Mr, you really saved my life. I searched for textbooks and internet pdfs only to understand that concept. You really solved my problem within 7minutes and I found it really hard not to smashed the red button and 👍button. Wish to see more of kind of explanation.
Dude... Thank you for actually going slow and taking the time to explain the thought process behind this stuff instead of just spitting words at me like everyone else.
Wow! This was so helpful! I do have a question about if the vector to be projected from and normal vector are on opposite sides of the plane. To test this I simply inverted the normal vector to . It seems to work out alright until the last subtraction step where the resulting projection becomes . If I don't know both vectors are on the same side of the plane, do I need to do an angle test, and invert the normal vector if the angle is more than 90 degrees? Or is there some way to make it work out either way without inverting the normal vector? Thanks!
The coefficients of plane come from the Determinate. They are the A,B,C values in the equation of the plane in standard form A(x-x0)+B(y-y0)+C(z-z)=0. From an earlier video, we know that coefficients of the plane also gives us the vector perpendicular to the plane. Which allows us to find the projection (is also perpendicular to the plane). I hope this helps.
Hey, if you look at the triangle formed, it should be a right triangle with sides a, b, c (c is the hypotenuse). Side a is the projection of of c onto an and side b is the projection of c onto b. I hope this helps.
![FNAF Into the Pit has A LOT of Unused Graphics | LOST BITS [TetraBitGaming]](http://i.ytimg.com/vi/6YhZCRdipl4/mqdefault.jpg)
Mr, you really saved my life. I searched for textbooks and internet pdfs only to understand that concept. You really solved my problem within 7minutes and I found it really hard not to smashed the red button and 👍button. Wish to see more of kind of explanation.
You legit just probably got me a job lol. Needed to know this for a programming test for an interview. Thanks!
How did it go?
@@Pokemonzeldabro I got the job! :D
@@monawoka97 CONGRATS!!
@@monawoka97 LETS FUCKING GOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO
This is like chapter 1 linear algebra stuff
Dude... Thank you for actually going slow and taking the time to explain the thought process behind this stuff instead of just spitting words at me like everyone else.
Thanks for this! My game engine lacked this feature. I needed to get the momentum of my player but only along a plane with a given normal.
Damn I just learned this in my multi class at Rice. Would’ve been completely confused if you didn’t explain this so well last year Mr. Wernau.
Thank you man, you video explains so well on the concept. Big appreciate!!!!
Thanks for explaining this within 7 minutes
Excellent explanation! Thank you very much!
Your video helped me understand a concept in a research paper. Thank you 😊
Thank you so much!! You rlly saved my test with this!
awesome dude. confusing at first but worked out perfectly. thank you
Why do all great math teacher have U and V LOOKING THE SAME
Great 👍🏻👍🏻 keep making such videos, really helpful 😇😇
Thanks a lot 😊
that vector addition part is what made it click for me
Wow! This was so helpful!
I do have a question about if the vector to be projected from and normal vector are on opposite sides of the plane. To test this I simply inverted the normal vector to . It seems to work out alright until the last subtraction step where the resulting projection becomes . If I don't know both vectors are on the same side of the plane, do I need to do an angle test, and invert the normal vector if the angle is more than 90 degrees? Or is there some way to make it work out either way without inverting the normal vector? Thanks!
thank you, i forgot this trick from school and needed a reminder :D
I really want to follow this but the constant changing of brightness keeps distracting me x...x
Thanks man! I never understood it better!🧡
Thank you dude thank you
This is a GEM!
i smashed that like button
Looks like you have a projection problem and can't focus right!
That is funny.
Thanks Thomas. Please what do you mean by 'coefficient of the plane'?
The coefficients of plane come from the Determinate. They are the A,B,C values in the equation of the plane in standard form A(x-x0)+B(y-y0)+C(z-z)=0. From an earlier video, we know that coefficients of the plane also gives us the vector perpendicular to the plane. Which allows us to find the projection (is also perpendicular to the plane). I hope this helps.
That's the coefficients in front of the equation of the plane.
i love you
u verky kewl -erika's firend
mr. wernau, i need you in college :,(
I need him as well 😅🥺🥺
good luck on your classes man
how were you sure that the sum of multiple projections is also a projection in this case ? , you went fast on the last bit ..😥
Hey, if you look at the triangle formed, it should be a right triangle with sides a, b, c (c is the hypotenuse). Side a is the projection of of c onto an and side b is the projection of c onto b. I hope this helps.
0:26 projection : shadow
1:27 projection of k on u vector; but we don't know "u"
Mr. Wernau, can you plzz share your Instagram, Facebook or email, so we can contact you additionally if we want some help?