(1) ||u × v|| = ||u|| ||v|| sin(θ), where θ is min{∠uv, ∠vu}, direction of u×v determined by the right-hand rule
is equivalent to:
(2) ||u × v|| = (u2v3-u3v2)i + (u3v1-u1v3)j
+ (u1v2-u2v1)k
Proof
Per (1), i × i = 0, i × j = k, j × k = i, i × k = -j,
Inoltre, come per il prodotto scalare, vale a×(b+c) = a×b+a×c. Quindi:
(u1i+u2j+u3k) ×(v1i+v2j+v3k) =
u1i×(v1i+v2j+v3k) + u2j×(v1i+v2j+v3k) + u3k×(v1i+v2j+v3k) =
= (u2v3-u3v2)i + (u3v1-u1v3)j
+ (u1v2-u2v1)k
In definitiva: | |||
u·v = | 0 if u=0 or v=0, | u·v = | ||u|| ||v|| cos(θ) altrimenti, |
||u × v|| = | 0 if u=0 or v=0, | ||u × v|| = | ||u|| ||v|| sin(θ) altrimenti. |