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Math Help - Vector equation of a line and points of intersection

  1. #1
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    Vector equation of a line and points of intersection

    Dear forum members, I don't understand why I am getting the wrong answer

    A line goes through the points (1,-3,1) and (13,-12,4). Does it cross the x-axis? If so, at what point?

    My solution:

    I call the points A and P, respectively. I form the direction vector AP, which becomes ( -4 -5 -1) (sorry for the nonstandard vector representation, please pretend this is a column).

    The equation of the line is then

    r=(1,-3,1) +t(-4 -5 -1)

    I know that at the point where the line crosses the x-axis, the y and z -coordinates are zero.

    hence

    x= 1-4t
    y=-3-5t
    z=1-t

    y and z must equal zero, so I solve for t to see whether it will be the same for both. I get

    for y t=-3/5
    for z t=1

    Hence, I would conclude that the line does not cross the x-axis, but the answer in my book says it does. Please help me find, where I went wrong.

    Thank you. All help is much appreciated.
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  2. #2
    Junior Member
    Joined
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    Quote Originally Posted by Coach View Post
    Dear forum members, I don't understand why I am getting the wrong answer

    A line goes through the points (1,-3,1) and (13,-12,4). Does it cross the x-axis? If so, at what point?

    My solution:

    I call the points A and P, respectively. I form the direction vector AP, which becomes ( -4 -5 -1) (sorry for the nonstandard vector representation, please pretend this is a column).

    The equation of the line is then

    r=(1,-3,1) +t(-4 -5 -1)

    I know that at the point where the line crosses the x-axis, the y and z -coordinates are zero.

    hence

    x= 1-4t
    y=-3-5t
    z=1-t

    y and z must equal zero, so I solve for t to see whether it will be the same for both. I get

    for y t=-3/5
    for z t=1

    Hence, I would conclude that the line does not cross the x-axis, but the answer in my book says it does. Please help me find, where I went wrong.

    Thank you. All help is much appreciated.
    the DR's of line AP should be 12,-9,3
    hence the eqn. of line is r=(i-3j+k) +t(12i-9j+3k)
    x= 1-12t
    y=-3-9t
    z=1+3t
    on comparing y and z to zero we get t=-1/3
    so it crosses the x axis

    Last edited by sumit2009; March 5th 2009 at 09:55 AM. Reason: typing error
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  3. #3
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    Your mistake is \overrightarrow {AP}  = \left\langle {12, {\color{red}- 9},3} \right\rangle
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