# Why is a vector equal to the sum of its projection on other vectors? [closed]

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

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!

• Got it! Thank you! :D Oct 11, 2021 at 21:24