Linear Algebra - Forming planes that intersect a given plane

[Throwback]

Homework Statement

Given the following 2-dimensional plane in R^4 written in parametric form:

x_1 = 1 + (-1)s
x_2 = 0 + ( 1)s
x_3 = 1 + (-1)t
x_4 = 0 + ( 1)t

a) find a 2-dimensional plane that does not intersect it
b) find a 2-dimensional plane that intersects it to form a point
c) find a 2-dimensional plane that intersects it to form a line

Homework Equations

dot product

The Attempt at a Solution

a) Just change the position vector, so what results is a parallel line, shifted away from the original plane, and therefore cannot intersect:

x_1 = 0 + (-1)s + (0)t
x_2 = 0 + (1)s + (0)t
x_3 = 0 + (0)s + (-1)t
x_4 = 0 + (0)s + (1)t

b) The dot product between the original plane and my new plane must be equal to zero, and hence they are perpendicular, forming a point. I'm not quite sure how to calculate this... Is this even possible (i.e., trick question)? Help!

c) I need the same position vector (1,0,1,0), but what about the direction vectors? I guess they shouldn't be perpendicular here, so... I'm really not sure... What if I change just one number in one of the direction vectors?

x_1 = 1 + (-1)s + (0)t
x_2 = 0 + (0)s + (0)t
x_3 = 1 + (0)s + (-1)t
x_4 = 0 + (0)s + (1)t

 

[Bacle]

This may be helpful:

Given subsets (in this case, subspaces)U,V of n-dimensional space, of complementary
codimension (in general position), i.e., DimU+DimV=n , the dimension of the intersection of U,V is 0; in general, when the expression makes sense, the dimension of the intersection of subspaces within a space of dimension n is n-CodimU-CodimV , where Codim is the difference between n and the dimension of the subspace. Then, two lines in R^2 , since each has condimension 1, intersect in a subspace of dimension 2-(1+1)=0 , i.e., at a point.

Anyway, maybe you can use that , and maybe shift one of your planes around to avoid a collision .