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Krasjet / quaternion

A brief introduction to the quaternions and its applications in 3D geometry.
https://krasjet.github.io/quaternion/
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Quaternion 2.3节中的问题 #30

Closed Mud-B closed 3 years ago

Mud-B commented 3 years ago

这个等式的第二行是怎么推出来的，能否细讲下或给个Reference，我代数没啥基础

v′⊥=v′𝑣+v′𝑤
   =cos(θ)v⊥+sin(θ)w
   =cos(θ)v⊥+sin(θ)(u×v⊥)
Krasjet commented 3 years ago

这个是很常用的投影公式

project v onto u
projᵤv = (⟨u,v⟩/⟨u,u⟩)u

你也可以直接算出来

cos(θ) = 临边/斜边 = ∥v⟂'∥ / ∥vᵥ'∥ = ∥v⟂∥ / ∥vᵥ'∥
⇒ ∥vᵥ'∥ = ∥v⟂∥cos(θ)
⇒ vᵥ' = ∥vᵥ'∥ (v⟂/∥v⟂∥ )
       = (∥v⟂∥cos(θ)) (v⟂/∥v⟂∥ )
       = cos(θ)v⟂
Mud-B commented 3 years ago

原来是这么回事，我懂了。 感谢Krasjet！

---Original--- From: @.> Sent at: 2021年9月10日(Fri) Early Morning0:56 To: @.>; Cc: @.**@.>; Subject: Re: [Krasjet/quaternion] Quaternion 2.3节中的问题 (#30)

这个是很常用的投影公式 project v onto u projᵤv = (⟨u,v⟩/⟨u,u⟩)u
你也可以直接算出来 cos(θ) = 临边/斜边 = ∥v⟂'∥ / ∥vᵥ'∥ = ∥v⟂∥ / ∥vᵥ'∥ ⇒ ∥vᵥ'∥ = ∥v⟂∥cos(θ) ⇒ vᵥ' = ∥vᵥ'∥ (v⟂/∥v⟂∥ ) = (∥v⟂∥cos(θ)) (v⟂/∥v⟂∥ ) = cos(θ)v⟂
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