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Thread: Vector - straight lines

  1. #1
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    Vector - straight lines

    The straight lines $\displaystyle l_1$ and $\displaystyle l_2$ have vector equations r = (i + 4j + 2k) + t(8i + 5j + k) and r = (i + 4j + 2k) + s(3i + j) respectively. P is the point with coordinate (1, 4, 2) & Q is the point with coordinate (9, 9, 3).

    Find the possible coordinates of the point R, such that R lies on $\displaystyle l_2$ and PQ = PR.



    How could I do this?
    Last edited by geton; Nov 12th 2008 at 09:15 AM.
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  2. #2
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    Hello, geton!

    Are there more parts to this problem?
    There is too much information . . .


    The line $\displaystyle L$ has vector equation: .$\displaystyle r \;=\;(i + 4j + 2k) + s(3i + j)$

    $\displaystyle P$ is the point $\displaystyle (1, 4, 2)$ and $\displaystyle Q$ is the point $\displaystyle (9, 9, 3).$

    Find the possible coordinates of the point $\displaystyle R$ such that $\displaystyle R$ lies on $\displaystyle L$ and $\displaystyle PQ = PR.$

    Distance $\displaystyle PQ$ is: .$\displaystyle \sqrt{(9-1)^2 + (9-4)^2 + (3-2)^2} \;=\; \sqrt{8^2+ 5^2 + 1^2} \:=\:\sqrt{90}$ .[1]


    Line $\displaystyle L$ has parametric equations: .$\displaystyle \begin{array}{c}x \:=\:1 + 3s \\ y \:=\:4 + s \\ z \:=\:2 \end{array}$ .(a)

    Point $\displaystyle R$ has coordinates: .$\displaystyle (1+3s,\:4+s,\:2)$

    Distance $\displaystyle QR$ is: .$\displaystyle \sqrt{(1+3s-4)^2 + (4+s-9)^2 + (2-3)^2} \;=\;\sqrt{10s^2 - 58s + 90} $ .[2]


    Equate [2] and [1]: .$\displaystyle \sqrt{10s^2 - 58s + 90} \;=\;\sqrt{90} \quad\Rightarrow\quad 10s^2 - 58s + 90 \;=\;90$

    . . $\displaystyle 10s^2 - 58s \:=\:0 \quad\Rightarrow\quad 2s(5s-29) \:=\:0 \quad\Rightarrow\quad s \;=\;0,\:\tfrac{29}{5}$


    Substitute into (a): . $\displaystyle \begin{array}{c|c}
    s=0 & s = \frac{29}{5} \\ \\[-4mm] \hline \\[-3mm] x = 1 & x = \frac{92}{5} \\ \\[-3mm] y = 4 & y = \frac{49}{5} \\ \\[-4mm] z = 2 & z = 2 \end{array}$


    Therefore: . $\displaystyle R \;=\;(1,4,2)\:\text{ or }\:\left(\tfrac{92}{5},\:\tfrac{49}{5},\:2\right) $

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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, geton!

    Are there more parts to this problem?
    There is too much information . . .
    Yes, I tried to short the question. But ...


    Distance $\displaystyle PQ$ is: .$\displaystyle \sqrt{(9-1)^2 + (9-4)^2 + (3-2)^2} \;=\; \sqrt{8^2+ 5^2 + 1^2} \:=\:\sqrt{90}$ .[1]


    Line $\displaystyle L$ has parametric equations: .$\displaystyle \begin{array}{c}x \:=\:1 + 3s \\ y \:=\:4 + s \\ z \:=\:2 \end{array}$ .(a)

    Point $\displaystyle R$ has coordinates: .$\displaystyle (1+3s,\:4+s,\:2)$

    Distance $\displaystyle QR$ is: .$\displaystyle \sqrt{(1+3s-4)^2 + (4+s-9)^2 + (2-3)^2} \;=\;\sqrt{10s^2 - 58s + 90} $ .[2]


    Equate [2] and [1]: .$\displaystyle \sqrt{10s^2 - 58s + 90} \;=\;\sqrt{90} \quad\Rightarrow\quad 10s^2 - 58s + 90 \;=\;90$

    . . $\displaystyle 10s^2 - 58s \:=\:0 \quad\Rightarrow\quad 2s(5s-29) \:=\:0 \quad\Rightarrow\quad s \;=\;0,\:\tfrac{29}{5}$


    Substitute into (a): . $\displaystyle \begin{array}{c|c}
    s=0 & s = \frac{29}{5} \\ \\[-4mm] \hline \\[-3mm] x = 1 & x = \frac{92}{5} \\ \\[-3mm] y = 4 & y = \frac{49}{5} \\ \\[-4mm] z = 2 & z = 2 \end{array}$


    Therefore: . $\displaystyle R \;=\;(1,4,2)\:\text{ or }\:\left(\tfrac{92}{5},\:\tfrac{49}{5},\:2\right) $

    Thank you so much for giving the method to do. But the distance QR will be PR.

    Thanks a lot.
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