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Math Help - line equation in vectors

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    line equation in vectors

    "find the equation of the line passing through B, with positiion vector b, which is perpendicular to the line r=a + \lambdac, c=/=0, given that B is not a point on the line". hi, i dont even get this at all to be honest, how is r=a + \lambdac a line? in terms of normal line equations, which is the gradient, and the intercept? because r and c are not axis they are vectors, i dont know what this is asking.
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    Re: line equation in vectors

    A line is essentially a vector (which determines the direction) that is of infinite length, which is then positioned somewhere.

    So basically, what has happened to get the line \displaystyle \begin{align*} \mathbf{r} \end{align*} is that \displaystyle \begin{align*} \mathbf{c} \end{align*} determines the direction of the line, it is multiplied by some parameter \displaystyle \begin{align*} \lambda \end{align*} to make it of infinite length, and then the point \displaystyle \begin{align*} a \end{align*} has been added (or rather, each component of \displaystyle \begin{align*} \mathbf{c} \end{align*} is adjusted numerically by each component of \displaystyle \begin{align*} a \end{align*} ) to position it in the correct spot.

    Do you think you can get a vector-form equation for the line passing through B in the direction of \displaystyle \begin{align*} \mathbf{b} \end{align*} ?
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    Re: line equation in vectors

    hi,
    when you say vector form equation, does this mean i am looking for another line like the r=a+ lc? so it will again be in these terms.

    but it must be perpendicular. my immediate thought is find to reciprocate and negate the lambda? so it would be r = b -c/l? a is written in the same style as b and r so it is a vector too but is it really just a point? how can a vector (position vector?) be a point, i thought all vectors are lines.
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    Re: line equation in vectors

    Quote Originally Posted by learning View Post
    hi,
    when you say vector form equation, does this mean i am looking for another line like the r=a+ lc? so it will again be in these terms.

    but it must be perpendicular. my immediate thought is find to reciprocate and negate the lambda? so it would be r = b -c/l? a is written in the same style as b and r so it is a vector too but is it really just a point? how can a vector (position vector?) be a point, i thought all vectors are lines.
    You need to start thinking of lines as VECTORS. How do you know if two VECTORS are perpendicular?
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    Re: line equation in vectors

    hi, i know this measn the dot product is zero for two perpendicular vectors. but what is the vector of r=a + lc? how can i dot this with something?
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    Re: line equation in vectors

    Quote Originally Posted by learning View Post
    hi, i know this measn the dot product is zero for two perpendicular vectors. but what is the vector of r=a + lc? how can i dot this with something?
    The vector IS \displaystyle \begin{align*} \mathbf{r} \end{align*}. This is extremely difficult to illustrate without knowing anything about the vectors you've actually been given. Is this part of a larger problem?
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    Re: line equation in vectors

    hi,

    this is the entirety of this problem
    there is no 'specific' vector given. i have writen the whole question. one thing i am not sure of is even what should the form be of this perpendicular vector? would it also be r (what is the meaning of the name r?) or should i call it r2 to show its a different vector? so maybe r2 = b + something d? or would it also have c in it? would it also have the lambda or a different parameter? should the parameter be perpendicular? -1/lamba? if i have an equation to solve i can do it no problem at all but i dont know what this is meaning
    Last edited by learning; January 31st 2013 at 04:25 AM.
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    Re: line equation in vectors

    Quote Originally Posted by learning View Post
    "find the equation of the line passing through B, with positiion vector b, which is perpendicular to the line \ell_1:~a +\lambda c, c\ne , given that B is not a point on the line".


    First write the equation of the plane, \Pi that contains the point B with normal \lambda .

    Now the given line \ell_1 is perpendicular to \Pi.

    Let \{C\}=\ell_1\cap\Pi.

    Now the line \overleftrightarrow {BC} is the answer to your question.
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    Re: line equation in vectors

    you have introducted 4 or 5 symbols and notations which are not in the question. it doesnt help if you go way too technical like that. what you say may be true but its not helpful..
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    Re: line equation in vectors

    Quote Originally Posted by learning View Post
    you have introducted 4 or 5 symbols and notations which are not in the question. it doesnt help if you go way too technical like that. what you say may be true but its not helpful..
    I am sorry that you find a detailed way to answer the question not helpful. If those symbols confuse you, then perhaps you don't understand the question in the first place.
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    Re: line equation in vectors

    why would i be posting it if i did? :s
    you may have the technical expertise but you dont have much common sense. being intelligent isnt all thats required of a good teacher..

    now i prefer someone to help who actually wants to help rathr than just 'show off' their knowledges.. prove it is giving good hints to steer in the right direction and you just spout out jargon.. if i knew what to do with that information i probably wouldnt have to ask for help on it now would i
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    Re: line equation in vectors

    someone explain the answer please in terms of the vectors

    for instance, i dont even know what the format required is? something like the r = a + lc, so r = b + md or something? just with different coefficeint and intercept?
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    Re: line equation in vectors

    ya steal need help in this please
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    Re: line equation in vectors

    someone better please help
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    Re: line equation in vectors

    No threatening learning. Keep it up and no one will want to come anywhere near you. And of course I won't allow it....Rethink how you want to communicate.

    -Dan

    Edit: After talking to learning the above comment was not meant as a threat. My apologies.

    -Dan
    Last edited by topsquark; February 6th 2013 at 12:47 AM.
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