# Thread: X circle plus Y = R^3

1. ## X circle plus Y = R^3

Let $a,b,x\in\mathbb{R}$ not all equal to zero and $X=\{(x,y,z)\in\mathbb{R}^3: ax+by+cz=0\}$

Find a subspace Y such that $\mathbb{R}^3= X\oplus Y$ and deduce X is 2 dimensional.

I am lost on this one.

Note:

$\oplus$ is the external direct sum define as $X\oplus Y=(x_1,y_1)+(x_2,y_2)=(x_1+x_2,y_1+y_2)$
$\lambda (x_1,y_1)=(\lambda x_1,\lambda x_2)$

If $X \cap Y =\{0\}$, then $\oplus$ can be define as $\text{dim}(X+Y)=\text{dim}X+\text{dim}Y$ that is the internal direct sum $X\oplus Y$

2. ## Re: X circle plus Y = R^3

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

3. ## Re: X circle plus Y = R^3

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Originally Posted by HallsofIvy
It should be obvious that this is a plane and that the vector $a\vec{i}+ b\vec{j}+ c\vec{k}$ 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?

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

4. ## Re: X circle plus Y = R^3

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