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Thread: Vector problem:

  1. #1
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    Vector problem:

    "The vectors v = i + 2j - 3k, u = 2i + 2j + 3k, and w = i + (2 - t)j + (t + 1)k
    are given. Find the value of t such that the three vectors u, v and w are coplanar. "



    I have no idea how to solve this one!Where do I start?
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  2. #2
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    Smile

    let's take $\displaystyle v=(1,2,-3),u=(2,2,3),w=(1,2-t,t+1)$
    Using Gauss method on vectors,we should have
    $\displaystyle \begin{pmatrix}
    1\\
    2\\
    -3
    \end{pmatrix}$$\displaystyle ,\begin{pmatrix}
    2\\
    2\\
    3
    \end{pmatrix}$$\displaystyle ,\begin{pmatrix}
    1\\
    2-t\\
    t+1
    \end{pmatrix}$ $\displaystyle \Rightarrow $
    $\displaystyle \begin{pmatrix}
    1\\
    2\\
    -3
    \end{pmatrix}$$\displaystyle ,\begin{pmatrix}
    0\\
    -2\\
    9
    \end{pmatrix}$$\displaystyle ,\begin{pmatrix}
    0\\
    0\\
    \frac{-7t+8}{2}
    \end{pmatrix}$
    for those three victors to be coplanar we must have $\displaystyle rang(\begin{pmatrix}
    1\\
    2\\
    -3
    \end{pmatrix}$$\displaystyle ,\begin{pmatrix}
    0\\
    -2\\
    9
    \end{pmatrix}$,$\displaystyle \begin{pmatrix}
    0\\
    0\\
    \frac{-7t+8}{2}
    \end{pmatrix} )$$\displaystyle =2 $ that is $\displaystyle \frac{-7t+8}{2}=0$ which means $\displaystyle t=\frac{7}{8}$
    hope that's right
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  3. #3
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    Smile

    Quote Originally Posted by Raoh View Post
    let's take $\displaystyle v=(1,2,-3),u=(2,2,3),w=(1,2-t,t+1)$
    Using Gauss method on vectors,we should have
    $\displaystyle \begin{pmatrix}
    1\\
    2\\
    -3
    \end{pmatrix}$$\displaystyle ,\begin{pmatrix}
    2\\
    2\\
    3
    \end{pmatrix}$$\displaystyle ,\begin{pmatrix}
    1\\
    2-t\\
    t+1
    \end{pmatrix}$ $\displaystyle \Rightarrow $
    $\displaystyle \begin{pmatrix}
    1\\
    2\\
    -3
    \end{pmatrix}$$\displaystyle ,\begin{pmatrix}
    0\\
    -2\\
    9
    \end{pmatrix}$$\displaystyle ,\begin{pmatrix}
    0\\
    0\\
    \frac{-7t+8}{2}
    \end{pmatrix}$
    for those three victors to be coplanar we must have $\displaystyle rang(\begin{pmatrix}
    1\\
    2\\
    -3
    \end{pmatrix}$$\displaystyle ,\begin{pmatrix}
    0\\
    -2\\
    9
    \end{pmatrix}$,$\displaystyle \begin{pmatrix}
    0\\
    0\\
    \frac{-7t+8}{2}
    \end{pmatrix} )$$\displaystyle =2 $ that is $\displaystyle \frac{-7t+8}{2}=0$ which means $\displaystyle t=\frac{7}{8}$
    hope that's right
    in any case,u should wait until someone else confirm my post or the value of $\displaystyle t$.
    (i don't trust my answer anyway )
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  4. #4
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    Thanks for the answer, but I was hoping for a more "high-school level" answer!
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  5. #5
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    Quote Originally Posted by karldiesen View Post
    "The vectors v = i + 2j - 3k, u = 2i + 2j + 3k, and w = i + (2 - t)j + (t + 1)k
    are given. Find the value of t such that the three vectors u, v and w are coplanar. "
    Here is another way.
    Solve $\displaystyle W\cdot(V\times U)=0$.
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