Results 1 to 5 of 5

Math Help - A Question with vectors

  1. #1
    Junior Member
    Joined
    Sep 2008
    Posts
    58

    A Question with vectors

    Sorry, I didn't know where to post this. But here is the question.

    A surveyor measures the location of point and determines that . He wants to determine the location of a point so that and . Assuming that the point lies to the east of the point , what are the (a) - and (b) - coordinates of point ?




    Any help would be great!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Here's one way to do it. Suppose the location of B is given by \mathbf{r}_{\textsf{\tiny OB}} = p\mathbf{i} + q\mathbf{j}. Then Pythagoras' theorem tells you that p^2+q^2=2400^2 and (p-297)^2 + (q-743)^2 = 1647^2.

    Multiply out the brackets in the second of those equations, subtract the first equation, and you'll get a simple linear relation between p and q. Solve this for q, substitute that into the equation p^2+q^2=2400^2, and you'll have a quadratic equation for p. That will have two solutions, and you want the larger one (because B has to lie to the east of A).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2008
    Posts
    58
    I can't seem to figure it out, it looks easy. But the numbers are getting to big and stuff. Is there an easier way to do it?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Sep 2008
    Posts
    58
    bump, anyone please?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    I agree, the numbers are not nice.

    We have the equations

    p^2+q^2=2400^2,
    p^2-594p+297^2 + q^2-1486q+743^2 = 1647^2.

    Subtract the first of these from the second, getting

    594p + 1486q = 2400^4+743^2+297^2-1647^2 = 3687649. (*)

    Now multiply both sides of the first equation by 1486^2:

    1486^2p^2 + (3687649-594p)^2 = 2400^2\times1486^2,

    which simplifies(?) to 2561032p^2 -4.380927\times10^9p + 8.7954619\times10^{11} = 0.

    Plug those coefficient numbers into the formula for solving a quadratic, and you'll find that the larger root is (approximately)

    p = \frac{4.380927\times10^9 + 3.1909775\times10^9}{5122064} \approx 1478.

    Having found p, substitute its value into (*) to get q\approx1891.

    On second thoughts, if I came across that problem in real life, I would solve it on a scale diagram, drawing a circle with a radius representing 2400m, centred at O, and one with radius repesenting 1647m, centred at A, and measuring the coordinates of the point(s) where the circles intersect. That would be quicker than doing the calculation, and would probably give you a sufficiently accurate answer for practical purposes.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: September 21st 2011, 07:21 PM
  2. Question regarding Vectors
    Posted in the Calculus Forum
    Replies: 2
    Last Post: July 24th 2009, 06:06 AM
  3. Another vectors question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: July 5th 2009, 01:23 PM
  4. vectors question
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: February 9th 2009, 07:15 AM
  5. Need help with vectors question
    Posted in the Calculus Forum
    Replies: 4
    Last Post: March 21st 2008, 07:45 AM

Search Tags


/mathhelpforum @mathhelpforum