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Math Help - Subspace

  1. #1
    Junior Member
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    Subspace

    Again not sure what to do here.

    Subspace of R3 spanned by the vectors u1 = (1,1,1) and u2 = (2, 0, -1) is a plane passing through the origin. Express w = (1, 2, 3) in the form w = w1 + w2 where w1 lies in the plane and w2 is perpendicular to the plane.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi
    Quote Originally Posted by chadlyter View Post
    Again not sure what to do here.

    Subspace of R3 spanned by the vectors u1 = (1,1,1) and u2 = (2, 0, -1) is a plane passing through the origin. Express w = (1, 2, 3) in the form w = w1 + w2 where w1 lies in the plane and w2 is perpendicular to the plane.
    Let u_3=u_1\times u_2 and let P be the plane spanned by u_1 and u_2 which passes through the origin.

    (u_1,u_2,u_3) is a basis of \mathbb{R}^3 hence w=\alpha u_1+\beta u_2+\gamma u_3 for some real numbers \alpha,\,\beta and \gamma. As \alpha u_1+\beta u_2\in P and \gamma u_3 \perp P , one can also write w=w_1+w_2 with \begin{cases}w_1=\alpha u_1+\beta u_2, & w_1\in P \\w_2=\gamma u_3, & w_2\perp P\end{cases}

    Hence solving this problem boils down to finding the values of the three unknowns \alpha,\,\beta and \gamma.

    Does it help ?
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  3. #3
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    The normal to the given plane is N = u_2  \times u_1
    Now you want w = w_1  + tN \Rightarrow \quad w_1  = w - tN\,\& \,w_1  \cdot N = 0.
    The last bit is to insure w_1 is in the plane.
    Now solve for t.
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  4. #4
    Newbie Catherine Morland's Avatar
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    Hi. I did it this way, which is a bit tedious but at least I understand what I'm doing.

    First find the equation of the plane. Any point (x,y,z) on it is of the form au_1+bu_2, a,b\in\mathbb{R}. So

    x=a+2b
    y=a
    z=a-b

    whence eliminating a and b gives x-3y+2z=0.

    Now let w=w_1+w_2 where w=\left(\begin{array}{r}1\\2\\3\end{array}\right) and w_1=\left(\begin{array}{c}3y-2z\\y\\z\end{array}\right).

    Note that u_1 and u_2 lie in the plane. So take the dot product of w with u_1 and u_2, noting that u_1\cdot w_2=u_2\cdot w_2=0:

    w\cdot u_1=w_1\cdot u_1\ \implies\ 6=4y-z
    w\cdot u_2=w_1\cdot u_2\ \implies\ -1=6y-5z

    Solving gives y=\frac{31}{14}, z=\frac{40}{14}

    Hence

    \left(\begin{array}{r}1\\2\\3\end{array}\right)\ =\ \left(\begin{array}{r}\frac{13}{14}\\\\\frac{31}{1  4}\\\\\frac{40}{14}\end{array}\right)+\left(\begin  {array}{r}\frac{1}{14}\\\\-\frac{3}{14}\\\\\frac{2}{14}\end{array}\right)
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