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About a triangle

  1. Jun 11, 2006 #1
    About a triangle....

    Hey guys, this is my second posts here, last week was my first week of uni. I've done most of my Phys homework but I'm stuck with the last 3 questions

    1.Given three points A,B and C which have coordinates (1,2,3); (2,-1,5); (-3,4,1). Find the area of triangle which has these three pints at its corners.

    What I've tried is I drew these three points on a X,Y,Z plane and work out the area by finding the surface area by drawing each in 2D X,Y , X,Z and Y,Z. But that didn't seem to work.:tongue2:

    2.The three vectors shown below, have magnitudes a=3.00, b=4.00 and c=10.0. a) Calculate the x and y components of these vectors. b) Find the numbers p and q such that c= pa + qb.

    I have been doing research on google and Wiki for probably three hrs and i stil don't know anything about it.(I'm sorry, pls help, I'm just too dumb for it.:uhh: )

    3.Show that, the vectors A = 3i + (-2j) + k, B = i - 3j + 5k and C = 2i + j - 4k can form the rectangular triangle.

    Like, I want to know what to do first and what formulas to use to solve these questions. I got absolutely no ideas how to handle these question, from highschool I have learnt nothing about 3D coordinates things, only 2D..:frown:

    Thanks a lot guys.
     

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  3. Jun 11, 2006 #2
    Have you learnt vectors properly ?
    You must also have some knowledge of 3D geometry to attempt these questions , so just find some good textbook and start revising ( it's not that hard :D ). You will find that the above questions are basic problems, it only requires some effort on your part to go through the theory .
    If you still have doubts feel free to post .

    PS: For the last question, I think it must mean a right angled triangle .
    Oh, and welcome to PF !
     
  4. Jun 11, 2006 #3
    Thanks, I'll try to find some Textbooks.. I'm not to sure what is the last question suppose to b, between a right angle triangle and rectangular triangle I'll ask my teacher when I see him next but thanks alot anyway....
     
  5. Jun 11, 2006 #4

    Hootenanny

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    A rectangular triangle is an archaic name for a right angled triagle. It is given this name I believe, since you can make two identical right angled triangles from any rectangle.
     
  6. Jun 11, 2006 #5

    nrqed

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    That's indeed a very difficult way to do it!
    Have you learned about the cross product?
    Have you seen that the magnitude of the cross product gives the area of the paralleliped spanned by the two vectors? So that half that value is the area of the triangle spanned by the two vectors?
    Are you ok with calculating the x and y components? This simply uses the diagram and the definitions of sin and cos.
    For the second part, I am assuming that p and q must be integers? (otherwise, there is an infinite number of solutions). Here the numbers are so simple that one can see the solution just by trial and error very quickly. I am not sure if they want you to do it in a certain way.

    Patrick
     
  7. Jun 11, 2006 #6

    HallsofIvy

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    These are physics problems? Looks like your physics teacher is poaching on the math teacher's territory! :rolleyes:
    I see two ways to do this problem:
    (a) Find the lengths of the three line segments, a, b, c, then use "Hero's formula": [itex]A= \sqrt{s(s-a)(s-b)(s-c)}[/itex] where s is half the perimeter, (a+b+c)/2.
    (b) Find parametric equations for one side, say AB as well as it's length, the "base" of the triangle. Now find the equations for the line through C, perpendicular to that line, and solve the two equations simultaneously to find D, the point where they interesect. The length of CD is the height and the area is 1/2 height* base.

    [/quote]2.The three vectors shown below, have magnitudes a=3.00, b=4.00 and c=10.0. a) Calculate the x and y components of these vectors. b) Find the numbers p and q such that c= pa + qb.

    I have been doing research on google and Wiki for probably three hrs and i stil don't know anything about it.(I'm sorry, pls help, I'm just too dumb for it.:uhh: )[/quote]
    "The three vectors shown below"?? Where? Presumably, you are given some information about them that you can use together with the fact that, for a, [itex]x^2+ y^2= 9[/itex], for b,[itex]x^2+ y^2= 16[/itex] and, for c, [itex]x^2+ y^2= 100[/itex].

    I would have thought, as others said, that a "rectangular triangle" is probably a peculiar way of saying "right triangle. However, these three vectors do not form a right triangle- they don't satisfy the Pythagorean theorem. It is true that A and C are perpendicular- their dot product is 0- but then B is too short to form the hypotenuse.
     
  8. Jun 12, 2006 #7
    Hey thanks alot for ur help HallsofIvy, that really gave me alot of info to finish my HW(wel havn't yet), "The three vectors shown below" is in the picture @ the bottom of my post. Thanks alot again anyway.
     
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