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Math Help - Lines question

  1. #1
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    Lines question

    If anyone could explain how the following is done without a grapher, it would be greatly appreciated:

    Determine if the two lines A and B are parallel, skew, or intersecting. If they are intersecting, find the point at which they intersect.

    A: x = 13 + 12t, y = -7 + 20t, z = 11 - 28t
    B: x = 22 + 9s, y = 8 + 15s, z = -10 - 21s
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by meebo0129 View Post
    If anyone could explain how the following is done without a grapher, it would be greatly appreciated:

    Determine if the two lines A and B are parallel, skew, or intersecting. If they are intersecting, find the point at which they intersect.

    A: x = 13 + 12t, y = -7 + 20t, z = 11 - 28t
    B: x = 22 + 9s, y = 8 + 15s, z = -10 - 21s
    the direction vector to A is <12,20,-28>
    the direction vector to B is <9,15,-21>

    note that <12,20,-28> = (12/9)*<9,15,-21>

    since the direction vectors are proportional, the lines are parallel
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  3. #3
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    Quote Originally Posted by meebo0129 View Post
    If anyone could explain how the following is done without a grapher, it would be greatly appreciated:

    Determine if the two lines A and B are parallel, skew, or intersecting. If they are intersecting, find the point at which they intersect.

    A: x = 13 + 12t, y = -7 + 20t, z = 11 - 28t
    B: x = 22 + 9s, y = 8 + 15s, z = -10 - 21s
    These are in fact the same line. Put t=(3/4)(1+s) and substitute into the
    equations for the first line, and you should find that the equations become
    those for the second.

    RonL
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  4. #4
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    Hello, meebo0129!

    Determine if the two lines A and B are parallel, skew, or intersecting.
    If they are intersecting, find the point at which they intersect.

    A:\;\begin{array}{ccc}x &= &13 + 12t \\ y & = & \text{-}7 + 20t \\ z & = & 11 - 28t\end{array}

    B:\;\begin{array}{ccc}x & = & 22 + 9s \\ y & = & 8 + 15s \\ z & = & \text{-}10 - 21s\end{array}

    The direction vector of A is: . \langle 12,\,20,\,\text{-}28\rangle \:=\:\langle3,\,5,\,\text{-}7\rangle

    The direction vector of B is: . \langle9,\,15,\,\text{-}21\rangle \:=\:\langle3,\,5,\,\text{-}7\rangle

    . . Therefore: .  A \parallel B


    The point (22,\,8,\,-10) is on line B.
    . . Is it on line A?

    We have: . \begin{Bmatrix}13 + 12t & = & 22 \\ \text{-}7 + 20t & = & 8 \\ 11 - 28t & = & \text{-}10\end{Bmatrix}\quad\Rightarrow\quad t \,=\,\frac{3}{4} . . . . Yes!


    Therefore, lines A and B are coincident.

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