Find the equation of the plane in the canonical basis

[Toncho1]

< Mentor Note -- thread moved to HH from the technical math forums, so no HH Template is shown >[/color]

Hi!
Can help me with this problem with my exercise?I don´t know if i did it okay or i have to do anymore
Is there another form to do it?
Be the π plane, whose equation with the base B (with coordinates (x ', y', z')) is z'= 0.
B={(1,1,0),(0,1,1),(1,-1,1)}.
a)Find the equation of the plane in the canonical basis. Prove analytically, how to get the same result.
b)Could you give a base B´(coordinate (x'', y'', z'')) in which the equation of the plane is y''= 0?

So, i did this:
a)
B = { u1, u2, u3} and canonic base K = { i, j, k }
u1 = i + j, u2 = j + k
u1 x u2 = u3
the equation of plane π is z ' = 0
therefore : this plane passes through origin and is perpendicular to u3 = i - j + k
the equation of plane π in K is :
OM . (i - j - k) = 0
giving :
x - y + z = 0

b)the equation of plane π is : third coordinate = 0
it is enough to keep the same vectors as those in U provided the SECOND is u3 :
in basis :
V = { u1, u3, u2 }
the equation of plane π is : second coordinate : y" = 0

 

[andrewkirk]

Hello and welcome to physicsforums!

Your answers look correct to me.