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Thread: Vectors

  1. #1
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    Vectors

    The line r=(2i+3j-k) +t(-i+3j+k) passes through A (where t=1) and B (where t=5). Find the position vector of point C where AC:BA=-1:4
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  2. #2
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    Re: Vectors

    Did you not even try to do this yourself? What are A and B? BA is the distance between them. AC is the signed between this new point, C, and A. I would just write it as CA: BA= 1:4 C is 4 times as far from A as B is. Rather than calculating the direct line distance from A to B or A to C, by "similar triangles" it is sufficient to find C such that the difference of each coordinate satisfies that condition.
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    Re: Vectors

    Actually I did try, and I asked because I couldn't go any further. FYI, I did eventually get the answer.

    A: <1,6,0> B: <-3,18,4> BA= <4, -12,-4> AC:BA= -1:4 CA:BA=1:4 Dividing BA by 4, <1, -3, -1> CA=<1,6,0> - OC
    Therefore, OC=9j+k

    PS, please don't feel obliged to reply if it's such trouble
    Last edited by APK911; Oct 2nd 2017 at 06:05 AM.
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  4. #4
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    Re: Vectors

    Quote Originally Posted by APK911 View Post
    [/B]PS, please don't feel obliged to reply if it's such trouble
    The point is that if you show your work then we can help you better. Just giving you a full solution is not nearly as effective.

    -Dan
    Thanks from SlipEternal and skeeter
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  5. #5
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    Re: Vectors

    Yes, I understand, I'm sorry, i don't really use this site a lot, but I would have added working out, it's just that I was in a rush as I have an exam tomorrow, but thank you for putting it in a more polite way.
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