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Math Help - 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 A\;\&\;B: l(t) = B + t\left( {\overrightarrow {AB} } \right).
    Be sure to note that \overrightarrow {AB}  = B - A. The point you want is \color{blue}P = l(1)
    Last edited by Plato; March 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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  4. #4
    LHS
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    My text says that P= (-7i, -2j, -5k)
    Is this correct?
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  5. #5
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    If you use this equation l(t) = B + t\left( {\overrightarrow {AB} } \right), then l(1) = \left\langle { - 7, - 2, - 5} \right\rangle , as in your textbook.
    That means that the order of the points is [LaTeX ERROR: Convert failed] , 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 \overrightarrow{OP}

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

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

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

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

    AB = 2BP + PB \equiv 3PB

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

    \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

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

    from above

    (-i + j - 2k) + (-6i - 3j - 3k) = -7i -2j -5k
    Last edited by sammy28; March 14th 2009 at 11:23 AM.
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  8. #8
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    Quote Originally Posted by sammy28 View Post
    you are asked to find position vector OP
    \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:

    P_1 \ at\  t = \frac13

    P_2 \ at\  t =- \frac12
    Attached Thumbnails Attached Thumbnails Vectors question!-zweipunkt_abstand.png  
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  11. #11
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    Hello, LHS!

    Here's an "eyeball" solution . . .


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

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

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


    Going from A to B, we see:

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


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

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


    Threfore: . 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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  13. #13
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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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