Download Vector Practice

Vectors
Quantities that have direction as well as magnitude are called as vectors. Examples of
vectors are velocity, acceleration, force, momentum etc.
Vectors can be added and subtracted. Let a and b be two vectors. To get the sum of the two
vectors, place the tail of b onto the head of a and the distance between the tail of a
and b is a+b.
Multiplication of a vector by a positive scalar k multiplies the magnitude but leaves the
direction unchanged. If k=2 then the magnitude of a doubles but the direction
remains the same.
Dot product of two vectors is the product of a vector to the projection
of the other vector on the vector. a. b is called the dot product
of the two vectors.
a. b = a b cos . If the two vectors are parallel, then a. b = a b
And if the two vectors are perpendicular to each other,
then a. b = 0
Cross Product of any two vectors is defined by a  b= c = a b sin  n̂ ,
where n̂ is a unit vector (vector of length 1) pointing
perpendicular to the plane of a and b. But as there are two
directions perpendicular to any plane, the ambiguity is
resolved by the right hand rule: let your fingers point in the
direction of the first vector and curl around (via the smaller
angle) towards the second; then your thumb indicates the
direction of n̂ .
Unit vector is a vector whose magnitude is 1 and point is a particular direction. Without loss
of generality, we can assume iˆ, ˆj , kˆ to be three distinct unit vectors along the x, y, and
z-axis relatively.
Then,
iˆ  iˆ  ˆj  ˆj  kˆ  kˆ  1 and
iˆ  ˆj  ˆj  kˆ  kˆ  iˆ  0
Also,
iˆ  iˆ  ˆj  ˆj  kˆ  kˆ  0
iˆ  ˆj   ˆj  iˆ  kˆ
ˆj  kˆ  kˆ  ˆj  iˆ
kˆ  iˆ  iˆ  kˆ  ˆj

In cylindrical coordinate systems, a vector A  As sˆ  A ˆ  Az zˆ , where sˆ,ˆ, zˆ are the unit
vectors of the coordinate system. 
In spherical coordinate system, a vector A  Ar rˆ  A ˆ  A ˆ , where rˆ,ˆ,ˆ are the unit
vectors of the coordinate system.
Vector Practice
1. Consider three vectors:

A  3iˆ  0 ˆj

B  2 3iˆ  2 ˆj

C  5iˆ  5 3 ˆj
a. Draw the three vectors.
 

b. What is the length or magnitude of A , B and C ?

Length of the vector A 

Length of the vector B 

Length of the vector C 

A  Ax2  Ay2 = 32  02 = 3

B  B x2  B y2 = (2 3) 2  2 2 = 4

C  C x2  C y2 = (5) 2  (5 3) 2 =10
 


 
c. What is the angle between A and C , A and B , B and C ?
   
We know that A  C = A C cos ,
 

AC
So, cos    
So,   cos 1 

AC

 
ˆ
ˆ
ˆ
ˆ
Now, A  C = ( Ax i  Ay j )  (C x i  C y j )
  
AC 
 
A C

= Ax C x  Ay C y
= 3  (5)  0  5 3
 15
 
A C  3  10  30
 
1
A  C  15
     either 120 or 240 .
So, cos     
30
2
AC

But notice that the y-component of C is positive and the x-component negative. So it

is in the 2nd quadrant and the A vector is in the first quadrant. So, the angle between


them cannot be more than 180 . So the angle between A and C is 120 .


A
B
Similarly, the angle
between
and
is
 
 
 
A B
A  B  3  (2 3 )  0  2  6 3 ,
A  B  3  4  12
cos    ,
AB
 
3
A B 6 3
   either 30 or 330 . But   330 , as none of the

cos    
12
2
AB


components of A and B are negative. So,    30


B and C is
Similarly, the angle
between
 
 
B C
cos     ,
B  C  (2 3)(5)  2  (5 3)  0 ,
BC
 
B   C  4  10  40
 
0
B C
 0    either 90 or 270 . But   270 , as none of the
cos     
B C 40


components of B and C are negative. So,    90
2. Consider three vectors:

A  4iˆ  6 ˆj  2kˆ

B  2iˆ  7 ˆj  1kˆ

C  0iˆ  3 ˆj  5kˆ


a. What is the length or magnitude of A , also written as A ?

Length of the vector A 
Ax2  Ay2  Az2 = 4 2  6 2  (2) 2 = 56 = 2 14

b. Write the expression for 2 A .


2 A = 2( A ) = 2( 4iˆ  6 ˆj  2kˆ ) = 8iˆ  12 ˆj  4kˆ
 
c. What is A  B ?
 
A  B = ( Ax iˆ  Ay ˆj  Az kˆ)  ( Bx iˆ  B y ˆj  Bz kˆ) = ( Ax  Bx )iˆ  ( Ay  B y ) ˆj  ( Az  Bz )kˆ
= (4  2)iˆ  (6  7) ˆj  (2  1)kˆ
= 6iˆ  13 ˆj  3kˆ
 
d. What is C  A ?
 
C  A = (Cxiˆ  C y ˆj  Cz kˆ)  ( Axiˆ  Ay ˆj  Az kˆ) = (C x  Ax )iˆ  (C y  Ay ) ˆj  (C z  Az )kˆ
= (0  4)iˆ  (3  6) ˆj  (5  (2)) kˆ
=  4iˆ  3 ˆj  7kˆ
 
e. What is C  A ?
 
C  A = (C x iˆ  C y ˆj  C z kˆ)  ( Ax iˆ  Ay ˆj  Az kˆ)
=
kˆ
iˆ
ˆj
Cx
Cy
Cz =
Ax
Ay
Az
Cy
Ay
Cz
Cx
iˆ 
Az
Ax
Cz
Az
ĵ 
Cx
Cy
Ax
Ay
k̂
= (C y Az  C z Ay )iˆ  (C x Az  C z Ax ) ˆj  (C x Ay  C y Ax )kˆ
= (3  (2)  5  6)iˆ  (0  (2)  5  4) ˆj  (0  6  3  4)kˆ
=  36iˆ  20 ˆj  12kˆ
 
f. What is the magnitude of C  A ?
 
 
Magnitude of C  A = C  A
 

Let us name C  A = M


Then magnitude of M = M  M x2  M y2  M z2
= (36) 2  20 2  (12) 2 = 1840 = 4 115
 
g. What is B  C ?
 
B  C = ( Bx iˆ  B y ˆj  Bz kˆ)  (C x iˆ  C y ˆj  C z kˆ)
= Bx C x  B y C y  Bz C z
= 2  0  7  3  (1)  5
=16
Note:
 
B  C is the dot product of the two vectors.

   

B  C = B C cos  , where  is the angle formed by B and C when they are place tailto-tail.


h. What is the angle between A and C ?
 
 
We know that A  C = A C cos  ,
 
AC
So, cos    
AC
   
AC
So,   cos 1    
A C




 2 
8
  cos 1 

So   cos 1 
 2 14  34 
 119 
 
 
i. Does B  C equal C  B ?
 
Yes. We can see that B  C is a scalar (hence the alternative name scalar product).
   
So the dot product is commutative. That is B  C = C  B
 
 
j. How is C  A and A  C related?
Using the formula for the cross product,
 
C  A = (C x iˆ  C y ˆj  C z kˆ)  ( Ax iˆ  Ay ˆj  Az kˆ)
iˆ
ˆj
kˆ
= Cx
Cy
Cz
Ax
Ay
Az
= (C y Az  C z Ay )iˆ  (C x Az  C z Ax ) ˆj  (C x Ay  C y Ax )kˆ
---1
 
And A  C = ( Ax iˆ  Ay ˆj  Az kˆ)  (C x iˆ  C y ˆj  C z kˆ)
=
iˆ
ˆj
kˆ
Ax
Ay
Az
Cx
Cy
Cz
= (C z Ay  C y Az )iˆ  (C z Ax  C x Az ) ˆj  (C y Ax  C x Ay )kˆ
---2
Equating 1) and 2), we can clearly see that
 
 
( C  A ) = -( A  C )
k. Give an example of the use of dot product in Physics and explain.
Mechanical work is a dot product between the force and the displacement.
 
W  F r
It means that the work is amount of displacement times the projection of force along
the displacement.
l. Give an example of the use of cross product in Physics and explain.
Magnetic force is a good example of a cross product of velocity of a charged particle
times the magnetic field.

 
F  q (v  B )


It means that the force is perpendicular to both v and B , and that the magnitude of
the force is equal to the area of the parallelogram they span.

m. Imagine that the vector A is a force whose units are given in Newtons. Imagine vector

B is a radius vector through which the force acts in meters. What is the value of the
  
torque (  r  F ) , in this case?
  
Torque is the cross product between the radius vector and the force vector, (  r  F ) .
  
So, in this case   B  A
 
B  A = ( Bx iˆ  B y ˆj  Bz kˆ)  ( Ax iˆ  Ay ˆj  Az kˆ)
iˆ
ˆj
kˆ
= Bx
By
Bz
Ax
Ay
Az
= ( B y Az  Bz Ay )iˆ  ( Bx Az  Bz Ax ) ˆj  ( Bx Ay  B y Ax )kˆ
= (7  (2)  (1)  6)iˆ  (2  (2)  (1)  4) ˆj  (2  6  7  4)kˆ
=  8iˆ  0 ˆj  16kˆ
=  8(iˆ  2kˆ) N  m


n. Now imagine that A continues to be a force vector and C is a displacement vector whose


units are meters. What is the work done in applying force A through a displacement C ?


Work done in applying force A through a displacement C is
 
W  AC
= ( Ax iˆ  Ay ˆj  Az kˆ)  (C x iˆ  C y ˆj  C z kˆ)
= Ax C x  Ay C y  Az C z
= 4  0  6  3  (2)  5
=8 joules


o. What is the vector sum of a vector D given by 40 m, 30 degrees and a vector E given by
12 m, 225 degrees? Use the method of resolving vectors into their components and then
adding the components.
Let us start by resolving the vectors into their x-components and y-components.
It can be assumed that the angles given in the question are w.r.t the x-axis.

For D :


The magnitude of D = D is 40 and the angle is 30o


So, as the projection of D along the x-axis = D cos 
3
D x = 40 cos 30 = 40 
 20 3
2


The projection of D along the y-axis = D sin 
1
D y  40 sin 30  40   20
2

Similarly for E :
 1 
E x  12 cos 225  12  
  6 2
 2
 1 
E y  12 sin 225  12  
  6 2
 2
 
So, D + E = ( D x xˆ  D y yˆ )  ( E x xˆ  E y yˆ ) = ( D x  E x ) xˆ  ( D y  E y ) yˆ
= (20 3  6 2 ) xˆ  (20  6 2 ) yˆ
3. Consider three vectors:

A  3iˆ  3 ˆj  2kˆ

B  2iˆ  4 ˆj  2kˆ

C  2iˆ  3 ˆj  1kˆ
  
a. Find A  ( B  C ) .
First we shall find
 
B  C = ( B x iˆ  B y ˆj  B z kˆ)  (C x iˆ  C y ˆj  C z kˆ) = ( Bx  C x )iˆ  ( B y  C y ) ˆj  ( Bz  C z )kˆ
= (( 2)  2)iˆ  (( 4)  3) ˆj  (2  1)kˆ
= 0iˆ  1 ˆj  3kˆ
  
Let us name B  C  E
 
  
Now, A  ( B  C ) = A  E  ( Ax iˆ  Ay ˆj  Az kˆ)  ( E x iˆ  E y ˆj  E z kˆ)
= Ax E x  Ay E y  Az E z
= (3)  0  3  (1)  2  3
=3
  
b. Find A  ( B  C ) .
First we shall find
 
B  C = ( Bx iˆ  B y ˆj  Bz kˆ)  (C x iˆ  C y ˆj  C z kˆ)
=
iˆ
ˆj
kˆ
Bx
By
Bz
Cx
Cy
Cz
= ( B y C z  Bz C y )iˆ  ( Bx C z  Bz C x ) ˆj  ( Bx C y  B y C x )kˆ
= (( 4)  1  2  3)iˆ  (( 2)  1  2  2) ˆj  (( 2)  3  (4)  2)kˆ
=  10iˆ  6 ˆj  2kˆ
  
Let us name B  C  D
    
Now, A  ( B  C ) = A  D  ( Ax iˆ  Ay ˆj  Az kˆ)  ( Dx iˆ  D y ˆj  Dz kˆ)
= Ax Dx  Ay D y  Az D z
= (3)  (10)  3  6  2  2
= 52

 
c. Find A  ( B  C ) .
First we shall find
 
B  C = ( B x iˆ  B y ˆj  B z kˆ)  (C x iˆ  C y ˆj  C z kˆ) = ( Bx  C x )iˆ  ( B y  C y ) ˆj  ( Bz  C z )kˆ
= (( 2)  2)iˆ  (( 4)  3) ˆj  (2  1)kˆ
= 0iˆ  1 ˆj  3kˆ
  
Let us name B  C = F
 

 
Now A  ( B  C ) = A  F
 
A  F = ( Ax iˆ  Ay ˆj  Az kˆ)  ( Fx iˆ  Fy ˆj  Fz kˆ)
=
iˆ
ˆj
kˆ
Ax
Ay
Az
Fx
Fy
Fz
= ( Ay Fz  Az Fy )iˆ  ( Ax Fz  Az Fx ) ˆj  ( Ax Fy  Ay Fx )kˆ
= (3  3  2  (1))iˆ  (( 3)  3  2  0) ˆj  (( 3)  (1)  3  0)kˆ
= 11iˆ  9 ˆj  3kˆ