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Geometry Help

  1. Jul 22, 2006 #1
    The vertices of a triangle are P(-3,1,2), Q(1,-3,-1) and R(3,-1,-1)

    -Verify that it is a right angled
    -Determine the area
    -Determine the perimeter
    -What are the coordinates of S(x,y,z) such that PQRS is a rectangle

    I'm not sure how to start to go about solving this problem. Could someone give me a nudge forward? =/

    Thank you.
  2. jcsd
  3. Jul 22, 2006 #2


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    Find the distance between P and Q, the distance between Q and R, and the distance between P and R. Do those three sides satisfy the Pythagorean equation?

    Once you know that this is a right triangle and you know the lengths of the sides, the area is (1/2)hb where h and b are the lengths of the two shorter sides (the longest side is, of course, the hypotenuse). Of course, knowing the lengths of the three sides makes it easy to find the perimeter.

    Once you decided which of P, Q, R is the right angle vertex, S is just its reflection in the hypotenuse. You might do that by finding the equation of the lines parallel to each side by passing through the other vertex. Where do those two lines intersect?
  4. Jul 22, 2006 #3
    They would intersect on the midpoint of the hypotenuse, right?
  5. Jul 23, 2006 #4
    1) Let PQ and QR be two vectors. If two vectors are perpendicular, the cross product of those two vectors will equal to 1.
    2) Find the length of PQ and QR. I think the area should be pretty easy to find.
    3) Again, self-explanatory. Use the pythagorean theorem to find the length of the remaining side.
    4) Now you have PQ and QR as two vectors. The addition of another point S give you two more, RS and SP (in terms of x,y,z). Since it's a rectangle, don't you think they should be perpendicular? Use the cross product to find the equations, and you're home free.
    Last edited: Jul 23, 2006
  6. Jul 23, 2006 #5


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    ?? Since they are drawn from ends of the hypotenuse, no, they can't intersect "on the midpoint of the hypotenuse"!
    Draw a picture! You have a right triangle. If you construct line segments at each (non-right) angle, parallel to the opposite leg, you will have opposite sides of a rectangle. The point where they intersect is precisely the point you wanted: "S(x,y,z) such that PQRS is a rectangle".
  7. Jul 23, 2006 #6


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    Not true. You're thinking of the dot product, which will equal 0.
  8. Jul 23, 2006 #7
    Woops. My bad.
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