Introduction to projections | Matrix transformations | Linear Algebra | Khan Academy

Determining the projection of a vector on s line

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[chongdak]

nice teaching with very basic concept unlike many just simply dump out formula without explaining :D thumbs up for that

[royminsukkim]

This lecture is really better than my university professor. I really appreciate it.

[eqwerewrqwerqre]

Amazing video. Every 13 years later this has explained projections better than my linear algebra book (admittedly skipped all calculation instruction in favor of waiting until inner products were defined, but then it just went straight to subspaces!). Thank you for your amazing information, my ancestors praise you

[squared]

Love, love, love these videos!!! Can't express how grateful I am for them. Can't wait to finish these and pursue more and more. So generous of Sal and the team! One thing that I hope people find interesting is that there's another way of arriving at the formula for the scalar multiplier of v for the projection of x on L. I think of it as first determine the length of the projection, then ensure it's directed appropriately. The length to me is simply x cos(theta) where theta is the angle between x and v. But we know that x cos(theta) is then just v dot x divided by the length of v, which in the numerical example yields 7/sqrt(5) as the required length of the projection using his example vectors for v and x. Then to find the right direction of the projection vector, it must be some scalar c times the starting vector of L, v, to yield the projection. And c times v must have that length v dot x divided by the length of v which we calculated as being 7/sqrt(5) using his example vectors as stated above. Solving for c gives you v dot x divided by (the length of v times the length of v) which is just (v dot x) divided by (v dot v) which is exactly what Sal arrived at as well with the resulting vector being [14/5, 7/5]. He just got there using the difference of vectors and leveraging the resulting normal vector to the line and this version outlined above doesn't, felt a bit more intuitive to me, just focusing on the projection itself using cosines.

[hannabethhansen]

I succeeded in my uni only thanks to Khan Academy!!! Ty so much! Love, Hannabeth

[behrampatel]

Love it !
Its worth mentioning that V-hat is a normalized vector and that the scalar c when multiplied by a normalized vector (v hat) gives us the vector projection of x onto v.
cheers,
b

[jarbi]

thanks sal it was killing me not knowing where the equation for a projection came form

[BoZhaoengineering]

Linear algebra is my favorite topic. As I am engineer, this topic is the fundamental of infinite element method or called an application of it. Thinking of the car we drive daily, we actually benefit from this topic .

[udayshankars]

Thank you for the series of these amazing lectures. This helped me a lot to brush up and understand more about linear algebra, so a big thank you to you Sal!!

[gurra]

You're like one million times better at explaining this than my teachers!!
Thank you alot, I'd be so lost if it wasn't for Khan Academy :D

[Murf_Workshop]

Thank for the upload. My instructor's explanation was a little tough to follow, and I am usually able to understand him pretty well, but your video helped out a lot.

[leafslizer]

I don't think I have ever been so surprised how easy a concept could be explained. I've struggled with understanding formulas and the books explanation of projections and with info from this video noted down. I'll never forget it :) Thanks for guiding us all toward a more rich understanding of math Sal & khan team <3

[tedtdu]

Never buy books of projection matrix, just listen what the man are talking about!!Thanks again.

[creesrees]

Dont forget to mention about youtube n khan academy for your graduation speech!

[-gouthamkumarreddy]

I watch explanations of you you explain what I want exactly, but even one video is there like this....which explains what I want

[theghost]

Thank you Khan Academy for your free online courses!

[jonathannielsen]

Just for clearity it's not (x-cv)*v. Because you start in cv and go up to x, so it is (cv-x)*v however you will get the same result because they are both orthogonal to v :)

[LAnonHubbard]

I'm going through Eric Lengyel's book " Mathematics for 3D game programming & computer graphics". I've been staring at his explanation of vector projection and it's not very well explained. This video has been a great additional resource. Thanks Sal!

[canetguy]

Thank you for the video, it is great. My only suggestion would be to add the tag: orthogonal.

[jsquaredm]

Could you also find the angle between the two vectors and then just multiply the vectors magnitude by Cos(theta)?