The dot product of two vectors and the angle between them

[@xavnqesh1416]

BPRP’s board should get a world record for the cleanest and the neatest board.

[@bigchungusdriplord2301]

If you clean the board with wet paper towels it leaves it completely spotless, so it isn't exactly that special

[@cyrusyeung8096]

We can also use the fact that |v|² = v·v and say |u - v|² = (u - v) · (u - v) = u·u - 2u·v + v·v = |u|² - 2u·v + |v|² Substituting this result into the cosine law gives cosθ = u·v/(|u||v|)

[@Ninja20704]

I feel that might be circular reasoning because a•a=|a|^2 comes from the dot product being a•b=|a||b|cos(theta). Maybe im wrong if theres a way you can prove the fact by another way.

[@cyrusyeung8096]

@@Ninja20704 For v = [v1 v2 v3 ...], you can show |v|² = (v1)(v1) + (v2)(v2) + (v3)(v3) + ... = v·v

[@Ninja20704]

@@cyrusyeung8096 ok thank you for the clarification. My teacher did a•a = |a||a|cos(0) = |a|^2 to show us.

[@MarioFanGamer659]

I was tought a different method where one of the vectors is scaled in such a way that it creates a right triangle with another vector: cos(θ) = adj / hyp, where adj = k|↑a| (with ↑a being a vector and k being a scalar) and hyp=|↑b| (with ↑b being a vector) That means, cos(θ) = k · |↑a| / |↑b| The hard part is getting the scalar factor k but that one takes advantage of a quirk with dot products: For ↑a ·↑ b, you can split b into an orthagonal (↑o) and parallel (↑p) component (relative to ↑a, that is): ↑b = ↑o + ↑p (Incidentally, you could easily calculate the cosine if you know the value of ↑p.) ↑a · ↑b is therefore ↑a · (↑o + ↑p) = ↑a · ↑o + ↑a · ↑p. By definition, ↑a · ↑o always equals to 0 so a dot product of ↑a · ↑b really is just that of ↑a · ↑p: ↑a · b = ↑a · ↑p But because ↑p is parallel to ↑a as well as a component of ↑b, you can rewrite ↑p as ↑a scalar of a (i.e. ↑p = ↑k · ↑a) so the dot product as a whole results in this: ↑a · ↑b = ↑a · ↑p = k · ↑a · ↑a Multiplying a vector by itself yields the square of its length: k · ↑a · ↑a = k · |↑a|² This means, the dot product of is the square of one of the vectors' length times a certain scalar factor: ↑a · ↑b = k · |↑a|² This k is the same one we're searching. Solve for k: k = ↑a · ↑b / |↑a|² And now you can plug it into the cosine formula: cos(θ) = k · |↑a| / |↑b| = ((↑a · ↑b) / |↑a|²) · (|↑a| / |↑b|) = (↑a · ↑b) / (|↑a| · |↑b|) QED

[@phill3986]

Bit of trivia about unit vectors, they were sometimes called directional vectors or cosine vectors... Where a is the angle between the vector and the x axis, same with b and y, c with z, etc

[@a.kofficial6140]

Please make video about stochastic differential equation

[@Mediterranean81]

Do the integral of inverse csc (x)

[@Gremriel]

How exciting!

[@phill3986]

Nice 👍

[@phill3986]

Problem I've always had with normal vs orthogonal is that normal can mean a vector has length 1 (a unit vector).

[@cyrusyeung8096]

Bprp has a video on his main channel, explaining the difference between perpendicular, normal and orthogonal.

[@stephenbeck7222]

It all depends on definition. They are just words and your text should define the words before using them.

[@jperez7893]

can you show the Coriolis vector of the earth from first principles pls

[@mindblast1011]

very good 👍. More proofs please 😊.