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Math Help - lines in 3 dimensional space

  1. #1
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    lines in 3 dimensional space

    i have 2 questions here which probably aren't as hard as i think but my teacher never taught me how to do it, so can anyone help?

    1) determine if the vector [1,-1,5] is perpendicular to [x,y,z] = [3,-1,5] + t [2,-3,-1]

    2) Determine if these two equations intersect
    [x,y,z] = [4,-3,2] + t [1,8,-3]
    [x,y,z] = [2,-19,8] + t [4,-5,-9]

    ...actually i got it.. i figured it out..thanks anyways
    Last edited by imthatgirl; May 26th 2008 at 03:27 PM.
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  2. #2
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    Hello, imthatgirl!

    1) Determine if the vector \vec{u} \:=\:[1,\text{-}1,5] is perpendicular
    to the line: . \vec{L} \:=\:[3,\text{-}1,5] + t[2,\text{-}3,\text{-}1]
    Two vectors are perpendicular if their dot product is zero.

    We have vector \vec{u} \:=\:[1,\text{-}1,5]
    The line has vector: . \vec{v} \:=\:[2,\text{-}3,\text{-}1]

    Then: . u\cdot v \:=\:[1,\text{-}1,5]\cdot[2,\text{-}3,\text{-}1] \:=\:(1)(2) + (\text{-}1)(\text{-}3) + (5)(\text{-}1) \:=\:2 + 3 - 5 \;=\;0

    Therefore: . \vec{u} \perp \vec{L}



    Determine if these two lines intersect.

    . . [x,y,z] \;= \;[4,\text{-}3,2] + t[1,8,\text{-}3]
    . . [x,y,z] \;=\; [2,\text{-}19,8] + u[4,\text{-}5,\text{-}9]

    We have: . \begin{Bmatrix}x & = & 4 + t \\ y &=&\text{-}3 + 8t \\ z &=& 2 - 3t\end{Bmatrix} \quad \begin{Bmatrix}x &=& 2 + 4u \\ y &=& \text{-}19 - 5u \\ z &=& 8 - 9u \end{Bmatrix} .(A)

    Equate components: . \begin{array}{cccccccc}4 + t &=& 2 + 4u & \Rightarrow & t - 4u &=& \text{-}2 &{\color{blue}[1]} \\ \text{-}3 + 8t &=& \text{-}19 - 5u & \Rightarrow & 8t + 5u &=&\text{-}16 & {\color{blue}[2]}\\ 2 - 3t &=& 8-9u & \Rightarrow & t - 3u &=& \text{-}2 & {\color{blue}[3]}\end{array}

    Subtract [1] from [3]: . u \:=\:0
    Substitute into [1]: . t \:=\:-2

    Substitute into (A) and we get: . (x,y,x) \;=\;(2,-19,8) . . . intersection!

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