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Math Help - Vector question 2-

  1. #1
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    Vector question 2-

    The line L passes through the point A, whose position vector is i-j-5k and is parrallel to the vector i-j-4k. The line l2 passes through the point B, whose position vector is 2i-9j-14k, and is parrallel to the vector 2i+5j+6k. The point L and the point Q on L2 are such that PQ is perpendicular to both L and L2.

    a. Find the length of PQ
    b. Find a vector perpendicular to the plane which contains PQ and L2.
    c. Find the perpendicular distance from A to the plane.
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  2. #2
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    part a) only

    Quote Originally Posted by kingkaisai2 View Post
    The line L passes through the point A, whose position vector is i-j-5k and is parrallel to the vector i-j-4k. The line l2 passes through the point B, whose position vector is 2i-9j-14k, and is parrallel to the vector 2i+5j+6k. The point L and the point Q on L2 are such that PQ is perpendicular to both L and L2.

    a. Find the length of PQ...
    Hi, have a look at the attached image.

    I used the writing of vectors I'm used to. So maybe you have to transscript the vectors into the form you should use:
    Attached Thumbnails Attached Thumbnails Vector question 2--abstd_zwei_geraden.gif  
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  3. #3
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    If l_1 :A + tD\quad \& \quad l_2 :B + tE are two skew lines, then the minimum distance between them is \frac{{\left| {\left( {B - A} \right) \cdot \left( {D \times E} \right)} \right|}}{{\left\| {D \times E} \right\|}}. That is the length of the PQ in your question.

    The plane that contains l_2 and PQ is \left[ {D \times \left( {D \times E} \right)} \right] \cdot \left( { < x,y,z >  - D} \right) = 0.
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  4. #4
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    part b) only

    Quote Originally Posted by kingkaisai2 View Post
    ...
    b. Find a vector perpendicular to the plane which contains PQ and L2.
    ...
    I've attached an image with my calculations:
    Attached Thumbnails Attached Thumbnails Vector question 2--point_on_plane.gif  
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  5. #5
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    Hello, kingkaisai2!

    Here's my approach to part (a).
    . . But I'm sure there's a better way!


    The line L_1 passes through the point A(1,\text{-}1,\text{-}5)
    . . and is parallel to the vector \vec{u} \:=\:\langle1,\text{-}1,\text{-}4\rangle
    The line L_2 passes through the point B(2,\text{-}9,\text{-}14)
    . . and is parallel to the vector \vec{v}\:=\:\langle 2,5,6\rangle.
    The point P on L_1 and the point Q on L_2 are such that
    . . PQ \perp L_1 and PQ \perp L_2.

    (a) Find the length of PQ

    Line L_1 has parametric equations; . \begin{Bmatrix}x \,= \,1 + t \\ y \,= \,\text{-}1 - t \\ z \,= \,\text{-}5 - 4t\end{Bmatrix}

    Line L_2 has parametric equations: . \begin{Bmatrix}x\,=\,2 + 2u \\ y \,=\,\text{-}9 + 5u \\ z\,=\,\text{-}14 + 6y\end{Bmatrix}

    The vector perpendicular to both \vec{u} and \vec{v} is given by:
    . . \vec{n}\;=\;\begin{vmatrix}i & j & k \\ 1 & \text{-}1 & \text{-}4 \\ 2 & 5 & 6\end{vmatrix} \;=\;\langle14,-14,7\rangle \;=\;\langle2,-2,1\rangle


    We have: . P(1+t,\,\text{-}1-t,\,\text{-}5,-4t) and Q(2+2u,\,\text{-}9+5u,\,\text{-}14+6u)

    . . and we want: . \overrightarrow{PQ} \parallel \vec{n}

    So we have: . \overrightarrow{PQ} \;=\;\langle\,(2\!+\!2u) - (1\!+\!t),\;(\text{-}9\!+\!5u) - (\text{-}1\!-\!t),\;(\text{-}14\!+\! 6u) -(\text{-}5 - 4t)\,\rangle
    . . . . . . . . . . PQ\;=\;\langle 2u - t + 1,\;5u + t- 8,\;6u + 4t- 9\rangle

    \text{If }\overrightarrow{PQ} \parallel \vec{n}\text{, then: }\;\langle 2u-t+ 1,\:5u+t-8,\:6u + 4t=9\rangle \:=\:k\langle2,-2,1\rangle

    And we have: . \begin{Bmatrix}2u - t +1= 2k \\ 5u + t -8= \text{-}2k \\ 6u + 4t -9 = k\end{Bmatrix} \;\Rightarrow\;\begin{Bmatrix}2u - t - 2k = \text{-}1 \\ 5u + t + 2k = 8 \\ 6u + 4t - k = 9\end{Bmatrix}

    Solve the system of equations: . k = 1,\:t = 1,\:u = 1


    Hence: . P(2,-2,-9) and Q(4,-4,-8)

    Therefore: . \overline{PQ}\;=\;\sqrt{(4-2)^2 + (-4+2)^2 + (-8+9)^2} \;=\;\sqrt{4 + 4 + 1} \;=\;\sqrt{9}\;=\;\boxed{3}

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