Recent content by jay_jay_lp

Thanks a lot, I'll reply again if I need more help Jay

At what points on the curve (y-2)^3 + y(x-6) = 0 does dy/dx not exist? So I've tried a few approaches: 1. find point where dx=0, since -Fx/Fy = dy/dx, therefore also Fy=0. So I took the partial derivative wrt y and got another equation, 3(y-2)^2 + (x-6) = 0. From here I don't know where to go...

Oh wow that's much more simplified. Thanks so much! :)

So this is going to be very general but: r = p+tv (x,y,z) = (Xo,Yo,Zo)-(aXo+bYo+cZo)/(a^2+b^2+c^2)(a,b,c) so x=Xo-(aXo+bYo+cZo)/(a^2+b^2+c^2)a y=Yo-(aXo+bYo+cZo)/(a^2+b^2+c^2)b z=Zo-(aXo+bYo+cZo)/(a^2+b^2+c^2)c Now is that it? hah Thanks again.

Thanks for helping so much! so a(Xo+ta)+b(Yo+tb)+c(Zo+tc)=0 t(a^2+b^2+c^2)+aXo+bYo+cZo=0 t=-(aXo+bYo+cZo)/(a^2+b^2+c^2) t=-(v.p)/(a^2+b^2+c^2) - don't know if this is necessary? So are the points Xo,Yo,Zo now the points we are interested in? Thanks! Jay

Would this be ax+by+cz=p+tv? Can I break this into components? ax=x+at by=y+bt cz=z+ct Is it okay to assume that point P is the point we are interested in?

This will give ax+by+cz=0, where x,y,z are components of the normal vector and a,b,c components of the direction vector of the line. but this is not unique?

Thanks for replying, Yea the shortest distance will be a normal to the line that passes through the origin, so will that will be a normal to the direction vector of the line,r, that also passes through the origin. But how do I find this? Thanks Jay

Find an expression for the distance between the origin and the line L given by r = p + tv, where t is an element of R and determine the point on L closest to the origin So the first part of the question i took as just the magnitude of p since the distance will just be...