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Thread: Vectors question!

  1. #1
    LHS
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    Exclamation Vectors question!

    Hi, having a little trouble on this vector question, any help appreciated!
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  2. #2
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    Write the equation of the line determined by $\displaystyle A\;\&\;B$: $\displaystyle l(t) = B + t\left( {\overrightarrow {AB} } \right)$.
    Be sure to note that $\displaystyle \overrightarrow {AB} = B - A$. The point you want is $\displaystyle \color{blue}P = l(1)$
    Last edited by Plato; Mar 14th 2009 at 06:27 AM.
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  3. #3
    LHS
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    Why is the point I want where t=1? I thought as AP=2BP I would want to set t=(2/3)?
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    LHS
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    My text says that P= (-7i, -2j, -5k)
    Is this correct?
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    If you use this equation $\displaystyle l(t) = B + t\left( {\overrightarrow {AB} } \right)$, then $\displaystyle l(1) = \left\langle { - 7, - 2, - 5} \right\rangle $, as in your textbook.
    That means that the order of the points is $\displaystyle A-B-P$, B is between A & P.
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  6. #6
    LHS
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    Ok, thank you, I think I see now, they could be in that order from the way the question its written. Not sure why I assumed P was between them.
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    you are asked to find position vector $\displaystyle \overrightarrow{OP}$

    $\displaystyle \overrightarrow{OP} = \overrightarrow{OB} + \overrightarrow{BP} = \overrightarrow{OA} + \overrightarrow{AP}$

    $\displaystyle \overrightarrow{OA} = 5i + 4j + k$

    $\displaystyle \overrightarrow{OB} = -i + j - 2k$

    $\displaystyle \overrightarrow{AB} = -\overrightarrow{OA} + \overrightarrow{OB} \equiv (-i + j - 2k) - (5i + 4j + k) \equiv -6i - 3(jk)$

    $\displaystyle AB = 2BP + PB \equiv 3PB$

    $\displaystyle \overrightarrow{AP} = \frac{2}{3} (-6i -3(jk)) \equiv -4i - 2(jk)$

    $\displaystyle \overrightarrow{OP} = \overrightarrow{0A} + \overrightarrow{AP} \equiv (5i + 4j + k) + (-4i - 2(jk)) \equiv i + 2j - k$

    i think this is load of rubbish but i try

    corrected for LHS and Plato's comments

    $\displaystyle \overrightarrow{OP} = \overrightarrow{OB} + \overrightarrow{BP} \text{ where } \overrightarrow{BP} = \overrightarrow{AB}$

    from above

    $\displaystyle (-i + j - 2k) + (-6i - 3j - 3k) = -7i -2j -5k $
    Last edited by sammy28; Mar 14th 2009 at 11:23 AM.
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    Quote Originally Posted by sammy28 View Post
    you are asked to find position vector OP
    $\displaystyle \color{red}OP = OA + PB = OA + AP$
    Sammy, you need to rethink your reply.
    You assume alot in the above.
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  9. #9
    LHS
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    Sammy I think you've done what I've done and assumed that P is in between A and B, I think P is one more lot of AB from B.
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    Quote Originally Posted by LHS View Post
    Hi, having a little trouble on this vector question, any help appreciated!
    There are 2 points satisfying the given conditions:

    $\displaystyle P_1 \ at\ t = \frac13$

    $\displaystyle P_2 \ at\ t =- \frac12$
    Attached Thumbnails Attached Thumbnails Vectors question!-zweipunkt_abstand.png  
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    Hello, LHS!

    Here's an "eyeball" solution . . .


    The position vectors of points $\displaystyle A$ and $\displaystyle B$ relative to an origin $\displaystyle O$
    . . are: .$\displaystyle \vec A \:=\: 5i + 4j + k,\;\;\vec B \:=\: \text{-}i + j - 2k$.

    Find the position vector of the point $\displaystyle P$ which lines on $\displaystyle AB$ produced
    . . such that: .$\displaystyle \overrightarrow{AP} \:=\:2\cdot\overrightarrow{BP}$.

    . . $\displaystyle \begin{array}{ccccc}
    (5,4,1) & & (\text{-}1,1,\text{-}2) & & (?,\,?,\,?) \\
    * & \longrightarrow & *& \longrightarrow & * \\
    A & & B & & P \end{array}$


    Going from $\displaystyle A$ to $\displaystyle B$, we see:

    . . $\displaystyle \begin{array}{ccc}
    \text{variable} & \text{moves} & \text{change} \\ \hline
    x & \text{5 to -1} & \text{-}6 \\
    y & \text{4 to 1} & \text{-}3 \\
    z & \text{1 to -2} & \text{-}3 \end{array}$


    Going from $\displaystyle B$ to $\displaystyle P$, the change would be the same:

    . . $\displaystyle \begin{array}{ccc}
    B & \text{change} & P \\ \hline
    x=\text{-}1 & \text{-}6 & x = \text{-}7 \\
    y = 1 & \text{-}3 & y = \text{-}2 \\
    z = \text{-}2 & \text{-}3 & z = \text{-}5 \end{array}$


    Threfore: .$\displaystyle P \:=\:(\text{-}7,\text{-}2,\text{-}5) $

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  12. #12
    LHS
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    Ah, thank you, that is much clearer!
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    Quote Originally Posted by LHS View Post
    Ah, thank you, that is much clearer!
    Your textbook gives the answer that Soraban & I both gave you.
    Reading the question as a mathematician, from the wording I assumed that if order made a difference. That is the point P could not be between A & B. B would be between A & P.
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