I was watching video about 3D rotation and Cross product (link at below of the post), at this moment of the video, he claimed the projected vector by v’ onto v and v’ onto n^×v are cos(θ)v and sin(θ)(n^×v) respectively (Timestamp: 7:44)

  1. What is the proof of that?
  2. Why is v’ equal to the sum of the vertically projected vector and horizontally projected vector? Proof? (Timestamp: 7:11)

Please explain like I’m five :) Thank you!

This is link to video: https://youtu.be/UaK2q22mMEg?t=548


1 Answer 1


Let's start by calling $w = n \wedge v$. It is a fact that the vector product of two vectors will give as a result a new vector $w$ orthogonal to the plane defined between vectors $n, v$ (you can see it graphically at the video), hence, the angle between $w$ and $n$ is $\pi \over 2$, and the same for the angle between $w$ and $v$.

In the image you included to the question, there is a vector $v'$ in the same plane defined by $w$ and $v$. Since $w$ and $v$ are orthogonal, the angles $\angle wv'$ and $\angle v'v$ are complementary, and now, using some trigonometry, you can represents the components of $v'$ related with $w,v$.

One way to verify that the vector addition

$$cos(\theta) \cdot |v| + sin(\theta)\cdot |w| = v'$$

... is using the pytagoras theorem. You can imagine the components of vector $v'$ as the catetus of the hypotenuse $|v'|$, and write that

$$(cos(\theta) \cdot |v|)^2+ (sin(\theta)\cdot |w|)^2 = |v'|^2$$

Hope I explained it clearly!

  • $\begingroup$ Got it! Thank you! :D $\endgroup$ Oct 11, 2021 at 21:24

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