# For any two vector u and v show that: (u.v)²-[(u×v)×v].u=u²v²

1
by ItzBittuPaul

2015-04-01T15:23:25+05:30

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Let  U = Ux i + Uy j + Uz k ,
i,  j and k are unit vectors in x, y and z directions respectively.
let  V = Vx i + Vy j + Vz k

Now,    U . V = Ux Vx + Uy Vy + Uz Vz  = dot product
U Χ V = cross product = (Ux i + Uy j + Uz k) X (Vx i + Vy j + Vz k)
= Ux Vy k - Ux Vz j - Uy Vx k + Uy Vz i + UzVx j - Uz Vy i
= (Uy Vz - Uz Vy) i + (Uz Vx - Ux Vz) j + ((Ux Vy - Uy Vx) k

(U Χ V) Χ V
=
[ (Uy Vz-Uz Vy) i + (Uz Vx-Ux Vz) j + (Ux Vy-Uy Vx) k] Χ (Vx i + Vy j + Vz k)
= (Uy Vz - Uz Vy)Vy k - (Uy Vz - Uz Vy) Vz j - (Uz Vx - Ux Vz) Vx k
+ (Uz Vx - Ux Vz) Vz i + (Ux Vy - Uy Vx) Vx j - (Ux Vy - Uy Vx) Vy i
= (Uy Vy Vz + Ux Vx Vz - Uz Vy² - Uz Vx² ) k
+ (Uz Vy Vz + Ux Vx Vy - Uy Vz² - Uy Vx²) j
+ (Uz Vx Vz + Uy Vx Vy - Ux Vz² - Ux Vy² ) i

[ (U Χ V) Χ V ] . U =
= (Ux Vx Uz Vz + Ux Vx Uy Vy - Ux² Vz² - Ux² Vy²)
+ (Uy Uz  Vy Vz + Ux Uy Vx Vy - Uy² Vz² - Uy² Vx²)
+ (Uy UzVy Vz + Ux Uz Vx Vz - Uz² Vy² - Uz² Vx²)

(U . V)² = (Ux² Vx² + Uy² Vy² + Uz² Vz² + 2 Ux Vx Uy Vy + 2 Uy Vy Uz Vz
+ 2 Ux Vx Uz Vz)

(U . V)² - [ (U Χ V) Χ V ] . U =
= Ux² Vx² + Uy² Vy² + Uz² Vz² + Ux² Vz² + Ux² Vy²  + Uy² Vz² + Uy² Vx²
+ Uz² Vy² + Uz² Vx²          --- other terms cancel each other.
= Ux² (Vx² +Vy²+Vz²) + Uy² (Vx² +Vy² + Vz²) + Uz² (Vx² + Vy² + Vz²)
= (Ux² + Uy² + Uz²) (Vx² + Vy² + Vz²)
= | U |² * | V

This is a long and detailed method.. a little difficult.
========================================
A simpler method.
We use the Triple product expansion or Lagrange's formula for vector products:
A X (B X C) =  (C . A) B  -  (A . B) C
(A X B) X C = ( C . A) B - (B . C) A

So  [ (U Χ V) Χ V ] . U
= [ (V . U) V - (V . V) U ] . U
= (V . U) (V . U)  -  (V . V) (U . U)
= (V . U)² - | V |²  * | U

Hence,
(U . V)² - [ (U Χ V) Χ V ] . U
= (U . V)² - [ (V . U)² - | V |²  * | U |² ]
= | V |²  *  | U