Distance between two polar coordinates

[@RajSandhu-gm8iz]

If you draw a simple diagram, angle between lines (theta2-theta1), then I used used cosine rule from the start, got you straight to the answer.

[@nanamacapagal8342]

Congratulations! You just proved the cosine rule for triangles: just mark the point opposite the line you want to measure as the origin, and there you go!

[@arjunkapoor5653]

Why not cos rule, it’s literally the Pythagorean theorem, but with an extra -(2 x r1 x r2 x cos(Theta1-Theta2)) at the end

[@xenumi]

Same thing

[@gregorysadofyev3981]

It can be calculated easier than that. Use cosine theorem for triangle that are composed of r1 and r2

[@mridu24_]

Hey Prime Newtons..I actually like the starting of your videos, by the way this video was adorable...learned a lot, thanks.

[@dirklutz2818]

Great video! Like always.

[@5Stars49]

Lovely Sir

[@dereklenzen2330]

There is nothing wrong with this solution. However, when I visualized this problem in my head, one of the things that popped into my head is the Law of Cosines, since we know the lengths of two sides r_1 and r_2 as well as the angle between them theta_2 - theta_1. Thus, d^2 = (r_1)^2 + (r_2)^2 - 2*(r_1)*(r_2)*cos(theta_2 - theta_1), which in turn gives us d = sqrt[(r_1)^2 + (r_2)^2 - 2*(r_1)*(r_2)*cos(theta_2 - theta_1)].

[@yasirabdulhakeem]

This looks kind of similar to the metric tensor in a polar coordinate system, is that a coincidence or are they related?

[@rachidtanan3229]

How to make same thing betwine two points on cylindre

[@charlesgodswill6161]

my observation prof, please add some lighting to brighten the board!

[@ThePhotonMan110]

Would love to see the 3D solution as well!

[@TR_Arial]

So this is basically a proof for the cosine rule Neat!

[@Archimedes_Notes]

Thank you that is great. We can also use Alkashi law for the cosine if we forgot how to get this result.

[@PrimeNewtons]

I'll have to research that

[@الر2ماني]

Use Alkashi law or cosinus law and the result is straightforward

[@surendrakverma555]

Excellent explanation Sir. Thanks 👍

[@claytonbenignus4688]

Elegant! I could see the problem work itself!

[@adw1z]

Indeed if z1 and z2 are complex numbers (isomorphism to polar co-ordinate geometry), |z1 - z2| = distance between z1 and z2: -> theta 1 = theta 2 ==> z1 and z2 are co-linear on the line through the origin in the same quadrant, and so the distance between them should be |r2 - r1|, as expected by the formula. -> theta 1 = -theta 2 ==> z1 and z2 are co-linear on the line in opposite quadrants, with distance |r1 + r2| = r1 + r2 between them, also given by the formula. -> If theta 1 = theta 2 + (2n+1)pi/2 for n integer, then z1 and z2 are orthogonal, and the distance between them is sqrt(r1^2 + r2^2) by Pythagoras. -> Of course, if any other angle, just use the cosine rule on the triangle formed (of which Pythagoras is a special case), and that's why the cos term even appears.

[@yplayergames7934]

This is basically a Law of Cosines

[@BartBuzz]

Nice!

[@rujon288]

i just draw a picture and used cosine rule /:

[@marshallmanz123]

Simple!

[@royluke4762]

Lighting on the board is very little to see the figures clearly

[@marcelocampos665]

Muito bom!

[@xgx899]

Using elementary vector algebra: d^2=|a-b|^2=a^2+b^2-2(a,b)=r_1^2+r_2^2-2r_1r_2 cos(theta_1-theta_2) This channel is an example of how not to do math.

[@punditgi]

Prime Newtons has all the answers! 🎉😊

[@harandianr]

Just use cos law. Lengths are r1 and r2 and the and in between is theta2 minus theta1 . Since you are using cos it makes no difference to have theta 1 minus their 2.

[@7ymke]

kinda dissapointed i like your videoes but i thought this one will not be just simple substitution but we will use some trick