May be this derivation is more simple.
The plane equation is
(r-r_0,n)=0
n is normal to plane.
(r,n)-(r_0,n)=0 this is your plane equation.
Q) Let H be a subset of R^n defined by one single linear equation,
i.e; H={(x1,x2,...,xn) | a1x1+...+anxn+d=0}.Then show that H is a hyperplane of R^n..
[ I was able to prove that "A hyperplane H in R^n is given by a single linear equation,a1x1+...+anxn+d=0"..
(What I did was as H is a hyperplane in R^n, it is an (n-1) dimension affine subspace of R^n. This implies H=W+p,for some p in R^n and W is a linear subspace of R^n of dim(n-1)..Then choose a basis {w1,...,wn-1} of W and extend it a basis {w1,...,wn} of R^n..Then applyin Gram-Schmidth we obtain orthonormal basis {u1,...,un} of R^n..Hence {u1,...,un-1} is an orthonormal basis of W..
This implies W={v in R^n | <v,un>=0}..
Let x belong to H <=>x-p is in W <=> <x-p,un>=0 <=><x,un>-<p,un>=0..
Let un=(a1,...,an) in R^n, Then we get <(x1,...,xn) , (a1,...,an)> - <p,un>=0
This implies a1x1+...+anxn+d=0, where d+ -<p,un>..]
I need help in provin the above posted question on the same lines of the proof i gave in one way..
Thank you.
Q) Let H be a subset of R^n defined by one single linear equation,
i.e; H={(x1,x2,...,xn) | a1x1+...+anxn+d=0}.Then show that H is a hyperplane of R^n..
..Ok i define a linear transformation T:R^n-->R by T(x)=Ax,where A=[a1...an]..then as atleast one of the ai is nonzero,so rank(A)=1..and the ker(a)={x in R^n |a1x1+....+anxn=0}..so dim(ker(A))=n-1..then what should i do?