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Use vectors to form a right triangle on a circle

  1. Oct 10, 2013 #1
    1. The problem statement, all variables and given/known data
    Use vectors to demonstrate that on a circle any two diametrically opposed points along with an arbitrary third point(on the circle) form a right triangle


    2. Relevant equations

    Hint: assume without a loss of generality that the circle is centered at the origin and let v, -v, and w denote the three points in question. show that the vector connecting w to -v is orthogonal to the vector connecting w to v

    3. The attempt at a solution

    i think i have a grasp on how to achieve this. to show that they are orthogonal the dot product of the two vectors must be 0.

    I am confused with the v and -v. In my mind that makes a straight line for example say that v is (1,0) then -v would be (-1,0).

    I dont see how those two points along with a third make a right triangle.
     
  2. jcsd
  3. Oct 10, 2013 #2

    Mark44

    Staff: Mentor

    That's the object of this exercise. Take any arbitrary point w on the circle, and form displacement vectors between it and v and -v. What are the coordinates of w?
     
  4. Oct 10, 2013 #3
    can w be (1,1)?
     
  5. Oct 10, 2013 #4

    Mark44

    Staff: Mentor

    No. That point is not on the unit circle, which I assume is the one you're working with since you are using the points (1, 0) and (-1, 0). You need the equation of whatever circle you're using so that you can write the coordinates of your third point on the circle.
     
  6. Oct 10, 2013 #5
    my apologies, lets make w (0,1). so then I see that if you connect the lines you get a right triangle. so then to show that I must show that wv is orthogonal to w-v.

    so for wv it would be v - w = <1,-1>
    and w-v would be -v - w = <-1,-1>

    then i take the dot product and if its 0 theyre orthogonal meaning they do form a right triangle, is that right?
     
  7. Oct 10, 2013 #6

    LCKurtz

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    But the problem says that for any point v and its opposite on the circle, and any third point w that the vectors are perpendicular. You can't just do it for (1,0),(-1,0), and (0,1).
     
  8. Oct 10, 2013 #7
    could it just be with general terms?
    v = (x1,y1)
    -v = (-x1,-y1)
    w = (x2,y2)
     
  9. Oct 10, 2013 #8

    LCKurtz

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    Yes. What do you get when you use those. Remember they are on the circle.
     
  10. Oct 10, 2013 #9
    So for wv = v-w = (x1-x2, y1-y2)
    and for w*-v = -v - w = (-x1-x1, -y1-y2)

    after doing the dot product i end up with

    -x12+ x22 - y12 + y22

    but then thats not 0
     
  11. Oct 10, 2013 #10

    Mark44

    Staff: Mentor

    It should be (-x1-x2, -y1-y2)
    You're not using the fact that all three points are on a circle. What's the equation of your circle?
     
  12. Oct 10, 2013 #11
    sorry that was a typo, i copy and paste to make the sub and sup easier

    the equation of the circle is x2 + y2 = r2 i believe
     
  13. Oct 10, 2013 #12

    Mark44

    Staff: Mentor

    OK, so what can you say about the expression you ended up with for the dot product?
     
  14. Oct 10, 2013 #13
    well if x1 = 1 and x2 = 1 then that would cancel out and same for y but im not sure if that would work in the general sense.
     
  15. Oct 10, 2013 #14

    Mark44

    Staff: Mentor

    No, that won't work. Your point w (x2, y2) is a point on the circle whose equation is x2 + y2 = r2, right? So are v and what you're calling -v.

    What characteristic do points on this circle have that all other points don't have?
     
  16. Oct 10, 2013 #15
    all the points on the circle will have the same radius
     
  17. Oct 10, 2013 #16

    Mark44

    Staff: Mentor

    That's true, but irrelevant. The radius of a point is 0.

    Let me see if I can make this simpler. Consider the line whose equation is 3x + 2y = 6. Is the point (2, 1) on this line? Why or why not? Is the point (0, 3) on this line? Why or why not?
     
  18. Oct 10, 2013 #17
    (2,1) isnt because when you plug it into the equation it doesnt equal 6, (0,3) is because it does equal 6. so then if they are all on the same circle they would be the same when plugged into the equation, is that what youre hinting at?
     
  19. Oct 10, 2013 #18

    Mark44

    Staff: Mentor

    Yes. Any point on the line has to satisfy the equation, and any solution (a pair of numbers) to the equation represents a point on the line. Same for points on the circle.

    With that in mind, what does that say about the value you got for your dot product back in post #9?
     
  20. Oct 10, 2013 #19
    im not entirely sure, im a little thrown off because of the x1 and x2 and y1 and y2
     
  21. Oct 10, 2013 #20

    Mark44

    Staff: Mentor

    Then maybe you're not able to work this problem...
     
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