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Math Help - Vector Product ( Area of a triangle )

  1. #1
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    Vector Product ( Area of a triangle )

    Q: A triangular shade cloth is to be suspended by 3 poles of different height as shown. The feet of the poles are positioned 10m from each other. Calculate the area of the shade cloth needed.

    http://i.imgur.com/FU9E3d8.png

    Could someone please draw in (or explain) the x,y,z axis so i can understand how to express AB--> and AC--> in i,j,k form.
    thanks.
    Last edited by rqyytrf; February 28th 2013 at 07:48 PM.
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    Re: Vector Product ( Area of a triangle )

    Well the big hint is that you've been given the position of the origin. Notice that point A is directly above it, so has coordinates (0, 0, 2.5).

    We can orient the triangle any way we like, so we may as well have the base along one of the x or y axes. Let's say it's axis x. The bottom triangle extends out horizontally 10m, and then another corner of your shade sail is 2m above that. So the coordinate of point C is (10, 0, 2).

    As for the co-ordinate of B, you'll need to use some trigonometry. Since the bottom triangle is equilateral, it can be split into two right-angle triangles by drawing in the perpendicular bisector of the base, to get the height of the triangle. Since the hypotenuse of the RAT is 10m, and the base of the RAT is 5m, by Pythagoras that gives the height of the triangle as \displaystyle \begin{align*} 5\sqrt{3}\textrm{ m} \end{align*}. Point B is positioned 3m above it, so the coordinate of B is \displaystyle \begin{align*} \left( 5, 5\sqrt{3}, 3 \right) \end{align*}.

    Now that you have the three co-ordinates, I would evaluate the area by evaluating the length of each side of your shade sail and applying Heron's Formula. I'm sure there are other methods too.
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    Re: Vector Product ( Area of a triangle )

    Thanks for replying although not quite what im looking for.

    The original diagram only had the shade and the 3 poles, so i added the origin, points(A,B,C) and the equilateral triangle by myself.

    The coordinate of OA = (0 2.5 0) and OB = (10 3 0) and OC = (5 2 8.66) ----------- OC = (10cos60i +2j + 10sin60k) according to my teacher.

    But why if the distance between each pole is 10m, shouldn't coordinate OC be (10 2 0) where k(z) is 0 like the others?

    That's what i currently don't understand
    Last edited by rqyytrf; February 28th 2013 at 08:14 PM.
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    Re: Vector Product ( Area of a triangle )

    Quote Originally Posted by rqyytrf View Post
    Thanks for replying although not quite what im looking for.

    The original diagram only had the shade and the 3 poles, so i added the origin, points(A,B,C) and the equilateral triangle by myself.

    The coordinate of OA = (0 2.5 0) and OB = (10 3 0) and OC = (5 2 8.66) ----------- OC = (10cos60i +2j + 10sin60k) according to my teacher.

    But why if the distance between each pole is 10m, shouldn't coordinate OC be (10 2 0) where k(z) is 0 like the others?
    No, it isn't. The length of (10, 2, 0) is \sqrt{104}, not 10.

    That's what i currently don't understand
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