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Use vectors and the dot product to prove the midpoint

  1. Jan 31, 2015 #1
    1. The problem statement, all variables and given/known data

    Use vectors and the dot product to prove that the midpoint of the hypotenuse of a right triangle is equidistant to all three vertices.

    2. Relevant equations

    I know the dot product is A⋅B = |A||B|cosΘ ...... or .... A1B1 + A2B2 + A3B3 ... + AnBn

    I know the magnitude of a vector is its length, and is given by√(x^2 + y^2) I have omitted the z^2 since I want to solve this in R^2.

    3. The attempt at a solution

    I figure doing this in R^2 is easier, so I drew a right triangle with points P(0,0), R(0, y), S(x, 0)

    I know the midpoint would be given by T = (x/2, y/2)

    I constructed vectors using these points:

    PS = <x, 0>
    PR = <0, y>
    TS = <x/2, -y/2>
    RT = <x/2, -y/2>
    PT = <x/2, y/2>

    From this I can show that the magnitude of TS, RT, and PT are all the same, thus their distances are the same. I can also use this to show that TS = RT. I can then take the dot product of TS and RT and work it down to show the angle between them is 0 (so cosΘ = 1) and thus they are the same vector.

    What I cannot do for the life of me is use the dot product to prove that PT has the same magnitude as the other two vectors, TS and RT. I just can't figure it out. I've tried finding ways to express PT as other vectors but everything I've tried has failed. I am tearing my hair out!

    Here is a diagram of how I am trying to layout my triangle:

    5FKRlNH.png

    I would greatly appreciate any insight into this very troubling problem. :(
     
    Last edited: Jan 31, 2015
  2. jcsd
  3. Jan 31, 2015 #2

    Stephen Tashi

    User Avatar
    Science Advisor

    I don't know what the exercise considers "using the dot product". Computing the magnitude of a vector can be regarded as using the dot product. [itex] |A| = \sqrt{ A \cdot A} [/itex].

    There is a vector version of the law of cosines:

    If [itex] C = A + B [/itex] then [itex] |C|^2 = |A|^2 + |B|^2 + 2( A \cdot B) [/itex].

    If you set [itex] C = A - B [/itex] then it looks like the law of cosines from trigonometry. [itex] |C|^2 = |A|^2 + |B|^2 - 2( A \cdot B) [/itex].

    Whether it is acceptable to use the law of cosines for vectors depends on what's been covered in your course.
     
  4. Jan 31, 2015 #3
    Hey, thank you for your reply. I went about it a little differently than I originally posted and was able to prove it. Thanks for mentioning that a vector's magnitude is equal to the square root of that vector dotted by itself. Using that idea, I was able to prove that the distance to every vertex was the same. I completely forgot about that property, and it turns out it's quite useful, haha. :)
     
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