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Math Help - Vector - straight lines

  1. #1
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    Vector - straight lines

    The straight lines l_1 and 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 l_2 and PQ = PR.



    How could I do this?
    Last edited by geton; November 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 L has vector equation: . r \;=\;(i + 4j + 2k) + s(3i + j)

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

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

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


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

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

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


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

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


    Substitute into (a): . \begin{array}{c|c}<br />
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: . 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 PQ is: . \sqrt{(9-1)^2 + (9-4)^2 + (3-2)^2} \;=\; \sqrt{8^2+ 5^2 + 1^2} \:=\:\sqrt{90} .[1]


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

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

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


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

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


    Substitute into (a): . \begin{array}{c|c}<br />
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: . 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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