Dvojni vektorski produkt - Student Info
Dvojni vektorski produkt - Student Info
Dvojni vektorski produkt - Student Info
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Algebra I <strong>Dvojni</strong> <strong>vektorski</strong> <strong>produkt</strong> 4<br />
6. Splošni primer<br />
Zdaj se lahko lotimo tudi reševanja sistema (7,8) v splošnem.<br />
(⃗a × ⃗ b,⃗c,⃗a) ⃗a ⃗ ∣ b ∣∣∣<br />
∣ (⃗a ×<br />
α =<br />
⃗ b,⃗c, ⃗ ⃗a⃗a (⃗a × ⃗ ∣<br />
b,⃗c,⃗a) ∣∣∣<br />
b) ⃗ b ⃗ b ∣⃗ b⃗a (⃗a × ⃗ b,⃗c, ⃗ b) ⃗a⃗a ⃗a ⃗ ∣<br />
b ∣∣∣<br />
, β =<br />
⃗a⃗a ⃗a ⃗ ∣<br />
b ∣∣∣<br />
. (15)<br />
∣⃗ b⃗a ⃗ b ⃗ b ∣⃗ b⃗a ⃗ b ⃗ b<br />
Preoblikujmo mešana <strong>produkt</strong>a. Prvega,<br />
in drugega,<br />
(⃗a × ⃗ b,⃗c,⃗a) = (⃗a,⃗a × ⃗ b,⃗c) (16)<br />
= (⃗a × (⃗a × ⃗ b))⃗c (17)<br />
= ((⃗a ⃗ b)⃗a − |⃗a| 2 ⃗ b)⃗c (18)<br />
= (⃗a ⃗ b)(⃗a⃗c) − |⃗a| 2 ( ⃗ b⃗c), (19)<br />
(⃗a × ⃗ b,⃗c, ⃗ b) = ( ⃗ b,⃗a × ⃗ b,⃗c) (20)<br />
= ( ⃗ b × (⃗a × ⃗ b))⃗c (21)<br />
= −( ⃗ b × ( ⃗ b × ⃗a))⃗c (22)<br />
= −(( ⃗ b⃗a) ⃗ b − | ⃗ b| 2 ⃗a)⃗c (23)<br />
= −(⃗a ⃗ b)( ⃗ b⃗c) + | ⃗ b| 2 (⃗a⃗c). (24)<br />
Pripravljeni smo na poenostavitev izrazov za skalarja α in β, podanih v (15): spričo zveze med<br />
<strong>vektorski</strong>m in skalarnim <strong>produkt</strong>om (38) in enakosti (19,24) sklepamo<br />
Podobno,<br />
α = ((⃗a⃗ b)(⃗a⃗c) − |⃗a| 2 ( ⃗ b⃗c))( ⃗ b ⃗ b) − (−(⃗a ⃗ b)( ⃗ b⃗c) + | ⃗ b| 2 (⃗a⃗c))(⃗a ⃗ b)<br />
|⃗a × ⃗ b| 2 (25)<br />
= −|⃗a|2 ( ⃗ b⃗c)( ⃗ b ⃗ b) + (⃗a ⃗ b) 2 ( ⃗ b⃗c)<br />
|⃗a × ⃗ b| 2 (26)<br />
= (−|⃗a|2 | ⃗ b| 2 + (⃗a ⃗ b) 2 )( ⃗ b⃗c)<br />
|⃗a × ⃗ b| 2 (27)<br />
= −|⃗a ×⃗ b| 2 ( ⃗ b⃗c)<br />
|⃗a × ⃗ b| 2 (28)<br />
= − ⃗ b⃗c. (29)<br />
β = (⃗a⃗a)(−(⃗a⃗ b)( ⃗ b⃗c) + | ⃗ b| 2 (⃗a⃗c)) − ( ⃗ b⃗a)((⃗a ⃗ b)(⃗a⃗c) − |⃗a| 2 ( ⃗ b⃗c))<br />
|⃗a × ⃗ b| 2 (30)<br />
= |⃗a|2 | ⃗ b| 2 (⃗a⃗c) − (⃗a ⃗ b) 2 (⃗a⃗c)<br />
|⃗a × ⃗ b| 2 (31)<br />
= |⃗a ×⃗ b| 2 (⃗a⃗c)<br />
|⃗a × ⃗ b| 2 (32)<br />
= ⃗a⃗c. (33)<br />
Izračunali smo iskana skalarja, dokaza je konec.<br />
Ljubljana 24. oktober 2001 c○AM in BZ