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Thread: Major trouble on vector calculus

  1. #1
    Newbie TheTideCaller's Avatar
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    Major trouble on vector calculus

    We barely even touched projection vectors (?) and I am completely lost on how to even do this problem because I have no idea what to do.

    The distance d of a point P to the line through points A and B is the length of the component of APŻŻŻŻŻŻŻŻ that is orthogonal to ABŻŻŻŻŻŻŻŻ, as indicated in the diagram. So the distance from P=(2,2,3) to the line through the points A=(1,4,2) and B=(5,4,3) is ...

    What do I even do here? Pythagorean Theorem at some point?
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    Re: Major trouble on vector calculus

    What diagram?
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    Re: Major trouble on vector calculus

    what this is saying is that the vector $\vec{AB}$ is orthogonal to the vector $\vec{PP^\prime}$ where $P^\prime$ is the point on the line $\overline{AB}$ closest to the point $P$

    $\vec{AB} = B-A = (−5,4,−3) -(1,4,−2) = (-6,0,-1)$

    let $P^\prime = A + \lambda \vec{AB} = (1-6 \lambda ,4,2-5 \lambda)$ (do you understand why we can write it this way?)

    then $\vec{PP^\prime} = P^\prime - P =(3-6 \lambda ,6,-5 \lambda -1)$

    and we want

    $\vec{PP^\prime}\cdot \vec{AB} = 0$

    $\vec{PP^\prime}\cdot \vec{AB} = 61 \lambda -13$

    $\lambda = \dfrac{13}{61}$

    $P^\prime = (1-6 \lambda ,4,2-5 \lambda) =\left( -\dfrac{17}{61},4,\dfrac{57}{61}\right)$

    $d = | \overline{PP^\prime} | =\left |\left( -\dfrac{17}{61},4,\dfrac{57}{61}\right)\right| = 3 \sqrt{\dfrac{293}{61}}$
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    Re: Major trouble on vector calculus

    Quote Originally Posted by TheTideCaller View Post
    We barely even touched projection vectors (?) and I am completely lost on how to even do this problem because I have no idea what to do.

    The distance d of a point P to the line through points A and B is the length of the component of APŻŻŻŻŻŻŻŻ that is orthogonal to ABŻŻŻŻŻŻŻŻ, as indicated in the diagram. So the distance from P=(−2,−2,3) to the line through the points A=(1,4,−2) and B=(−5,4,−3) is ... ?
    You learn about vectors. That is way one does these questions.
    If $\ell(t)=A+tD$ is a line and $P\notin\ell$ then the distance from $P$ to $\ell$, $\mathscr{D}(P;\ell)=\dfrac{\|\overrightarrow {AP}\times D\|}{\|D\|}$
    SEE HERE

    In this case $\overrightarrow {AP} = \left\langle { - 3, - 6,5} \right\rangle~\&~D=\left\langle { - 1,0,-6} \right\rangle $

    Now you have at it.
    Last edited by Plato; Aug 31st 2017 at 08:13 PM.
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