Niech 𝐚⃗ =2𝐩⃗⃗ −3𝐪⃗⃗ , gdzie 𝐩⃗⃗ ⊥𝐪⃗⃗ oraz |𝐩⃗⃗ |=4 i |𝐪⃗⃗ |=2 . Oblicz | zzzzzzzzzzz: Niech 𝐚⃗ =2𝐩⃗⃗ −3𝐪⃗⃗ , gdzie 𝐩⃗⃗ ⊥𝐪⃗⃗ oraz |𝐩⃗⃗ |=4 i |𝐪⃗⃗ |=2 . Oblicz |𝐚⃗ | oraz 𝒂⃗⃗ °𝒑⃗⃗ i 𝒂⃗⃗ °𝒒 i o dziwo wychodzi mi cosinus pomiedzy a i p równy 4 . Jak to sie stało ?

Adamm: a² = a*a = (2p−3q)(2p−3q) = 4p²−6p*q+9q² = 4p²+9q² = 4*25 = 100 |a| = 10 a*p = (2p−3q)p = 2p²−3q*p = 2p² = 32 a*q = (2p−3q)q = 2p*q−3q² = −3q² = −12