# Prove that the set of all vectors is a subspace

Gold Member
Let $$\vec{a} \ne 0$$ be a fixed vector in R3.

(a) Prove that the set of all vectors $$\vec{x} \in R^3$$ satisfying $$\vec{a}.\vec{x}=0$$ is a subspace of R^3. Describe this set geometrically.

(b) Is the set of all vectors $$\vec{x} \in R^3$$ satisfying $$\vec{a} \times \vec{x}=0$$ a subspace of R3?

For part (a), i thought about the problem geometrically and id be inclined to say that it is a subspace. The zero dot product implies that a and x are orthogonal, and so the set of all x will be a plane through the origin (since <0,0>.a=0 for any a) which is perpendicular to a.

Analytically, i proved it by letting p and q be vectors such that p.a=q.a=0. I then showed that any linear combination of these two vectors is also perpendicular to a.

(kp+lq).a=k(p.a)+l(q.a)=0

Is that correct?

Its part (b) where im a bit stuck. If the cross product of two vectors is the zero vector, this implies that they are parallel, yea? So x is some element of the set of all vectors which are parallel to a? Since the linear combination of any two vectors parallel to a will also be parallel to a, can we conclude that the set of all x is a subspace of R3?

Now, if my reasoning is correct, how could i go about showing this analytically? I guess it comes down to showing that the cross product being zero implies that the vectors are parallel; once i can show this i can do the rest myself.

Dan.

Defennder
Homework Helper

You can do (b) the same way you did (a). Just use the properties of cross product to do so.

Gold Member

You can do (b) the same way you did (a). Just use the properties of cross product to do so.

Oh...ofcourse..haha dont know why i didnt do that. Guess its time for bed :tongue2: Thanks for the quick reply