- #1

danago

Gold Member

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**Let [tex]\vec{a} \ne 0[/tex] be a fixed vector in R**

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

(b) Is the set of all vectors [tex]\vec{x} \in R^3[/tex] satisfying [tex]\vec{a} \times \vec{x}=0[/tex] a subspace of R

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

(b) Is the set of all vectors [tex]\vec{x} \in R^3[/tex] satisfying [tex]\vec{a} \times \vec{x}=0[/tex] a subspace of R

^{3}?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**.

(k

**p**+l

**q**).

**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 R

^{3}?

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.