Re: X circle plus Y = R^3

It should be obvious that this is a plane and that the vector $\displaystyle a\vec{i}+ b\vec{j}+ c\vec{k}$ is perpendicular to the plane.

Re: X circle plus Y = R^3

{ Quote:

Originally Posted by

**HallsofIvy** It should be obvious that this is a plane and that the vector $\displaystyle a\vec{i}+ b\vec{j}+ c\vec{k}$ is perpendicular to the plane.

Ok so I just need to add the subspace of the normal vector to the plane. How can I describe the set Y to encompass all the normal vectors <a,b,c> to the plane?

$\displaystyle Y=\{(x,y,z)\in\mathbb{R}^3: \text{???}\}$

Re: X circle plus Y = R^3

Y is the subspace spanned by $\displaystyle a\vec{i}+ b\vec{j}+ c\vec{k}$. That is the space of all vectors of the form $\displaystyle \alpha(a\vec{i}+ b\vec{j}+ c\vec{k})$ where $\displaystyle \alpha$ can be any real number.