Let u=(a,b,c) be the vector spanning the intersection of U and W.
First
we find the equation of the plane W spanned by the vectors v=(1,1,1) and
w=(1,2,3).
The normal vector, n=(nx,ny,nz), to this plane is the
cross-product n=[v,w].
This vector has the following
coordinates:
nx = det 1 1 = 3-2 = 1
2 3
ny
= det 1 1 = 1-3 = -2
3 1
nz = det 1 1 = 2-1 =
1
1 2
Thus n=(1,-2,1) and the equation of the plane
W:
x-2y+z=0
Thus u satisfies the equation:
a-2b+c=0.
Moreover, since u belongs to U, we have that c=0.
This gives
the relation
a-2b=0 => a=2b.
Hence we can take u=(2,1,0).
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