# Answer to Question #7979 in Linear Algebra for Muhammad Imran Qureshi

Question #7979

Find one vector in R^3 that spans the intersection of U and W where U is the xy-plane, U={(a,b,0)}, and W is the space spanned by the vectors (1,1,1) and (1,2,3).

Expert's answer

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).

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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