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goldmahkot

  • 3 years ago

I was comparing the areas of a triangle that could be obtained via the conventional method, dot product, and cross product. Strangely, the latter did not agree with the former two. Consider two vectors, A <1,1,1> and B<1,1,0>. Both these vectors form a right angled triangle with an angle of 45deg between both vectors, a hypotenuse of SQRT(3), and length of SQRT(2).

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  1. goldmahkot
    • 3 years ago
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    Now, the conventional method for finding the area of a triangle is (height X length)/2. In this case, it would be (SQRT(3)(sin45deg) X SQRT(2))/2. The answer obtained is 0.866.

  2. goldmahkot
    • 3 years ago
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    For the dot product, remember that we could find the area of a triangle with a "shortcut method" i.e. (|A'||B|sin(x))/2. Therefore, the area would be (SQRT(3)(sin45deg) X SQRT(2))/2 which gives an answer of 0.866.

  3. goldmahkot
    • 3 years ago
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    But when it comes to the cross product, things get a little strange.\[A \times B = <-1,1,0>\]To find its magnitude (which is equals to the area of a parallelogram): \[\sqrt{(-1)^{2}+(1^{2}) + 0^{2}} = \sqrt{2}\]To find the area of the triangle, we divide the answer above by 2: \[\frac{ \sqrt2 }{2} = 0.7071\] The area obtained using the cross product is different from the other two methods!

  4. Waynex
    • 3 years ago
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    The area you calculated via the cross product is the correct one. The problem you have on the first two is that the angle between A and B is not pi/4. The angle would be pi/4 if both vectors were in the xy plane, but one vector is shift upward in angle, and that changes the angle between them. The angle is actual arccos[2/sqrt(6)], or aprox. 0.61 radians. The first equation would be sqrt(3) * sqrt(2) * sin(0.615...) * 0.5 = 0.707....

  5. goldmahkot
    • 3 years ago
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    Waynex, Thanks for you input on my question. You were right; I was assuming the angle between A and B to be pi/4 (a HUGE mistake). After recalculating the equations, all three areas agree with each other. I can now sleep in peace!

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