VECTORS  

Vectors  in  2-­‐D  and  3-­‐D  in  Euclidean  space  or  flatland  are  easy  compared  to  vectors  in  non-­‐Euclidean  space.  In  Cartesian  coordinates  we  write  a  component  of  a  vector  as  

!!  

where   the   index   i   stands   for   either   x,   y   or   z.   For   instance,  !!  represents   the   y-­‐component   of   the  velocity  vector.  In  general,  we  identify  a  complete  vector  by  a  letter  with  an  arrow  on  top,    

!  

The  magnitude  of  vector  !  is  just  A  without  the  arrow,   ! = !.  

What  is  a  vector?  A  3-­‐D  vector  is  a  mathematical  object  that  has  three  independent  components  with  laws  that  distinguish  them  as  vectors.  Vectors  obey  laws  of  addition,  multiplication,  and  coordinate  transformations  such  as  rotations.  

Given  two  vectors  !  and  !  we  can  express  the  law  of  vector  addition  geometrically.    

 

 

 

 

 

Fig.  1  Geometrical  definition  of  vector  addition  

We  fix  one  vector  say  !,  we  bring  vector  !  to  vector  !  in  such  a  way  that  the  direction  and  magnitude  of  !  are  unchanged   (this   is   called  parallel   transport).  We  place   the   tail   of   vector  !  at   the   tip  of   the  arrow  of  vector  !.  The  vector  sum  is  the  vector  with  a  tail  at  the  tail  of  vector  !  and  with  the  arrow  at  the  arrow  of  vector  !,   as  we  can   see   in  Fig.  1.   If  we  would   like   to  add  more  vectors  we   follow   the  same  procedure  until  we  have  included  every  vector  in  any  order.  

The  analytical  approach   to  add  vectors   is  easiest   in  Cartesian  coordinates.  The  2-­‐D  case   is  best   for  showing  the  approach.  We  define  a  xy-­‐coordinate  system.  We  bring  vector  !  so  that  its  tail   is  at  the  origin.  We  draw  a  right  triangle  that  with  one  of  its  sides  on  the  x-­‐axis  or  y-­‐axis.  Out  of  the  sides  of  the  right  triangle  we  made  two  vectors  !!  and  !!  in  such  a  way  that  vector  !  is  the  sum  

! = !! + !!  

 

 

!⃗   !!⃗  !⃗+ !!⃗  

+   =  

!⃗  

!!⃗  

 

 

 

 

 

 

 

Fig.   2   Using   definition   of   vector   addition   we   break   vector   into   components  ! = !! + !! .  

Now,  we   introduce   the  useful  concept  of  unit  vector.  A  unit  vector   is   simple  a  vector  of   length  one  unit  in  a  given  direction.  We  notice  that  instead  of  an  arrow  we  use  a  hat  on  top  of  the  letter  assigned  to   the  vector.  For   instance,  we  call  !  the  vector  of   length  one  along   the  positive  x-­‐axis.   In   the  same  way  we  call  !  the  vector  of   length  one  along  the  positive  y-­‐axis  and  !  the  vector  of   length  one  along  the  positive  z-­‐axis  

 

 

 

 

 

 

 

 

Fig.  3  The  set  of  positive  x,  y  and  z  axes  and  their  corresponding  unit  vectors  

Using  unit  vectors  we  can  write  vector  !! = !!!  and  vector  !! = !!!.  Thus,  we  now  can  write  

! = !!! + !!!  

All  other  vectors  are  broken  in  the  same  way;  for  instance,  ! = !!! + !!!.  The  sum  of  vectors  is  now  simply  

! + ! = !! + !! ! + !! + !! !  

!  

!  

!  

!!  

!̂  

!!  

!⃗  

!  

!  

!⃗!  

!⃗!  

Vector   in   component   form   is   the   preferred  way   to   report   them.  The   alternative  way   to   describe   a  vector   using   magnitude   and   angle   is   less   useful   and   can   be   confusing.   If   we   are   required   to   give  magnitude  and  angle  we  must  include  a  sketch  of  the  vector,  the  angle  and  the  coordinate  system.  

Example:   Given   the   vectors  !:  ! = 10  and   angle   50   degrees  with   the   positive   x-­‐axis,  !:  ! = 15  and  angle  120  degrees  with  the  positive  x-­‐axis,  and  !:  ! = 6  and  angle  240  degrees  with  the  positive  x-­‐axis  find  their  vector  sum  ! = ! + ! + !.    

First  we   form   right   triangles  with   for   every   vector  with   either   x-­‐axis   or   y-­‐axis.   In  general  we  do  not  use  angles  grater  than  90  degrees  as  they  make  things  confusing.  It  is  better  to  use  the  sign  convention  for  the  axis:  +x  is  to  the  right,  +y  is  up  and  +z  is  out  of  the  page.  

 

 

 

 

 

 

 

 

 

 

 

Fig.  4  We  break  vectors  into  components  using  right  triangles  

We  use  trigonometry  to  find  the  sides  of  each  triangle  and  write  the  components  

!! = 10 cos 50° = 6.43;  !! = 10 sin 50° = 7.66  

!! = − 15cos 60° = −7.5;  !! = 15 sin 60° = 12.99  

!! = −6 sin 30° = −3;  !! = −6 cos 60° = −5.2  

We   notice   that   the   signs   of   the   components   are   place   by   hand   by  looking  at  the  side  of  the  x-­‐axis  and  y-­‐axis  where  the  components  are.    

!!⃗   !!⃗  

!!⃗  

!  

!  

50°  60°  

30°  

!!  

!!  

(link  for  the  sign  convention).  The  components  of  the  vector  sum  !  are  just  the  regular  addition  of  x  and  y  components:  

!! = −4.07;  !! = 15.45  

Dot  product  

The  dot  product   is  a   scalar.  A  scalar   is  a  number   that  does  not  change  under  common  coordinates  transformations   such  as   rotations.  An  example  of   a   scalar   is   the   length  of   a   vector  which  does  not  change  when  the  coordinates  are  rotated.  Given  vectors  !  and  !  the  dot  product  is  

! ∙ ! = !" cos !!"  

where  C  and  D  are  the  magnitudes  of  the  vectors  and  !!"  is  the  angle,  smaller  than  180°,  between  the  vectors.  It  is  useful  to  know  the  dot  product  between  unit  vectors;  for  instance,  

! ∙ ! = 1 1 cos 90° = 0  

! ∙ ! = 1 1 cos 0° = 1  

We  summarize  the  results  

! ∙ ! = ! ∙ ! = ! ∙ ! = 0  

! ∙ ! = ! ∙ ! = ! ∙ ! = 1  

As  an  example  let  us  compute  the  dot  product  ! ∙ !  given  ! = 3! − 2!  and  ! = ! − 4! + 2!  

! ∙ ! = 3! − 2!   ∙ ! − 4! + 2!= 3! ∙ ! − 12! ∙ ! + 6! ∙ ! − 2!   ∙ ! + 8!   ∙ ! − 4!   ∙ !  

Using  the  table  above  we  get  ! ∙ ! = −12 − 4 = −16.  

We  would  like  to  find  the  angle  between  to  !  and  !  using  the  dot  product  definition  ! ∙ ! = !" cos !!"  

We  use  Pythagoras  theorem  to  calculate  the  magnitudes  given  the  vectors  in  regular  coordinates;  so  we  get  ! = 13; ! = 21.  We  put  all  the  pieces  together  and  we  get  

−16 = 13 21 cos !!"  

We  use  the  inverse  cosine  and  get  

!!" = 165.55°  

Cross  Product  

Given  vectors  !  and  !  we  define  the  cross  product  !  as    

! = !×!  

which  is  a  new  vector  with  the  following  properties:    

The  magnitude  of  !  is  given  by  

! = !" sin !!"  

We  note  that  the  angle  !!"  is  less  than  180  degrees  so  the  magnitude  is  always  positive,  as  it  should  be.  The  direction  of  !  is  along  a  line  perpendicular  to  the  plane  formed  by  vectors  !  and  !.  Along  this  perpendicular   line,   the  direction  of  !  is  given  by  the  right  hand  rule:  we  place  vectors  !  and  !  with  their   tails   at   a   common   origin.  We   bring   our   right   hand   to   that   origin.  We   keep   our   four   fingers  together  and  the  thumb  at  90  degrees.  On  the  plane  expanded  by  the  vectors,  we  turn  our  four  fingers  from   the   first   vector   (!)   towards   the   second   vector   (!)   along   the   angle  !!" .   The   direction   of   the  thumb  is  the  correct  direction  of  !.  

 

 

 

 

 

 

 

 

 

 

Fig.  5  Right  had  rule:  the  four  fingers  of  our  right  hand  move  from  !  to  !;  the  thumb  is  along  !.  

As  an  example  let  us  compute  the  cross  product  of  unit  vectors  !×!.  First  the  magnitude  

!×! = 1 1 sin 90 = 1  

Since   the   vectors  !  and  !  expand   the   xy-­‐plane   we   know   that   the   direction   of   the   cross   product   is  along  the  z-­‐axis,  perpendicular  to  this  plane.  We  place  our  right  hand  at  the  origin  and  when  our  four  fingers  move  from  the  positive  x-­‐axis  to  the  positive  y-­‐axis  our  thumb  points  along  the  positive  z-­‐axis;  thus,  we  conclude  that  

!⃗  

!⃗  

!!⃗  

!!"  

!×! = !  

Similar  calculations  lead  us  to  conclude  that  

!×! = !;  !×! = !;  !×! = !  

We  note  that  the  order  is  important;  if  we  reverse  the  order  we  get  the  negative  

!×! = −!  

Finally,  the  cross  product  of  any  vector  with  itself  is  always  zero  

!×! = !×! = !×! = 0  

As  an  example  let  us  compute  the  cross  product  ! = !×!  given  ! = 3! − 2!  and  ! = ! − 4! + 2!  

!×! = 3! − 2!   × ! − 4! + 2!= 3!×! − 12!×! + 6!×! − 2!  ×! + 8!  ×! − 4!  ×!  

Using  the  table  above  we  get  

! = !×! = −3! + 6! − 2! − 8! = −2! − 2! − 3!  

To   check   our   answer   we   may   verify   that   the  !  is   perpendicular   (dot   product   is  zero)  to  !  and  !  

! ∙ ! = −2! − 2! − 3! ∙ 3! − 2! = −6 + 6 = 0  

! ∙ ! = −2! − 2! − 3! ∙ ! − 4! + 2! = −2 + 8 − 6 = 0  

We   would   like   to   find   the   angle   between   to  !  and  !  using   the   definition   of   cross  product  magnitude  

! = !" sin !!"  

We  use  Pythagoras  theorem  to  calculate  the  magnitude  of  vector  !,  ! = 17.  Thus,  we  get  

sin !!" =17

13×21  

Using  the  inverse  sine  we  get  

!!" = 14,45°  

Unfortunately,  this  is  not  the  answer  we  got  using  the  dot  product   !!" = 165,55° .  The   problem   is   that   the   sine   function   gives   the   exact   same   value   for  14,45°  and  165,55°  and  most  calculators  will  just  give  you  the  first  answer.  Therefore,  the  best  

way  to  avoid  complications  when  finding  the  angle  between  two  vectors  is  using  the  dot  product.    

There  are  other  ways  to  calculate  the  cross  product  of  two  vectors,  ! = !×!.  A  common  approach  is  to  calculate  the  determinant  of  the  array  

! =! ! !!! !! !!!! !! !!

= ! !!!! − !!!! + ! !!!! − !!!! + ! !!!! − !!!!  

 

BASIC  VECTOR  IDENTITIES  

! ∙ !×! = ! ∙ !×! = ! ∙ !×!  

!× !×! = ! ! ∙ ! − ! ! ∙ !  

To  show  that  these  identities  are  correct  we  use  Cartesian  coordinates   ! = !!! + !!! + !!!  for  all  the  vectors   involved.  Next,  we  write  all   the   terms  on  each  side  of   the  equation  and  make  sure   that  they  appear  on  both  sides.   In  cases  where  the  expressions  are  vectors,  such  as  the  second  identity,  we  only  need  to  show  that  one  of  the  three  components  (x,  y,  or  z)  works.    

Exercises:  

1. Given  the  vectors  !:  ! = 8  and  angle  20  degrees  with  the  positive  x-­‐axis,  !:  ! = 5  and  angle  150  degrees  with  the  positive  x-­‐axis,  and  !:  ! = 10  and  angle  250  degrees  with  the  positive  x-­‐axis  (a)  find  their  vector  sum  ! = ! + ! + !.  (b)  Find  the  angle  between  ! + !  and  !.  (c)  Find  the  cross  product  !×!.  

2. Given   the   vectors  !:  ! = 5  and   angle   60   degrees   with   the   positive   x-­‐axis,  !:  ! = 10  and  angle   120   degrees   with   the   positive   x-­‐axis,   and  !:  ! = 6  and   angle   290   degrees   with   the  positive  x-­‐axis   find   (a)   their  vector   sum  ! = ! + ! + !.   (b)  Find   the  angle  between  ! − !  and  !.  (c)  Find  the  cross  product  !×!.  

3. Given  the  vectors  !:  ! = 8  and  angle  20  degrees  with  the  positive  x-­‐axis,  !:  ! = 5  and  angle  150  degrees  with  the  positive  x-­‐axis,  and  !:  ! = 10  and  angle  250  degrees  with  the  positive  x-­‐axis  find  (a)  their  vector  sum  ! = ! + ! + !.  (b)  Find  the  angle  between  ! + !  and  !.  (c)  Find  the  cross  product  !×!.  

4. Verify  the  identity  ! ∙ !×! = ! ∙ !×! = ! ∙ !×!  5. Verify  the  identity  !× !×! = ! ! ∙ ! − ! ! ∙ !  Note:  only  one  component  is  enough.