Does the points P an Q lie on the given surface? r(u,v) = <2u+3v, 1+5u-v, 2+u+v> P(7,10,4) and Q(5,22,5)
can someone explain this question? :)
isn't r(u,x,v) or is it r(u,v) there is 3 number but we have 2 in r...
no it is r(u,v)
there is no variable given for the third component; what form is the equation in? cartesian or polar
i got it, zir confused me for a moment lol
yea i got it too but it needs 3 component anyway
its an exercise from parametric surface and their areas
r(u,v) = <2u+3v, 1+5u-v, 2+u+v> given a point (x,y,z); does it satisfy these parts? x = 2u+3v = 7 y = 1+5u-v=10 z = 2+u+v = 4 and x = 2u+3v = 5 y = 1+5u-v= 22 z = 2+u+v = 5
2u+3v = 7 5u - v = 9 u + v = 2 now solve the equation and see if u and v are consistent throut them
i have also those equations :) owhhh oke.. so i can put this in a matrix right
matrix is fine, or regular substitution since they is simple enough
i gotta run to class, have fun and good luck ;)
5u - v = 9 v = 2 - u 5u - (2-u) = 9 5u - 2+u = 9 6u = 11 u = 11/6 perhaps
im back :) if u = 11/6, then v would have to be: v = 2 - 11/6 = 1/6 so lets see if it fits the point: 2(11/6)+3(1/6) = 7 25/6 not = 7 ---- 5(11/6) - (1/6) = 9 54/6 = 9 ---- 11/6 + 1/6 = 2 12/6 = 2 well, at least we know those dont fit but we should try to solve it by using the 2 other combonations of equations just to dbl chk
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