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Homework Help: Calc III, Determing the coordinates of P

  1. Sep 12, 2005 #1
    Hello everyone. I'm stuck on the last part of this problem. It says, Draw two distinct nonzero postion vectors a = <x1,y1> and b = <x2,y2> such that the angle between them is less than pi/2. Draw the line L perpendicular with the line determined by the bector a such that l passes through the terminal point of b. Let P = (s,t) be the point where L intersects a. Determine the length of the line segment connecting the origin to the point P. Determine the coordinates of P. I drew a picture and I found the L = mx+b. The line determined by a = y1/x1 * x. m = -x1/y1 through (x2,y2). So now I have 2 lines, I need to find the intercept of these 2 lines to determine the coordinates of P (s,t). Any ideas on how i can do this? I'm also confused on what my 2 lines are exactly. Thanks.
    Last edited by a moderator: Apr 21, 2017
  2. jcsd
  3. Sep 12, 2005 #2


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    I missed your response on the other thread, sorry.

    You can think of the problem as one in vectors or one in coordinate geometry.

    OA is a vector starting at the origin O and ending at the point A. The vector OA has the position vector a<x1,y1>.
    OB is a vector starting at the origin O and ending at the point B. The vector OB has the position vector b<x2,y2>.
    OP is a vector starting at the origin O and ending at the point P. The vector OP has the position vector p<s,t>.


    treat it as a problem in coordinate geometry.
    With an origin O and the points A(x1,y1), B(x2,y2), P(s,t).
    Let m be the slope of OA.

    You have already worked this out as m = (y1 - 0)/(x1 - 0) = y1/x1

    So, the slope of the line OA is m = y1/x1
    Then the eqn of the line OA is y = mx, or
    y = (y1/x1)x ---------------------------------------------------(1)

    BP is perpindicular to OA, so you can write the slope of BP in terms of m.
    So, slope of BP = m' = -1/m = -x1/y1, which you have.
    Since you have the coords of B as (x2,y2), and the slope of BP, then you can write the eqn of BP as,
    (y - y2)/(x - x2) = m'
    y = m'(x-x2) + y2
    y = -(x1/y1)(x-x2) + y2 ----------------------------------------(2)

    Eqn (1) is the eqn of the line OA.
    Eqn (2) is the eqn of the line BP.

    You can find out where these lines intersect by equating (1) with (2) and solving for x and y, which will be s and t respectively, the coords of P.
  4. Sep 13, 2005 #3
    Thanks alot for your explanation,it was very helpful! :smile: but i'm stuck I think. When you said set equation 1 and 2 together and solve for x and y, did you mean x1 and y1 or did u mean x and y? When I set the two equations eqqual to eachother, y isn't there, y1 is though. I tried solving for x and got a really messy equation. Here is my work:
    http://img143.imageshack.us/img143/5492/gsfdg5pa.jpg [Broken]
    Last edited by a moderator: May 2, 2017
  5. Sep 13, 2005 #4


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    You're almost there.

    Yes, you have to solve for x and y. You will then get expressions for x and y in terms of x1, y1, x2, y2. The final expression you got can be rearranged to give,

    [tex]x = x_1\frac{(x_1x_2 + y_1y_2)}{(x_1^2 + y_1^2)}[/tex]

    Substitute this value for x into eqn (1) and get the value of y.

    These values of x and y are the x- and y-coords of the intersection of the line-eqns (1) and (2). And these coords are the coords of the point P. ie x = s and y = t.
    Last edited by a moderator: May 2, 2017
  6. Sep 13, 2005 #5
    Awesome, thanks so much for the help!! :) I was wondering how u got [tex]x = x_1\frac{(x_1x_2 + y_1y_2)}{(x_1^2 + y_1^2)}[/tex] I posted how far I got with rearranging but I didn't see how you did that exactly. http://img143.imageshack.us/img143/650/pointsss7gn.jpg [Broken]
    Last edited by a moderator: May 2, 2017
  7. Sep 13, 2005 #6


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    Staring from here,

    [tex]\left (\frac{y_1}{x_1}\right )x + \frac{x_1(x)}{y_1} = \frac{x_1x_2}{y_1} + y_2[/tex]

    [tex]x\left (\frac{y_1^2 + x_1^2}{x_1y_1}\right ) =\frac{x_1x_2 + y_1y_2}{y_1}[/tex]

    [tex]x = x_1\left (\frac{x_1x_2 + y_1y_2}{x_1^2 + y_1^2}\right )[/tex]
    Last edited by a moderator: May 2, 2017
  8. Sep 13, 2005 #7
    awesome, thanks alot for the help! :biggrin:
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