The dot product of two vectors and the angle between them

We will use the Law of Cosine to show the connection between the dot product of two vectors and the angle between them. This tutorial will help you with your precalculus or calculus 3 classes.
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#calculus #bprpcalculus #apcalculus #tutorial #math

[@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)

[@jperez7893]

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

[@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.

[@Gremriel]

How exciting!

[@mindblast1011]

very good 👍. More proofs please 😊.