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Math Help - IBV23 find point P the intersection of 2 vector equations of lines

  1. #1
    Super Member bigwave's Avatar
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    IBV23 find point P the intersection of 2 vector equations of lines

    r_1 (5 \ 1)+\lambda (3 \  -2) and r_2 (-2 \ 2)+t (4 \ 1)

    presume one way to do this is turn these into line equations

    y=-\frac{2}{3}x+6

    y=\frac{1}{4}x+\frac{3}{2}

    then x=\frac{46}{3} and y=\frac{16}{3}
    Last edited by bigwave; August 29th 2013 at 05:30 PM. Reason: removed image
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  2. #2
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    Re: IBV23 find point P the intersection of 2 vector equations of lines

    Are r1 and r2 constants or parameters?
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    Re: IBV23 find point P the intersection of 2 vector equations of lines

    Hello, bigwave!

    Your equations are wrong . . .


    r_1\!:\: (5, 1)+\lambda (3 ,-2)\,\text{ and }\,r_2\!:\:(-2,2)+t (4,1)

    \text{Find }P\text{, their point of intersection.}

    \text{We have: }\:r_1\!:\:\begin{Bmatrix}x &=& 5 + 3\lambda \\ y &=& 1 - 2\lambda \end{Bmatrix} \qquad r_2\!:\:\begin{Bmatrix}x &=& \text{-}2+4t \\ y &=& 2 +t \end{Bmatrix}

    \text{Equate }x's\text{ and }y's: \:\begin{Bmatrix}5+3\lambda &=& \text{-}2+4t & \;\;\Rightarrow\;\; & 3\lambda - 4t &=& \text{-}7 \\ 1 - 2\lambda &=& 2 + t & \Rightarrow & 2\lambda +t &=& \text{-}1 \end{Bmatrix}

    \text{Solve the system of equations: }\:\lambda = \text{-}1,\;t = 1

    \text{Therefore: }\:P(2,3)
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  4. #4
    Super Member bigwave's Avatar
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    Re: IBV23 find point P the intersection of 2 vector equations of lines

    Thanks for help
    and it was r=
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