I always liked this formula because if you look at it and try things like: Limit as m1m2 approach -1, you will get 90 degrees which makes sense since one of the rules for perpendicularl lines is that m1m2 = -1. Also you can do a similar observation when m1 = m2 (parallel lines since same slope) you will get arctan(0) or 0, which also makes sense since the lines are parallel.
It's very smart and interesting! However, does it work when beta is equal to pi over two? Moreover, I wonder how to justify to put the absolute value on the fraction.
Orthogonal lines have m₂ = -1/m₁, hence denominator is zero, so we have arctangent of infinity/undefined. 😉 Line y = m₁x+b₁ is collinear with vector r₁=(1; m₁), second one - with r₂=(1; m₂). Scalar product (r₁, r₂) = 1+m₁m₂ = |r₁||r₂|cos Θ ⇒ cos Θ = (1+m₁m₂)/(√[1²+m₁²]√[1²+m₂²]) ⇒ Θ = arccos((1+m₁m₂)/√[(1+m₁²)(1+m₂²)].
Solution: Since m describes the slope (m > 0) or dip (m < 0) per unit of x of the respective line, you can easily calculate the angle between that line and the x-axis with tan⁻¹(|m|). So the angle between two lines is: (with m₁ ≥ m₂) if m₁, m₂ > 0 then tan⁻¹(m₁) - tan⁻¹(m₂) if m₁ > 0 > m₂ then tan⁻¹(m₁) + tan⁻¹(-m₂) if 0 > m₁, m₂ then tan⁻¹(-m₂) - tan⁻¹(-m₁)
Curious why you didn't just move the x-y coordinates to where the two lines intersect and call the transformed coordinates x'-y''? That would eliminate having to draw the added parallel lines. The proof would be the same. I'm assuming you thought a coordinate transformation would confuse some of your viewers?
Would it not be simpler to say that alpha and gamma = arctan of m1 and m2, and beta = gamma -alpha, so beta = arctan m2 - arctan m1? Or is my reasoning flawed?
Get yourselves a girl who looks at you the same way Prime Newtons looks at his camera when there's an earthquake XD
🤣🤣🤣🤣🤣🤣🤣🤣
😂
And thanks the Lord.
Your lessons are so beautifully constructed and delivered. Great stuff! Please keep making them.
😊
Thank you! Will do!
You're very welcome.
Yes, very enthusiastic about math and very motivating. 👍
❤️🙏
Some day I gotta meet you and thank for you for inpsiring my mathematical passion!!!!!
Maybe one day!
Its called a math quake. It happens when chalk rubs harshly against the chalkboard when one gets too excited about doing math.
Your chalkboard handwriting is so pretty. I wish mine was that nice!
Ha, loved the cold open. Great video.
i was just trying to derive the formula and this video came up couple of minutes later, though ur proof is easier to understand😂
I always liked this formula because if you look at it and try things like:
Limit as m1m2 approach -1, you will get 90 degrees which makes sense since one of the rules for perpendicularl lines is that m1m2 = -1. Also you can do a similar observation when m1 = m2 (parallel lines since same slope) you will get arctan(0) or 0, which also makes sense since the lines are parallel.
gracias, por tu clara explicación paso a paso
You are so good at teaching. Keep it up!
Thank you! 😃
no problem@@PrimeNewtons
You are very good. Funny too and expressions entice your students. Well done, more please.
Thank you! 😃
Very interesting, never happened to see this formula inferred
God wanted to make sure you know that we don’t live under Euclidean geometry 😂😂
Enjoyed it very well, thanks dear!
I was waiting for you to say "...take the magnitude of the earthquake, er tangent...".
Wow, aweome man !
It's very smart and interesting! However, does it work when beta is equal to pi over two?
Moreover, I wonder how to justify to put the absolute value on the fraction.
The man does love his cap.
Excellent
Nice video :)
Orthogonal lines have m₂ = -1/m₁, hence denominator is zero, so we have arctangent of infinity/undefined. 😉
Line y = m₁x+b₁ is collinear with vector r₁=(1; m₁), second one - with r₂=(1; m₂).
Scalar product (r₁, r₂) = 1+m₁m₂ = |r₁||r₂|cos Θ
⇒ cos Θ = (1+m₁m₂)/(√[1²+m₁²]√[1²+m₂²])
⇒ Θ = arccos((1+m₁m₂)/√[(1+m₁²)(1+m₂²)].
asombroso Θ = arccos((1+m₁m₂)/√[(1+m₁²)(1+m₂²)].
11th std. Student
❤ From india❣️
The two parallel lines to the first and passing through the origin have their respective equations: y=mx and y=m'x (y-intercept zero).
Would it work to take the arctan(m1) = alpha and the arctan(m2) = gamma, then gamma - alpha = beta?
Thanks 🙏
Reminds me of problem 11 on AMC12A 2023
Solution:
Since m describes the slope (m > 0) or dip (m < 0) per unit of x of the respective line, you can easily calculate the angle between that line and the x-axis with tan⁻¹(|m|).
So the angle between two lines is: (with m₁ ≥ m₂)
if m₁, m₂ > 0 then tan⁻¹(m₁) - tan⁻¹(m₂)
if m₁ > 0 > m₂ then tan⁻¹(m₁) + tan⁻¹(-m₂)
if 0 > m₁, m₂ then tan⁻¹(-m₂) - tan⁻¹(-m₁)
The video only shows the solution for m₁, m₂ > 0 😕
Curious why you didn't just move the x-y coordinates to where the two lines intersect and call the transformed coordinates x'-y''? That would eliminate having to draw the added parallel lines. The proof would be the same. I'm assuming you thought a coordinate transformation would confuse some of your viewers?
Oh yeah! I was imagining 9th graders
@@PrimeNewtons There are some smart 9th graders if they have understood your Lambert W videos 😉
Hmm, didn't you record exactly the same thing just a few days ago, did you?
This is to derive the formula
no way an earthquake 😭 what are the chances of recording that
I was hoping to see the shaking in the video, but optical stabilization in the camera ruined it 😢
@@PrimeNewtons I did hear it
One. The chance was one. 😂
(Someone had to say it)
When these lines are translated to the origin, doesn't b1 and b2 both equal zero?
Would it not be simpler to say that alpha and gamma = arctan of m1 and m2, and beta = gamma -alpha, so beta = arctan m2 - arctan m1? Or is my reasoning flawed?
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Id say the spherical earth got jealous of the flat plane
Forgot mention what happens when denominator is zero
I love geometry
Never stop earthquake! 😂😀😉
Them fault lines though
Scary
y1=m1x+b1
y2=m2x+b2
y1'=m1=tan(alfa)
y2'=m2=tan(alfa+beta)
tan(alfa+beta)=(tan(alfa)+tan(beta)/(1-tan(alfa)•tan(beta)
m2=(m1+tan(beta))/(1-m1•tan(beta)
m2(1-m1•tan(beta)=m1+tan(beta)
m2-m1•m2•tan(beta)=m1+tan(beta)
tan(beta)+m1•m2•tan(beta)=m1-m2
tan(beta)(1+m1•m2)=m1-m2
tan(beta)=(m1-m2)/(1+m1•m2)
beta=arcotan(m1-m2)/(1+m1•m2)
qaqas bu ne emosiyadi verirsen zemlyatresenyaya gore da noooldu
asnwer=2 isit hmm gmm