Distance between two polar coordinates

In this video I showed how to compute the distance between two points with polar coordinates. The strategy is to convert to cartesian coordinates first before using the pythagorean formula

[RajSandhu-gmiz]

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.

[nanamacapagal]

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!

[gregorysadofyev]

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

[arjunkapoor]

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

[dereklenzen]

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)].

[mridu_]

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

[adwz]

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.

[الرماني]

Use Alkashi law or cosinus law and the result is straightforward

[TR_Arial]

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

[charlesgodswill]

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

[ThePhotonMan]

Would love to see the 3D solution as well!

[Archimedes_Notes]

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

[claytonbenignus]

Elegant! I could see the problem work itself!

[yasirabdulhakeem]

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

[rachidtanan]

How to make same thing betwine two points on cylindre

[xgx]

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.

[rujon]

i just draw a picture and used cosine rule /:

[royluke]

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

[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.

[ymke]

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