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Math Help - Intersection of lines,planes,vector spaces

  1. #1
    jht
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    Intersection of lines,planes,vector spaces

    hi i got some question that i dont understand,hope someone could help me with them,thanks so much.:'(

    i] Find intersection of two lines

    L1:={(1) + altha(1) :altha E R }
    (2) (1)
    (1) (2)


    L2:={(2) + mu(2) :mu E R }
    (2) (1)
    (1) (2)

    how come is (0) ?
    (1)
    (-1)

    -----------------------------------
    ii] Find intersection of two planes

    P1:={(2) + lamda1(2) +lamda2 (1) :lamda1,lamda2 E R }
    (2) (1) (2)
    (3) (2) (2)


    P2:={(3) + mu1(2) +mu2 (3) :mu1,mu2 E R }
    (1) (3) (2)
    (3) (2) (3)

    how come is {(1) + lamda1 (2) :lamda1 E R}??
    (0) (1)
    (1) (2)

    -------------------------------------------
    iii]V=R^3 and subspaces

    W1 = {(x) E R^3 :5x +2y -z =0 }
    (y)
    (z)
    W2=span:{(1) , (2) , (4) }
    (1) (1) (3)
    (1) (3) (5)
    find bases for W1,W2,W1+W2 and intersection of W1 and W2.


    ------------------

    Once again thank you so much for your time and attention,
    It would help alot if i understand the questions
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  2. #2
    MHF Contributor

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    Quote Originally Posted by jht View Post
    hi i got some question that i dont understand,hope someone could help me with them,thanks so much.:'(

    i] Find intersection of two lines

    L1:={(1) + altha(1) :altha E R }
    (2) (1)
    (1) (2)
    I think you mean the line given by the vector equation \begin{bmatrix}1 \\ 2 \\ 1 \end{bmatrix} + \alpha\begin{bmatrix}1 \\ 1 \\2 \end{bmatrix}
    which could also be written as the parametric equations x= 1+ \alpha, y= 2+ \alpha, and z= 1+ 2\alpha with \alpha as parameter.

    L2:={(2) + mu(2) :mu E R }
    (2) (1)
    (1) (2)
    This line is given by \begin{bmatrix}2 \\ 2 \\ 1\end{bmatrix}+ \mu\begin{bmatrix}2 \\ 1\\ 2\end{bmatrix} or the parametric equations x= 2+ 2\mu, y= 2+ \mu, and z= 1+ 2\mu

    how come is (0) ?
    (1)
    (-1)
    Of course, at the point of intersection, the x, y, and z values (or corresponding components of the vector) must be equal. Setting the x coordinates equal, we have x= 1+ \alpha= 2+ 2\mu. Setting the y coordinates equal, y= 2+ \alpha= 2+ \mu. That second equation clearly says that \mu= \alpha putting that into the first equation 1+ \alpha= 2+ 2\alpha so that \mu= \alpha= -1. But "most" lines, in three dimensions do not intersect at all. We need to check that these values also satisfy z= 1+ 2\alpha= z= 1+ 2\mu but it is obviously true that any \alpha= \mu will satisfy that. Putting \alpha= -1 into x= 1+ \alpha, y= 2+ \alpha, and z= 1+ 2\alpha gives x= 1- 1= 0, y= 2- 1= 1, and z= 1+ 2(-1)= -1.

    -----------------------------------
    ii] Find intersection of two planes

    P1:={(2) + lamda1(2) +lamda2 (1) :lamda1,lamda2 E R }
    (2) (1) (2)
    (3) (2) (2)
    So x= 2+ 2\lambda_1+ \lambda_2, y= 2+ \lambda_1+ 2\lambda_2, and z= 3+ 2\lambda_1+ 2\lambda_2.

    P2:={(3) + mu1(2) +mu2 (3) :mu1,mu2 E R }
    (1) (3) (2)
    (3) (2) (3)
    Here, x= 3+ 2\mu_1+ 3\mu_2, y= 1+ 3\mu_1+ 2\mu_2, and z= 3+ 2\mu_1+ 3\mu_2

    how come is {(1) + lamda1 (2) :lamda1 E R}??
    (0) (1)
    (1) (2)
    Again, on the line of intersection, x= 2+ 2\lambda_1+ \lambda_2= 3+ 2\mu_1+ 3\mu_2, y= 2+ \lambda_1+ 2\lambda_2=  1+ 3\mu_1+ 2\mu_2, and z= 3+ 2\lambda_1+ 2\lambda_2= 3+ 2\mu_1+ 3\mu_2. In the first problem, we had three equations in two unknown values which normally doesn't have a solution just a two line don't normally intersect in three dimensions. Here we have three equations in four unknown values which is "under determined"- there will be an infinite number of solutions just as two planes intersect on a line, not just at one point.

    If we subtract the second equation from the third we eliminate \lambda_2: 1+ \lambda_1= 2- \mu_1+ \mu_2 or \lambda_1= 1- \mu_1+ \mu_2. Putting that into the first equation we have 2+ 2- 2\mu_1+ 2\mu_2+ \lambda_2= 3+ 2 \mu_1+ 3\mu_2 or \lambda_2= -1+ 4\mu_1+ \mu_2.

    Now put those values for \lambda_1 and \lambda_2 into, say the third equation: 3+ 2(1- \mu_1+ \mu_2)+ 2(-1+ 4\mu_1+ \mu_2)= 3+ 2\mu_1+ 3\mu_2 or 3+ 5\mu_1+ 4\mu_2= 3+ 2\mu_1+ 3\mu_2 which reduces to \mu_2= -3\mu_1.

    We can replace \mu_2 with 3\mu_2 in the parametric equations for the second plane to get x, y, and z given in terms of the single parameter \mu_1, the parametric equations of a line. It might not be precisely the answer given- a single line can have an infinte number of different parametric representations depending upon how the parameter is found.

    -------------------------------------------
    iii]V=R^3 and subspaces

    W1 = {(x) E R^3 :5x +2y -z =0 }
    (y)
    (z)
    W2=span:{(1) , (2) , (4) }
    (1) (1) (3)
    (1) (3) (5)
    find bases for W1,W2,W1+W2 and intersection of W1 and W2.
    W1 consists of the vectors of the form \begin{bmatrix}x \\ y \\ z\end{bmatrix} such that 5x+ 2y- z= 0. From that equation, z= 5x+ 2y so that \begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}x \\ y \\ 5x+ 2y\end{bmatrix} = \begin{bmatrxi}x \\ 0 \\ 5x\end{bmatrix}+ \begin{bmatrix}0 \\ y \\ 2y\end{bmatrix} = x\begin{bmatrix}1 \\ 0 \\ 5\end{bmatrix}+ y\begin{bmatrix}0 \\ 1 \\ 2\end{bmatrix}.

    That should make it clear what a basis for W1 is.

    You are given vectors that span W2. If they are independent they will be a basis. To see if they are independent, look at a\begin{bmatrix}1 \\ 1\\ 1\end{bmatrix}+ b\begin{bmatrix}2 \\ 1 \\ 3\end{bmatrix}+ c\begin{bmatrix}4 \\ 3 \\ 5\end{bmatrix} = \begin{bmatrix}a+ 2b+ 4c \\ a+ b+ 3c \\ a+ 3b+ 5c\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}

    That gives the three equations a+ 2b+ 4c= 0, a+ b+ 3c= 0, and a+ 3b+ 5c= 0. If we subtract the second equation from the first, we eliminate a: b+ c= 0. If we subtract the first equation from the third, we also eliminate a: b+ c= 0 again. Those are the same equation so we can take any b and c, as long as b+ c= 0. For example, b= 1, c= -1. Putting those into the first equation, a+ 2b+ 4c= a+ 2- 4= 0 or a= 2. The given vectors are NOT independent because 2\begin{bmatrix}1 \\ 1\\ 1\end{bmatrix}+ \begin{bmatrix}2 \\ 1 \\ 3\end{bmatrix}- \begin{bmatrix}4 \\ 3 \\ 5\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}. In particular, we can write \begin{bmatrix}2 \\ 1 \\ 3\end{bmatrix}= \begin{bmatrix}4 \\ 3 \\ 5\end{bmatrix}- 2\begin{bmatrix}1 & 1 & 1\end{bmatrix}

    Since one vector can be written in terms of the other two, we don't need that vector. Since the two remaining vectors are not multiples of one another, they form a basis for this space. (In fact, any two of these three vectors will form a vector.)

    One way to form a basis for the direct sum of the subspaces, W1+ W2, is to combine the two bases. If that set of vectors is independent, then it is a basis. (They can't be because you have four vectors and this is a subspace of dimension 3 or less.) If they are dependent, at least one of the vectors can be writen in terms of the others and you can simply discard that vector.

    To find a basis for the intersection, you have to see what that intersection is: which vectors are in both spaces.

    Using the bases above, any vector in W1 can be written as a\begin{bmatrix}1 \\ 0 \\ 5\end{bmatrix}+ b\begin{bmatrix} 0 \\ 1 \\ 2\end{bmatrix}= \begin{bmatrix}a \\ b \\ 5a+ 2b\end{bmatrix} for some numbers a and b. Any vector in W2 can be written as c\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}+ d\begin{bmatrix}2 \\ 1 \\ 3\end{bmatrix}= \begin{bmatrix}c+ 2d \\ c+ d \\ c+ 3d\end{bmatrix}.

    Any vector in the intersection of the two subspaces can be written in both ways: we must have a= c+ 2d, b= c+ d, and 5a+ 2d= c+ 3d.


    ------------------

    Once again thank you so much for your time and attention,
    It would help alot if i understand the questions
    Last edited by HallsofIvy; August 14th 2010 at 06:14 AM.
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  3. #3
    jht
    jht is offline
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    sorry i dont understand this part

    ''W1 consists of the vectors of the form such that 5x+ 2y- z= 0. From that equation, z= 5x+ 2y so that [/tex]= x\begin{bmatrix}1 \\ 0 \\ 5\end{bmatrix}+ y\begin{bmatrix}0 \\ 1 \\ 2\end{bmatrix}[/tex].''
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  4. #4
    MHF Contributor

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    Was the difficulty that my LaTex didn't work? I have corrected that.

    If not, the problem is that W1 consists of all <x, y, z> such that 5x+ 2y- z= 0. From the equation, z= 5x+ 2y so <x, y, z>= <x, y, 5x+ 2y>= <x, 0, 5x>+ <0, y, 2y>= x<1, 0, 5>+ y<0, 1, 2>. That is, any vector in the set can be written as a linear combination of <1, 0, 5> and <0, 1, 2> so those span the subset. Since neither is a multiple of the other, they are also independent and so form a basis for the subspace.
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