Find Orthogonal Vector

[@rehmanmohammed9578]

Whenever the guy has a thick sharpie and a clean sheet of paper you know you gonna get what you came for

[@BestGermanLPs]

Thank you very much! Short, accurate and easily understandable.

[@ximm3039]

So fucking simple, god bless you.

[@princesspinkblue]

omg! i went to my teacher for help and he explained it in the most ****ed up way possible....now i see this andeverything has become so much more simple!!!! THANK YOU

[@Joshjowen]

Thank you, this is the best explanation I've seen

[@James310193]

Very clearly explained. Thank you.

[@eudaeu8345]

finally someone that explain clearly and fast

[@georgezenin2826]

Good explanation! Finally cracked the code. Thanks

[@MindfulnessKanalen]

thank you man!! just one thing, change orthogonal to perpendicular or simply add the word perpendicular to the description of the video

[@anonymous_king479]

Awesome 👍😊 explaination

[@UGPepe]

that doesn't work for v = (3, -1, 0) though. you need to solve for either a or b or c depending on how v is oriented. is there a more general way to find an orthogonal vector?

[@AnuragSharma7775]

Who are watching in 2021 👇

[@RealPhysicsFusion]

Thanks, that was helpful

[@calvinpks]

Very helpful, couldn't find anything on this anywhere!

[@asherb6095]

thats a great method and explanation,but lets say I have a vector with n=10 or more components . Can I really rely on this system ? is there another way other than figuring out how to zerorize one side of the equation by assuming the scalars ? lets say if one of the vectors has a variable ?

[@aranglancy]

This basic principle will work no matter what the dimensions of the vectors. The more dimensions, however, the more "degrees of freedom" for the answer that means there will be more and more components of the unknown vector that are free for you to specify. This was just a random problem taken from a textbook that I was using to illustrate to my students how to make simple videos. Typically, you'll want to find a vector that is orthogonal to another vector AND satisfies some other constraints. In that case, the same approach applies but you'll need to incorporate the other constraints as well.

[@apbeers]

Awesome video and straight to the point. thanks

[@NSDaishi]

very cleanly done.

[@aaronmatthews6066]

What if you're asked to find an orthogonal vector of a certain length/magnitude? For example, if the unknown perpendicular vector was length 5, could you simply multiply vector U by 5?

[@aranglancy]

Sort of. That only works if u is length 1. If it isn't though it is ok, you can still use that basic idea. First find the length of u, call it L, then multiply by 5/L

[@maindepth8830]

Thank you very much

[@McCillan2]

hey man can you make a video of finding all vectors using the same example in the video?

[@jacob-jm2xp]

there are unlimited vectors

[@ThemisTheotokatos]

Nice, Could we choose a and c and try to find a b? or any other combination?

[@jacob-jm2xp]

yes

[@mohammedmistarihi3275]

Awesome 👏 Thank you

[@grantboyd2883]

You're the man!

[@LetsJustBuildIt]

That helped alot thanks

[@nckkjohnson]

Very helpful.

[@l8870]

this is why i hate math. thank you for making it simple

[@bucketrance]

to the point.thanks.

[@Tom-gn2gb]

Can we assign the a and b any values so long as we can satisfy the u.v = 0?

[@aranglancy]

Yes, exactly. I chose values I thought would make it simple, but you can choose any values that ultimately satisfy the dot product = 0 requirement.

[@tayibmalik]

oh nice

[@shivam123qwe]

Amazing Thank you!

[@bobbyoctaviano2661]

Thanks Aran :)

[@yousufsyed844]

Thank you!

[@matthewanstead798]

Awesome, thanks

[@goplay5936]

What should I do to find a unit vector orthogonal to a given vector??

[@aranglancy]

Just scale your answer up or down so it is unit length. The answer I got in the video was u = [1,1,-1] which has length |u| = sqrt(1^2 + 1^2 + (-1)^2 ) = sqrt(3). So I can just divide u by sqrt(3) to get a unit vector: u' = u/sqrt(3) = [1/sqrt(3), 1/sqrt(3), -1/sqrt(3) ]

[@noumanshah4275]

thanks

[@brendakasale5559]

you da best

[@changling30]

the sound of the marker just so annoying thanks for you effort though

[@malisettypullaiah3258]

Can we take values of a,b,c as 0?

[@jacob-jm2xp]

sure, but the zero vector isn't very useful

[@CamViesky]

where did that 1 come from?

[@aranglancy]

It didn't have to be 1. That was a choice. The issue is that there are many, many answers to this question. Any 3 numbers, a,b,c, that satisfy the equation 3a - b + 2c = 0 will give a valid answer. I picked a = 1, just because I thought it would be easy. If I had picked a = 0, then we have 3*0 - b + 2c = 0 or -b + 2c = 0. I could rewrite that as 2c = b. From here we have another choice. There are lots of b's and c's that would make 2c = b true. b=2 and c=1 for example, or b = 4 and c = 2, or even b = 1 and c = 1/2. So all of this tells us that (0, 2, 1) or (0, 4, 2) or (0, 1, 1/2) are all vectors that are orthogonal to (3, -1, 2). The a = 1 (and the b =1) were just choices. Does that help?

[@CamViesky]

@@aranglancy oohhh! Yes it does thank you so much for the reply! Honestly didn’t expect a response so quickly. So this is a case where there are infinitely many solutions? Okay. Yes I understand thank you so much for taking the time to respond and explain

[@lukeskyzy5833]

You just got vectored

[@realnesss101]

cant they a b c equal 0

[@aranglancy]

Yes, they can but then U = (0,0,0) is the zero vector. The zero vector is (trivially) orthogonal to all vectors, so although it is technically correct it is probably not what the problem was looking for. A better stated problem would have been "find a non-zero vector orthogonal to ..."

[@jeromeherrera4482]

so can, a=0 and b=any real number?

[@aranglancy]

Sure. Any triple (a,b,c) that works in the equation 3a - b + 2c = 0 gives a solution. So if a = 0, b can be anything as long as c = b/2. (I got that from plugging a = 0 into 3a - b + 2c = 0)

[@jeromeherrera4482]

Thanks!! :)

[@noble64890]

this is wrong.

[@aranglancy]

In what way?

[@loyal0713]

It's not necessarily wrong. He gave an example to find a specific vector. There's a whole other process for finding all the vectors that are orthogonal to a given vector.

[@reinowarren7880]

Presenter was not prepared. There are a lot of "ah" and stops.

[@Mativer]

if the message was understood then no harm done. :)

[@michaelshope3319]

Do you want your money back? Oh wait

[@asherb6095]

Great this means hes not a robot thats spitting nonsense like me lecturers that read of paper and explain fast..