Linear Algebra for Machine Learning: Distance of a point from a plane, half-spaces Lecture 6

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[@YulycaNebunica]

Really well explained, congrats!

[@htascon]

Good video thanks. How would the formula change if w does not cross origin? If the hyperplane has an offset from origin?

[@wasiqueumar9916]

the distance can be proved using the projection of vector a.b/||b||

[@hasibulhaque9452]

Amazing. Sir, would you share the source such as books for learning this beautiful Linear Algebra?

[@deepthikiran8345]

Is d=P.W/|W| instead of W.P/|W| ? or please explain?

[@AppliedAICourse]

W.P=P.W So, both are correct.

[@guptamols]

How can we visualise a N dimensional plane...say a 5 Dimensional plane ?

[@wikd13]

we cant visualize anything above 3 dimension.

[@nishant9847]

u can not imagine anything above 3 dimentions
u can just operate in more than 3 dimentions using liner algerbra not more than that

[@pankajchoudhury3593]

WtP=||W||.||P|| cos (theta)
=>||P||=WtP/||W|| cos (theta) ------------------- (1)
||P|| is distance (d) of point P to the plane in perpendicular
and Theta is angle between W and P so it is zero because W and P are parallel
From (1) , d=WtP/||W||
proved ...

[@thebigmonstaandy6644]

you should not spoil the proof. i would not get this proof by myself , if i would see comments

[@tulikasarkar9122]

Sir Please provide the link to the webpage where the proof is done.

[@larss5530]

the link to the proof is not working, can you add a seperate new link?

[@holyprogrammer8335]

if u didnt get the formula of distance of a point from a plane...
then i would recommend u to get to the time stamp 7:20 and have a look at the plot...
the distance d looks more like a projection of vector p over vector w.....and finally it is the shadow of vector p over vector w when looked from the top....so its pretty easy to understand the formula now..its just a formula of projection

[@kids94594]

math.stackexchange.com/questions/1210545/distance-from-a-point-to-a-hyperplane/1210685#1210685

[@PhoenixRisingFromAshes471]

@@holyprogrammer8335 spot on i also thought of same and now i am pretty sure i was right and thanks for explaination though:)

[@DesertHash]

@@holyprogrammer8335 I see that, but I don't see how the formula for the distance of P dash from the plane is derived to be that shown in 7:00. Any ideas?

[@thebigmonstaandy6644]

@@DesertHash cos(fi dash)=cos(180°-fi)=-cos (fi)