Prove that the set of all vectors is a subspace

[danago]

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 I am 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.

Thanks in advance,
Dan.

 

[Defennder]

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

 

[danago]

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

Oh...ofcourse..haha don't know why i didnt do that. Guess its time for bed Thanks for the quick reply