Vectors u and v are vectors
where | u + v | = √5 and | u-v | = √3.
So | u |²+ | v |²?
Answers
Answer:
Step-by-step explanation:
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 |²
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 |²
Hence,
(U . V)² - [ (U Χ V) Χ V ] . U
= (U . V)² - [ (V . U)² - | V |² * | U |² ]
= | V |² * | U |²
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