.
~
u·~
v+~
u· ~
w = ux ·vx +uy ·vy +uz ·vz
+ ux ·wx +uy ·wy +uz ·wz
.
~
u·(~
v+ ~
w) = ~
u·~
v+~
u· ~
w (1)
2. The dot product is commutative:
~
v· ~
w = ~
w·~
v
Proof:
Assuming that the vectors are represented by ~
v = (vx,vy,vz) and ~
w = (wx,wy,wz), then we have:
~
v· ~
w = (vx,vy,vz)·(wx,wy,wz)
= vx ·wx +vy ·wy +vz ·wz
= wx ·vx +wy ·vy +wz ·vz .
~
w·~
v = (wx,wy,wz)·(vx,vy,vz)
= wx ·vx +wy ·vy +wz ·vz .
~
v· ~
w = ~
w·~
v (2)
3. The formula for the Euclidean length of the vector is:
~
v·~
v = k~
vk2
1

3. Proof:
Assuming that the vector ~
v is represented by ~
v = (vx,vy,vz) , then we have:
~
v·~
v = (vx,vy,vz)·(vx,vy,vz) = vx ·vx +vy ·vy +vz ·vz = (vx)2
+(vy)2
+(vz)2
.
k~
vk2
= k~
vk k~
vk =
q
(vx)2 +(vy)2 +(vz)2
q
(vx)2 +(vy)2 +(vz)2
= (vx)2
+(vy)2
+(vz)2
~
v·~
v = k~
vk2
(3)
4. Distributive property for scalar multiplication:
(c~
v)· ~
w =~
v·(c~
w) = c(~
v· ~
w)
Proof:
Assuming that the vectors~
v and ~
w are represented by~
v = (vx,vy,vz), ~
w = (wx,wy,wz) and c ∈ R, then
we have:
(c~
v)· ~
w = (c(vx,vy,vz))·(wx,wy,wz) = (cvx,cvy,cvz)·(wx,wy,wz)
= cvxwx +cvywy +cvzwz = (cvxwx +cvywy +cvzwz)
= cvxwx +cvywy +cvzwz .
~
v·(c~
w) = (vx,vy,vz)·(c(wx,wy,wz)) = (vx,vy,vz)·(cwx,cwy,cwz)
= vxcwx +vycwy +vzcwz = (vxcwx +vycwy +vzcwz)
= cvxwx +cvywy +cvzwz .
c(~
v· ~
w) = c((vx,vy,vz)·(wx,wy,wz)) = c(vxwx +vywy +vzwz)
= cvxwx +cvywy +cvzwz = (cvxwx +cvywy +cvzwz)
= cvxwx +cvywy +cvzwz .
(c~
v)· ~
w =~
v·(c~
w) = c(~
v· ~
w) (4)
5. Dot product between a vector different of zero and the zero vector:
~
v·~
0 = 0
Proof:
Assuming that the vectors~
v different of zero and~
0 are represented by~
v = (vx,vy,vz) and~
0 = (0x,0y,0z),
then we have:
~
v·~
0 = (vx,vy,vz)·(0x,0y,0z) = vx ·0x +vy ·0y +vz ·0z
= 0+0+0 = 0
~
v·~
0 = 0 (5)
2

4. 6. If ~
v·~
v = 0 , then ~
v =~
0
Proof:
Assuming that the vector ~
v is represented by ~
v = (vx,vy,vz) and that the vector ~
0 = (0x,0y,0z), then we
have:
~
v·~
v = (vx,vy,vz)·(vx,vy,vz) = vx ·vx +vy ·vy +vz ·vz = (vx)2
+(vy)2
+(vz)2
= 0
=
q
(vx)2 +(vy)2 +(vz)2
2
= 0 .
k~
vk2
=
q
(vx)2 +(vy)2 +(vz)2
2
=
q
(0)2 +(0)2 +(0)2
2
= 0
~
v = (0x,0y,0z) (6)
Part II
Cross product
2 Cross product properties
1. If θ is the angle between the vectors ~
u and ~
v then, the length of the cross product of two vector is:
k~
u×~
vk = kukkvksinθ
Proof:Assuming that the vectors ~
u, ~
v and ~
w are represented by ~
u = (ux,uy,uz), ~
v = (vx,vy,vz), ~
w =
(wx,wy,wz), then we have:
(a) The cross product between the vectors ~
u and ~
v is calculated:
~
u×~
v =
î ĵ k̂
ux uy uz
vx vy vz
~
u×~
v =
(uy)(vz)−(vy)(uz)
î−
(ux)(vz)−(vx)(uz)
ĵ+
(ux)(vy)−(vx)(uy)
k̂
(b) The length of the cross product between the vectors ~
u and ~
v is calculated:
k~
u×~
vk =
q
(uy)(vz)−(vy)(uz)
2
+ (ux)(vz)−(vx)(uz)
2
+ (ux)(vy)−(vx)(uy)
2
(c) The square of the length of the cross product between the vectors ~
u and ~
v is calculated:
k~
u×~
vk2
=
(uy)(vz)−(vy)(uz)
2
| {z }
part 1
+
(ux)(vz)−(vx)(uz)
2
| {z }
part 2
+
(ux)(vy)−(vx)(uy)
2
| {z }
part 3
3

5. (d) The squared binomials that appear within the radical are calculated separately:
(uy)(vz)−(vy)(uz)
2
| {z }
part 1
=
(uy)(vz)
2
−2
(uy)(vz)(vy)(uz)
+
(vy)(uz)
2
= (uy)2
(vz)2
−2
(uy)(uz)(vy)(vz)
+(vy)2
(uz)2
= (uy)2
(vz)2
−2(uy)(uz)(vy)(vz)+(vy)2
(uz)2
= (uy)2
(vz)2
+(vy)2
(uz)2
−2(uy)(uz)(vy)(vz)
(ux)(vz)−(vx)(uz)
2
| {z }
part 2
=
(ux)(vz)
2
−2
(ux)(vz)(vx)(uz)
+
(vx)(uz)
2
= (ux)2
(vz)2
−2
(ux)(vz)(vx)(uz)
+(vx)2
(uz)2
= (ux)2
(vz)2
−2(ux)(vz)(vx)(uz)+(vx)2
(uz)2
= (ux)2
(vz)2
+(vx)2
(uz)2
−2(ux)(uz)(vx)(vz)
(ux)(vy)−(vx)(uy)
2
| {z }
part 3
=
(ux)(vy)
2
−2
(ux)(vy)(vx)(uy)
+
(vx)(uy)
2
= (ux)2
(vy)2
−2
(ux)(vy)(vx)(uy)
+(vx)2
(uy)2
= (ux)2
(vy)2
−2(ux)(uy)(vx)(vy)+(vx)2
(uy)2
= (ux)2
(vy)2
+(vx)2
(uy)2
−2(ux)(uy)(vx)(vy)
(e) The square of the length of the cross product between the vectors ~
u and ~
v is calculated by replacing
the calculations of the item (d) and simplifying:
k~
u×~
vk2
=
(uy)2
(vz)2
+(vy)2
(uz)2
−2(uy)(uz)(vy)(vz)
+
+
(ux)2
(vz)2
+(vx)2
(uz)2
−2(ux)(uz)(vx)(vz)
+
+
(ux)2
(vy)2
+(vx)2
(uy)2
−2(ux)(uy)(vx)(vy)
k~
u×~
vk2
= (uy)2
(vz)2
+(vy)2
(uz)2
+(ux)2
(vz)2
+(vx)2
(uz)2
+(ux)2
(vy)2
+(vx)2
(uy)2
−2(uy)(uz)(vy)(vz)−2(ux)(uz)(vx)(vz)−2(ux)(uy)(vx)(vy)
(f) Now it can be verified by comparing components that:
k~
u×~
vk2
= k~
uk2
k~
vk2
−(~
u·~
v)2
(7)
k~
u×~
vk2
= (uy)2
(vz)2
+(vy)2
(uz)2
+(ux)2
(vz)2
+(vx)2
(uz)2
+(ux)2
(vy)2
+(vx)2
(uy)2
−2(uy)(uz)(vy)(vz)−2(ux)(uz)(vx)(vz)−2(ux)(uy)(vx)(vy)
k~
uk2
= (ux)2
+(uy)2
+(uz)2
4

6. k~
vk2
= (vx)2
+(vy)2
+(vz)2
k~
uk2
k~
vk2
=
(ux)2
+(uy)2
+(uz)2
(vx)2
+(vy)2
+(vz)2
= (ux)2
(vx)2
+(ux)2
(vy)2
+(ux)2
(vz)2
+(uy)2
(vx)2
+(uy)2
(vy)2
+(uy)2
(vz)2
+(uz)2
(vx)2
+(uz)2
(vy)2
+(uz)2
(vz)2
k~
uk2
k~
vk2
= (ux)2
(vx)2
+(ux)2
(vy)2
+(ux)2
(vz)2
+(uy)2
(vx)2
+(uy)2
(vy)2
+(uy)2
(vz)2
+(uz)2
(vx)2
+(uz)2
(vy)2
+(uz)2
(vz)2
k~
uk2
k~
vk2
= (ux)2
(vx)2
+(ux)2
(vy)2
+(ux)2
(vz)2
+(uy)2
(vx)2
+(uy)2
(vy)2
+(uy)2
(vz)2
+
(uz)2
(vx)2
+(uz)2
(vy)2
+(uz)2
(vz)2
−(~
u·~
v)2
= −
h
(ux,uy,uz)·(vx,vy,vz)
i2
= −
h
(ux)(vx)+(uy)(vy)+(uz)(vz)
i2
= −
h
(ux)(vx)+(uy)(vy)+(uz)(vz)
(ux)(vx)+(uy)(vy)+(uz)(vz)
i
= −
h
(ux)(vx)(ux)(vx)+(ux)(vx)(uy)(vy)+(ux)(vx)(uz)(vz)
+
+
(uy)(vy)(ux)(vx)+(uy)(vy)(uy)(vy)+(uy)(vy)(uz)(vz)
+
+
(uz)(vz)(ux)(vx)+(uz)(vz)(uy)(vy)+(uz)(vz)(uz)(vz)
i
= −(ux)2
(vx)2
−(uy)2
(vy)2
−(uz)2
(vz)2
−2
(ux)(uy)(vx)(vy)
−2
(ux)(uz)(vx)(vz)
−2
(uy)(uz)(vy)(vz)
(g) Expressing equality k~
u×~
vk2
= k~
uk2
k~
vk2
−(~
u·~
v)2
and simplifying, we have:
(uy)2
(vz)2
+(vy)2
(uz)2
+(ux)2
(vz)2
+(vx)2
(uz)2
+(ux)2
(vy)2
+(vx)2
(uy)2
−
2(uy)(uz)(vy)(vz)−2(ux)(uz)(vx)(vz)−2(ux)(uy)(vx)(vy) =
(ux)2
(vx)2
+(ux)2
(vy)2
+(ux)2
(vz)2
+(uy)2
(vx)2
+(uy)2
(vy)2
+(uy)2
(vz)2
+(uz)2
(vx)2
+
(uz)2
(vy)2
+(uz)2
(vz)2
−(ux)2
(vx)2
−(uy)2
(vy)2
−(uz)2
(vz)2
−2(ux)(uy)(vx)(vy)−
2(ux)(uz)(vx)(vz)−2(uy)(uz)(vy)(vz)
(uy)2
(vz)2
+(uz)2
(vy)2
+(ux)2
(vz)2
+(uz)2
(vx)2
+(ux)2
(vy)2
+(uy)2
(vx)2
−
2(uy)(uz)(vy)(vz)−2(ux)(uz)(vx)(vz)−2(ux)(uy)(vx)(vy) =
(uy)2
(vz)2
+(uz)2
(vy)2
+(ux)2
(vz)2
+(uz)2
(vx)2
+(ux)2
(vy)2
+(uy)2
(vx)2
−
2(uy)(uz)(vy)(vz)−2(ux)(uz)(vx)(vz)−2(ux)(uy)(vx)(vy)+(ux)2
(vx)2
−(ux)2
(vx)2
+
(uy)2
(vy)2
−(uy)2
(vy)2
+(uz)2
(vz)2
−(uz)2
(vz)2
5

7. (uy)2
(vz)2
−(uy)2
(vz)2
+(uz)2
(vy)2
−(uz)2
(vy)2
+(ux)2
(vz)2
−(ux)2
(vz)2
+(uz)2
(vx)2
−
(uz)2
(vx)2
+(ux)2
(vy)2
−(ux)2
(vy)2
+(uy)2
(vx)2
−(uy)2
(vx)2
−2(uy)(uz)(vy)(vz)+
2(uy)(uz)(vy)(vz)−2(ux)(uz)(vx)(vz)+2(ux)(uz)(vx)(vz)−2(ux)(uy)(vx)(vy)+
2(ux)(uy)(vx)(vy)−(ux)2
(vx)2
+(ux)2
(vx)2
−(uy)2
(vy)2
+(uy)2
(vy)2
−(uz)2
(vz)2
+
(uz)2
(vz)2
0 = 0
k~
u×~
vk2
= k~
uk2
k~
vk2
−(~
u·~
v)2
(h) So, as (~
u·~
v)2
= k~
uk2
k~
vk2
cos2
θ from the definition of dot product, we have:
k~
u×~
vk2
= k~
uk2
k~
vk2
−k~
uk2
k~
vk2
cos2
θ = k~
uk2
k~
vk2
1−cos2
θ
k~
u×~
vk2
= k~
uk2
k~
vk2
sin2
θ
q
k~
u×~
vk2
=
q
k~
uk2
k~
vk2
sin2
θ .
k~
u×~
vk = k~
uk k~
vk sinθ (8)
2. The length of the cross product of two vectors is equal to the area of the parallelogram determined by the
two vectors (see figure below).
Proof:
Assuming that the vectors ~
u and ~
v are represented by ~
u = (ux,uy,uz), ~
v = (vx,vy,vz), and the angle θ
between the vectors ~
u and ~
v, then we have for example:
6

8. (a) After verifying that the length of the cross product of two vectors is equal to area of parallelogram
determined by two vectors. The formula will be applied given the following vectors:
~
u = (3,1,−2), ~
v = (1,3,−4)
(b) The dot product between the vectors ~
u and ~
v is calculated:
~
u·~
v = (3,1,−2)·(1,3,−4) = 3·1+1·3+(−2)·(−4) = 3+3+8 = 14
~
u·~
v = 14
(c) The length of the vectors ~
u and ~
v is calculated:
k~
uk =
√
14, k~
vk =
√
26
(d) So, as (~
u·~
v) = k~
uk k~
vkcosθ and θ the angle between the vectors ~
u and ~
v from the definition of
dot product, we have:
cosθ =
~
u·~
v
k~
uk k~
vk
, θ = arccos
~
u·~
v
k~
uk k~
vk
= arccos
14
√
14
√
26
= arccos
14
√
364
θ = 42.79◦
(e) The cross product between the vectors ~
u = (3,1,−2) and ~
v = (1,3,−4) is calculated:
~
u×~
v =
î ĵ k̂
3 1 −2
1 3 −4
= (1)(−4)−(3)(−2)
î− (3)(−4)−(1)(−2)
ĵ+ (3)(3)−(1)(1)
k̂
~
u×~
v = 2î+10ĵ+8k̂
(f) The length of cross product between the vectors ~
u = (3,1,−2) and ~
v = (1,3,−4) is calculated:
k~
u×~
vk =
q
(2)2 +(10)2 +(8)2 =
√
4+100+64 =
√
168
k~
u×~
vk =
√
168 = 12.96
(g) (Previously verified property in literal 1). If θ is the angle between the vectors ~
u and ~
v then, the
length of the cross product of two vector is:
k~
u×~
vk = kukkvksinθ
(h) Replacing the values in the literal (g), we have:
k~
u×~
vk =
√
168 = 12.96
k~
u×~
vk = 12.96
k~
uk k~
vk sinθ =
√
14
√
26 sin(42.79◦
) =
√
364 sin(42.79◦
) = 12.96
k~
uk k~
vk sinθ = 12.96
12.96 = 12.96
k~
u×~
vk = k~
uk k~
vk sinθ
7

9. 3. Anticommutative property:
~
u×~
v = −(~
v×~
u)
Proof:
Assuming that the vectors ~
u and~
v are represented by ~
u = (ux,uy,uz) and~
v = (vx,vy,vz), then we have:
(a) The cross product between the vectors ~
u and ~
v is calculated:
~
u×~
v =
î ĵ k̂
ux uy uz
vx vy vz
~
u×~
v = (uy)(vz)−(uz)(vy)
î− (ux)(vz)−(uz)(vx)
ĵ+ (ux)(vy)−(uy)(vx)
k̂
(b) The cross product between the vectors ~
v and ~
u is calculated:
~
v×~
u =
î ĵ k̂
vx vy vz
ux uy uz
~
v×~
u = (uz)(vy)−(uy)(vz)
î− (uz)(vx)−(ux)(vz)
ĵ+ (uy)(vx)−(ux)(vy)
k̂
(c) The cross product between the vectors ~
v and ~
u multiplying by (-1) is calculated:
− ~
v×~
u
= (uy)(vz)−(uz)(vy)
î− (ux)(vz)−(uz)(vx)
ĵ+ (ux)(vy)−(uy)(vx)
k̂
(d) Comparing the expressions in the literals (a) and (c) we have:
~
u×~
v = (uy)(vz)−(uz)(vy)
î− (ux)(vz)−(uz)(vx)
ĵ+ (ux)(vy)−(uy)(vx)
k̂
− ~
v×~
u
= (uy)(vz)−(uz)(vy)
î− (ux)(vz)−(uz)(vx)
ĵ+ (ux)(vy)−(uy)(vx)
k̂
~
u×~
v = −(~
v×~
u) (9)
4. Distributive property for Multiplication by a constant:
(c~
u)×~
v = c(~
u×~
v) = ~
u×(c~
v)
Proof:
Assuming that the vectors ~
u and ~
v are represented by ~
u = (ux,uy,uz), ~
v = (vx,vy,vz) and c ∈ R, then
we have:
(a) The cross product between the vectors c~
u and ~
v is calculated:
(c~
u)×~
v =
î ĵ k̂
cux cuy cuz
vx vy vz
(c~
u)×~
v =
(cuy)(vz)−(cuz)(vy)
î−
(cux)(vz)−(cuz)(vx)
ĵ+
(cux)(vy)−(cuy)(vx)
k̂
8

10. (b) The multiplication by c of the cross product between the vectors ~
u and ~
v is calculated:
c(~
u×~
v) =
î ĵ k̂
ux uy uz
vx vy vz
c(~
u×~
v) = c
h
(uy)(vz)−(uz)(vy)
î−
(ux)(vz)−(uz)(vx)
ĵ+
(ux)(vy)−(uy)(vx)
k̂
i
c(~
u×~
v) =
(cuy)(vz)−(cuz)(vy)
î−
(cux)(vz)−(cuz)(vx)
ĵ+
(cux)(vy)−(cuy)(vx)
k̂
(c) The cross product between the vectors ~
u and c~
v is calculated:
~
u×(c~
v) =
î ĵ k̂
ux uy uz
cvx cvy cvz
~
u×(c~
v) =
(uy)(cvz)−(uz)(cvy)
î−
(ux)(cvz)−(uz)(cvx)
ĵ+
(ux)(cvy)−(uy)(cvx)
k̂
~
u×(c~
v) =
(cuy)(vz)−(cuz)(vy)
î−
(cux)(vz)−(cuz)(vx)
ĵ+
(cux)(vy)−(cuy)(vx)
k̂
(d) Comparing the results of the cross product in the literals a, b and c, it is verified that:
(c~
u)×~
v = c(~
u×~
v) = ~
u×(c~
v) (10)
5. Distributive property:
~
u×(~
v+ ~
w) = ~
u×~
v+~
u× ~
w
Proof:Assuming that the vectors ~
u, ~
v and ~
w are represented by ~
u = (ux,uy,uz), ~
v = (vx,vy,vz), ~
w =
(wx,wy,wz), then we have:
(a) The cross product between the vectors ~
u and ~
v+ ~
w is calculated:
~
u×(~
v+ ~
w) =
î ĵ k̂
ux uy uz
vx +wx vy +wy vz +wz
~
u×(~
v+ ~
w) =
(uy)(vz +wz)−(uz)(vy +wy)
î
−
(ux)(vz +wz)−(uz)(vx +wx)
ĵ
+
(ux)(vy +wy)−(uy)(vx +wx)
k̂
(b) The cross product between the vectors ~
u and ~
v is calculated:
~
u×~
v =
î ĵ k̂
ux uy uz
vx vy vz
~
u×~
v =
(uy)(vz)−(uz)(vy)
î−
(ux)(vz)−(uz)(vx)
ĵ+
(ux)(vy)−(uy)(vx)
k̂
9

11. (c) The cross product between the vectors ~
u and ~
w is calculated:
~
u× ~
w =
î ĵ k̂
ux uy uz
wx wy wz
~
u× ~
w =
(uy)(wz)−(uz)(wy)
î−
(ux)(wz)−(uz)(wx)
ĵ+
(ux)(wy)−(uy)(wx)
k̂
(d) The sum of the cross product between the vectors ~
u, ~
v and ~
u, ~
w is calculated:
~
u×~
v+~
u× ~
w =
(uy)(vz)−(uz)(vy)
+ (uy)(wz)−(uz)(wy)
î
−
(ux)(vz)−(uz)(vx)
+ (ux)(wz)−(uz)(wx)
ĵ
+
(ux)(vy)−(uy)(vx)
+ (ux)(wy)−(uy)(wx)
k̂
Simplifying we have:
~
u×~
v+~
u× ~
w = (uy)(vz +wz)−(uz)(vy +wy)
î
− (ux)(vz +wz)−(uz)(vx +wx)
ĵ
+ (ux)(vy +wy)−(uy)(vx +wx)
k̂
(e) Comparing the results in the literals (a) and (d) we have:
~
u×(~
v+ ~
w) = (uy)(vz +wz)−(uz)(vy +wy)
î
− (ux)(vz +wz)−(uz)(vx +wx)
ĵ
+ (ux)(vy +wy)−(uy)(vx +wx)
k̂
~
u×~
v+~
u× ~
w = (uy)(vz +wz)−(uz)(vy +wy)
î
− (ux)(vz +wz)−(uz)(vx +wx)
ĵ
+ (ux)(vy +wy)−(uy)(vx +wx)
k̂
(f) Equating the right side of the two previous equations component by component we have:
(uy)(vz +wz)−(uz)(vy +wy)
î = (uy)(vz +wz)−(uz)(vy +wy)
î
− (ux)(vz +wz)−(uz)(vx +wx)
ĵ = − (ux)(vz +wz)−(uz)(vx +wx)
ĵ
(ux)(vy +wy)−(uy)(vx +wx)
k̂ = (ux)(vy +wy)−(uy)(vx +wx)
k̂
(g) With the above it was possible to verify the distributive property of the cross product:
~
u×(~
v+ ~
w) = ~
u×~
v+~
u× ~
w (11)
6. The scalar triple product of the vectors ~
u, ~
v, and ~
w:
~
u·(~
v× ~
w) = (~
u×~
v)· ~
w
Proof:Assuming that the vectors ~
u, ~
v and ~
w are represented by ~
u = (ux,uy,uz), ~
v = (vx,vy,vz), ~
w =
(wx,wy,wz), then we have:
10

12. (a) The cross product between the vectors ~
v and ~
w is calculated:
~
v× ~
w =
î ĵ k̂
vx vy vz
wx wy wz
~
v× ~
w = (vy)(wz)−(vz)(wy)
î− (vx)(wz)−(vz)(wx)
ĵ+ (vx)(wy)−(vy)(wx)
k̂
(b) The cross product between the vectors ~
u and ~
v is calculated:
~
u×~
v =
î ĵ k̂
ux uy uz
vx vy vz
~
u×~
v = (uy)(vz)−(uz)(vy)
î− (ux)(vz)−(uz)(vx)
ĵ+ (ux)(vy)−(uy)(vx)
k̂
(c) The scalar triple product of the vectors ~
u·(~
v× ~
w) is calculated:
~
u·(~
v× ~
w) = (ux,uy,uz)·
·
h
(vy)(wz)−(vz)(wy)
î
− (vx)(wz)−(vz)(wx)
ĵ
+ (vx)(wy)−(vy)(wx)
k̂
i
.
~
u·(~
v× ~
w) = +ux · (vy)(wz)−(vz)(wy)
î
−uy · (vx)(wz)−(vz)(wx)
ĵ
+uz · (vx)(wy)−(vy)(wx)
k̂.
~
u·(~
v× ~
w) = + (ux)(vy)(wz)−(ux)(vz)(wy)
î
− (uy)(vx)(wz)−(uy)(vz)(wx)
ĵ
+ (uz)(vx)(wy)−(uz)(vy)(wx)
k̂
(12)
(d) The scalar triple product of the vectors (~
u×~
v)· ~
w is calculated:
(~
u×~
v)· ~
w =
h
(uy)(vz)−(uz)(vy)
î
− (ux)(vz)−(uz)(vx)
ĵ
+ (ux)(vy)−(uy)(vx)
k̂
i
·(wx,wy,wz).
(~
u×~
v)· ~
w = + (uy)(vz)−(uz)(vy)
·(wx)î
− (ux)(vz)−(uz)(vx)
·(wy)ĵ
+ (ux)(vy)−(uy)(vx)
·(wz)k̂.
(~
u×~
v)· ~
w = + (uy)(vz)(wx)−(uz)(vy)(wx)
î
− (ux)(vz)(wy)−(uz)(vx)(wy)
ĵ
+ (ux)(vy)(wz)−(uy)(vx)(wz)
k̂
(13)
11

13. (e) Comparing the results in the literals (c) and (d) we have:
~
u·(~
v× ~
w) = + (ux)(vy)(wz)−(ux)(vz)(wy)
î
− (uy)(vx)(wz)−(uy)(vz)(wx)
ĵ
+ (uz)(vx)(wy)−(uz)(vy)(wx)
k̂
(~
u×~
v)· ~
w = + (uy)(vz)(wx)−(uz)(vy)(wx)
î
− (ux)(vz)(wy)−(uz)(vx)(wy)
ĵ
+ (ux)(vy)(wz)−(uy)(vx)(wz)
k̂
(f) With the above it was possible to verify the scalar triple product of the vectors property of the cross
product:
~
u·(~
v× ~
w) = (~
u×~
v)· ~
w (14)
7. The volume of the parallelepiped determined by the vectors ~
u, ~
v and ~
w is the magnitude of their scalar
triple product.
|~
w·(~
u×~
v)| = volume of the parallelepiped
Proof:Assuming that the vectors ~
u, ~
v and ~
w are represented by ~
u = (ux,uy,uz), ~
v = (vx,vy,vz), ~
w =
(wx,wy,wz), then we have:
(a) The scalar triple product of the vectors ~
w·(~
u×~
v) is:
~
w·(~
u×~
v) =
wx wy wz
ux uy uz
vx vy vz
~
w·(~
u×~
v) = (uy)(vz)−(uz)(vy)
(wx)− (ux)(vz)−(uz)(vx)
(wy)+ (ux)(vy)−(uy)(vx)
(wz)
~
w·(~
u×~
v) = (uy)(vz)(wx)−(uz)(vy)(wx)−(ux)(vz)(wy)
+(uz)(vx)(wy)+(ux)(vy)(wz)−(uy)(vx)(wz)
~
w·(~
u×~
v) = (ux)(vy)(wz)+(uy)(vz)(wx)+(uz)(vx)(wy)
−(ux)(vz)(wy)−(uy)(vx)(wz)−(uz)(vy)(wx)
(b) The magnitude of their scalar triple product is determinate by:
~
w·(~
u×~
v) = (ux)(vy)(wz)+(uy)(vz)(wx)+(uz)(vx)(wy)
−(ux)(vz)(wy)−(uy)(vx)(wz)−(uz)(vy)(wx)
(c) After verifying that the volume of the parallelepiped determined by the vectors ~
u, ~
v and ~
w = ~
u×~
v
is the magnitude of their scalar triple. The formula will be applied given the following vectors:
~
u = (3,1,−2), ~
v = (1,3,−4), ~
w = (2,10,8)
12

14. (d) Replacing the values in the literal (b), we have:
~
w·(~
u×~
v) = (3)(3)(8)+(1)(−4)(2)+(−2)(1)(10)−(3)(−4)(10)−(1)(1)(8)−(−2)(3)(2)
72−8−20+120−8+12 = 168 = 168
8. The vector triple product of the vectors ~
u, ~
v, and ~
w is:
~
u×(~
v× ~
w) = (~
u· ~
w)~
v−(~
u·~
v)~
w
Proof: Assuming that the vectors ~
u, ~
v and ~
w are represented by ~
u = (ux,uy,uz), ~
v = (vx,vy,vz),
~
w = (wx,wy,wz), then we have:
(a) The cross product between the vectors ~
v and ~
w is calculated:
~
v× ~
w =
î ĵ k̂
vx vy vz
wx wy wz
~
v× ~
w = (vy)(wz)−(vz)(wy)
î− (vx)(wz)−(vz)(wx)
ĵ+ (vx)(wy)−(vy)(wx)
k̂
(b) The vector triple product of the vectors ~
u, ~
v, and ~
w is calculated:
~
u×(~
v× ~
w) =
î ĵ k̂
ux uy uz
(vy)(wz)−(vz)(wy)
− (vx)(wz)−(vz)(wx)
(vx)(wy)−(vy)(wx)
~
u×(~
v× ~
w) = +
h
(uy)
(vx)(wy)−(vy)(wx)
+(uz)
(vx)(wz)−(vz)(wx)
i
î
−
h
(ux)
(vx)(wy)−(vy)(wx)
−(uz)
(vy)(wz)−(vz)(wy)
i
ĵ
+
h
−(ux)
(vx)(wz)−(vz)(wx)
−(uy)
(vy)(wz)−(vz)(wy)
i
k̂
13

15. ~
u×(~
v× ~
w) = +
h
(uy)(vx)(wy)−(uy)(vy)(wx)+(uz)(vx)(wz)−(uz)(vz)(wx)
i
î
−
h
(ux)(vx)(wy)−(ux)(vy)(wx)−(uz)(vy)(wz)+(uz)(vz)(wy)
i
ĵ
+
h
−(ux)(vx)(wz)+(ux)(vz)(wx)−(uy)(vy)(wz)+(uy)(vz)(wy)
i
k̂
~
u×(~
v× ~
w) = +
h
(uy)(vx)(wy)+(uz)(vx)(wz)−(uy)(vy)(wx)−(uz)(vz)(wx)
i
î
+
h
(ux)(vy)(wx)+(uz)(vy)(wz)−(ux)(vx)(wy)−(uz)(vz)(wy)
i
ĵ
+
h
(ux)(vz)(wx)+(uy)(vz)(wy)−(ux)(vx)(wz)−(uy)(vy)(wz)
i
k̂
(15)
(c) The right side of the equation in the literal (a) is calculated:
(~
u· ~
w)~
v−(~
u·~
v)~
w
~
u· ~
w = (ux,uy,uz)·(wx,wy,wz) = (ux)(wx)+(uy)(wy)+(uz)(wz)
~
u·~
v = (ux,uy,uz)·(vx,vy,vz) = (ux)(vx)+(uy)(vy)+(uz)(vz)
(~
u· ~
w)~
v = (ux)(wx)+(uy)(wy)+(uz)(wz)
(vx,vy,vz)
+
h
(ux)(vx)(wx)+(uy)(vx)(wy)+(uz)(vx)(wz)
i
î
+
h
(ux)(vy)(wx)+(uy)(vy)(wy)+(uz)(vy)(wz)
i
ĵ
+
h
(ux)(vz)(wx)+(uy)(vz)(wy)+(uz)(vz)(wz)
i
k̂
(16)
(~
u·~
v)~
w = (ux)(vx)+(uy)(vy)+(uz)(vz)
(wx,wy,wz)
+
h
(ux)(vx)(wx)+(uy)(vy)(wx)+(uz)(vz)(wx)
i
î
+
h
(ux)(vx)(wy)+(uy)(vy)(wy)+(uz)(vz)(wy)
i
ĵ
+
h
(ux)(vx)(wz)+(uy)(vy)(wz)+(uz)(vz)(wz)
i
k̂
(17)
(~
u· ~
w)~
v−(~
u·~
v)~
w =
+
h
(ux)(vx)(wx)+(uy)(vx)(wy)+(uz)(vx)(wz)−(ux)(vx)(wx)−(uy)(vy)(wx)−(uz)(vz)(wx)
i
î
+
h
(ux)(vy)(wx)+(uy)(vy)(wy)+(uz)(vy)(wz)−(ux)(vx)(wy)−(uy)(vy)(wy)−(uz)(vz)(wy)
i
ĵ
+
h
(ux)(vz)(wx)+(uy)(vz)(wy)+(uz)(vz)(wz)−(ux)(vx)(wz)−(uy)(vy)(wz)−(uz)(vz)(wz)
i
k̂
(~
u· ~
w)~
v−(~
u·~
v)~
w = +
h
(uy)(vx)(wy)+(uz)(vx)(wz)−(uy)(vy)(wx)−(uz)(vz)(wx)
i
î
+
h
(ux)(vy)(wx)+(uz)(vy)(wz)−(ux)(vx)(wy)−(uz)(vz)(wy)
i
ĵ
+
h
(ux)(vz)(wx)+(uy)(vz)(wy)−(ux)(vx)(wz)−(uy)(vy)(wz)
i
k̂
(18)
(d) Comparing the equations 15 and 18 literals (b) and (c) we have:
~
u×(~
v× ~
w) = +
h
(uy)(vx)(wy)+(uz)(vx)(wz)−(uy)(vy)(wx)−(uz)(vz)(wx)
i
î
+
h
(ux)(vy)(wx)+(uz)(vy)(wz)−(ux)(vx)(wy)−(uz)(vz)(wy)
i
ĵ
+
h
(ux)(vz)(wx)+(uy)(vz)(wy)−(ux)(vx)(wz)−(uy)(vy)(wz)
i
k̂
14

16. (~
u· ~
w)~
v−(~
u·~
v)~
w = +
h
(uy)(vx)(wy)+(uz)(vx)(wz)−(uy)(vy)(wx)−(uz)(vz)(wx)
i
î
+
h
(ux)(vy)(wx)+(uz)(vy)(wz)−(ux)(vx)(wy)−(uz)(vz)(wy)
i
ĵ
+
h
(ux)(vz)(wx)+(uy)(vz)(wy)−(ux)(vx)(wz)−(uy)(vy)(wz)
i
k̂
(e) With the above it was possible to verify the vector triple product of the vectors ~
u, ~
v, and ~
w is:
~
u×(~
v× ~
w) = (~
u· ~
w)~
v−(~
u·~
v)~
w
15