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Math Help - X circle plus Y = R^3

  1. #1
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    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
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  2. #2
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    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.
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  3. #3
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    Re: X circle plus Y = R^3

    {
    Quote Originally Posted by HallsofIvy View Post
    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{???}\}
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  4. #4
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    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.
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