# Vector analysis ```Dot product: the scalar or dot product is defined by u  v  u v cos()
u  a1ˆi  b1ˆj c1kˆ , v  a 2ˆi  b2ˆj c2kˆ  uˆ  vˆ  a1a 2  b1b2  c1c2
ˆi ˆi  ˆjˆj kˆ kˆ 1 , ˆi ˆj ˆjkˆ  kˆ ˆi  0 , u  v  v u , u  a 2 b 2 c 2 ,
1
1
1
v  a 22 b22 c22 and the angle between the 2 vectors u &amp; v is defined
a1a 2 b1b2 c1c2
u

v

by cos() 
uv
a12 b12 c12 a 22 b22 c22
If p  f(t)iˆ g(t)jˆ h(t)kˆ , q  r(t)iˆ s(t)jˆ  w(t)kˆ are 2 vector functions,
then
d (p  q)  d (p)  q  p  d (q) = [f`(t)iˆ g`(t)jˆ  h`(t)k]
ˆ [r(t)iˆ s(t)jˆ  w(t)k]
ˆ
dt
dt
dt
ˆ [r`(t)iˆ s`(t)jˆ  w`(t)k]
ˆ
+ [f(t)iˆ g(t)jˆ  h(t)k]
If two vectors a &amp;b are orthogonal then a b  0
1
Example
Find the angle between the 2 vectors u  3iˆ  2jˆ  7kˆ , v ˆi 3jˆ 5kˆ
Solution
ˆ (ˆi 3jˆ 5k)
ˆ  3(1)  2(3)  7(5) 44
u  v  (3iˆ  2jˆ  7k)
u  (3)2 (2)2 (7)2  62 , v  (1)2 (3)2 (5)2  35
cos()  44
62 35
2
Example
Find the first derivative of the scalar product of the 2 vector functions
p  (t2)iˆ (2t 1)jˆ (e3t )kˆ , q  n(t)iˆ sin(3t)jˆ (1)kˆ
t
Solution
d (p  q)  d (p)  q  p  d (q) = [(2t)iˆ  (2)jˆ  (3e3t )k]
ˆ [ n(t)iˆ sin(3t)jˆ (1)k]
ˆ
t
dt
dt
dt
ˆ [(1)iˆ  3cos(3t)jˆ  ( 1 )k]
ˆ
+ [(t2)iˆ  (2t 1)jˆ (e3t )k]
t
t2
3t
3t
e
3e
= [2t n(t)  2sin(3t)  (
)] + [t 3(2t 1)cos(3t)  ( )]
t
t2
3
Cross product: the vector product of 2 vectors u  a1ˆi  b1ˆj c1kˆ and
v  a 2ˆi  b2ˆj c2kˆ is defined by
ˆi
u  v  a1
a2
ˆj
kˆ
c1  (b1c2  c1b2)iˆ  (a1c2  c1a2)jˆ + (a1b2  b1a 2)kˆ
b 2 c2
b1
ˆiˆi  ˆjˆj kˆ kˆ  0 , ˆiˆj kˆ &amp; ˆjkˆ  ˆi &amp; kˆ ˆi  ˆj , u v  (v u)
The area of the parallelogram having vectors u and v as sides equal
u v  (b1c2 c1b2)2 (a1c2 c1a 2)2 (a1b2 b1a 2)2
If two vectors a &amp;b are parallel, then a b  0 (zero vector)
4
The volume of a parallelepiped having vectors u , v and w as sides
a1
equal a 2
a3
If
b1
c1
b2 c2 , u  a1ˆi  b1ˆj c1kˆ , v  a 2ˆi  b2ˆj c2kˆ , w  a ˆi  b ˆjc kˆ
3
3
3
b3 c3
p  f (ˆt ) i gˆ (t ) j ˆ, hq(t r(t)i
) k ˆ s(t)jˆ  w(t)kˆ are 2 vector functions,
then d (p  q)  d (p)  q  p  d (q)
dt
dt
dt
5
Example
Find the volume of the parallelepiped spanned by the vectors
u = (1,0,2)
v = (0,2,3)
w = (0,1,3)
Solution
1 0 2
Volume = 0 2 3  3
0 1 3
6
Example
Find the area of the parallelogram formed by the vectors a and b , given
a = (1, -1, 3), b = (2, 7, -2)
Solution
ˆi
ˆj
a b  1 1
2
7
kˆ
3 19iˆ 8jˆ 9kˆ ,
2
Therefore the area of the parallelogram =
7
(19)2 (8)2 (9)2  506
Example
Find the first derivative of the vector product of the 2 vector functions
p  (t2)iˆ (2t 1)jˆ (e3t )kˆ , q  n(t)iˆ sin(3t)jˆ (1)kˆ
t
Solution
d (p  q)  d (p)  q  p  d (q) = [(2t)iˆ  (2)jˆ  (3e3t )k]
ˆ [ n(t)iˆ sin(3t)jˆ (1)k]
ˆ
t
dt
dt
dt
ˆi
ˆ  [(1)iˆ  3cos(3t)jˆ  ( 1 )k]
ˆ =
+ [(t2)iˆ (2t 1)jˆ (e3t )k]
2t
t
t2
n(t)
ˆi
+ t2
1
t
ˆj
kˆ
2t 1
e3t
3cos(3t)
 12
t
= ???? (complete)
8
ˆj
kˆ
2
3e3t
sin(3t)
1
t
For any continuous function f(x, y, z)
Grad (f) = f   f (x,y,z) ˆi+  f (x,y,z) ˆj+  f (x,y,z) kˆ
x
y
z
Divergence
For any vector function v(x,y,z)  f(x,y,z) ˆi  g(x,y,z) ˆj h(x,y,z) kˆ
Div( v(x,y,z) ) = f   f (x,y,z) +  g(x,y,z) +  h(x,y,z)
x
y
z
9
Curl
Curl ( v(x,y,z) ) = f 
ˆi
ˆj
kˆ

x

y

z
f (x,y,z) g(x,y,z) h(x,y,z)
= [  h(x,y,z) -  g(x,y,z)] ˆi - [  h(x,y,z) -  f (x,y,z)] ˆj
y
z
x
z
+ [  g(x,y,z) -  f (x,y,z)] kˆ
x
y
Exercise
1) Find the gradient of the scalar field f(x, y, z) = x2y + xcosh yz.
2) Find curl( u )