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Thread: Shortest distance between a point and a line - Vectors

  1. #1
    Junior Member Fnus's Avatar
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    Shortest distance between a point and a line - Vectors

    I'd just like it if someone could tell me where it is I'm going wrong in my calculations.. It might even be at the very beginning, I'm at a loss. ):

    'Find the distance from the point P(3,0,-1) to the line with equation $\displaystyle r=2i-j+4k+\lambda(3i+2j+k)$'

    Any point on the line I got to be $\displaystyle A(2+3\lambda,-1+2\lambda,4+\lambda)$.

    That means that $\displaystyle \overline{PA}=\left(\begin{array}{cc}-1+3\lambda\\-1+2\lambda\\5+\lambda\end{array}\right)$, and the direction vector of the line is $\displaystyle \overline{b}=\left(\begin{array}{cc}3\\2\\1\end{ar ray}\right)$

    Then I tried to calculate their scalar product to find when they are perpendicular and thus closest to each other, but then I got $\displaystyle \lambda$ to be 0...

    So would someone please tell me where it is I go wrong?
    Is it something at the very beginning? ):

    And then another question that is very simple!
    What is the direction vector of the line 3x-y=4?
    A fixed point would be (0,-4), right?

    Thanks in advance!
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  2. #2
    MHF Contributor

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    Suppose that $\displaystyle P$ is a point not on the line $\displaystyle \ell :Q + \lambda\overrightarrow d $.
    Then distance $\displaystyle D(\ell ,P) = \frac{{\left\| {\overrightarrow {QP} \times \overrightarrow d } \right\|}}{{\left\| {\overrightarrow d } \right\|}}$.
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