Quaternion 2.3节中的问题
Closed this issue · 2 comments
Mud-B commented
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
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
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⟂
—
You are receiving this because you authored the thread.
Reply to this email directly, view it on GitHub, or unsubscribe.