
Saeed Ebrahimi writes,
I have a question and I would be very grateful if you could help me.
Consider that I have three vectors r_{1},r_{2},r_{3} which have a common initial point and their endpoints form a triangle. So this triangle consists of three triangles which are formed by the r_{1},r_{2},r_{3} and the edges of that triangle and their corresponding areas are A_{1},A_{2},A_{3} (for example: A_{1} is the area of triangle formed by r_{2},r_{3} and the vector of endpoints of r_{2},r_{3}).
Would you please prove this relationship for me:
(r_{1},r_{2},r_{3} are vectors)
A_{1}*r_{1} + A_{2}*r_{2} + A_{3}*r_{3}
= 0 (Zero Vector)
If you think of the areas A_{1}, A_{2}, and A_{3} as signed numbers, then the sum A_{1}+A_{2}+A_{3}=A, the area of the triangle formed by the three endpoints of the vectors r_{1},r_{2},r_{3}. (If the angles between pairs of vectors, measured counterclockwise, are all less than 180�, then all the areas are positive numbers anyway, and the areas A_{1}, A_{2}, and A_{3} are nonoverlapping.)
In the xy coordinate plane, the signed areas are given by
A_{1} = (x_{3} y_{2}  x_{2} y_{3})/2
A_{2} = (x_{1} y_{3}  x_{3} y_{1})/2
A_{3} = (x_{2} y_{1}  x_{1} y_{2})/2
The xcoordinate of A_{1} r_{1} + A_{2} r_{2} + A_{3} r_{3} is given by
x_{1} (x_{3} y_{2}  x_{2} y_{3})/2 +
x_{2} (x_{1} y_{3}  x_{3} y_{1})/2 + x_{3}
(x_{2} y_{1}  x_{1} y_{2})/2 =
(x_{1} x_{3} y_{2}  x_{1} x_{2} y_{3})/2
+ (x_{2} x_{1} y_{3}  x_{2} x_{3} y_{1})/2
+ (x_{3} x_{2} y_{1}  x_{3} x_{1} y_{2})/2
= 0
Similarly, the ycoordinate of A_{1} r_{1} + A_{2} r_{2} + A_{3} r_{3} can be calculated, and everything cancels, leaving 0.
Another way to think of it is using the crossproduct. In this case, if r_{1},
r_{2}, and r_{3} are confined to the xy plane, then
A_{1} = (1/2) r_{2} � r_{3} is a vector aligned with the
z axis whose magnitude is equal to the area of the triangle formed by r_{2}
and r_{3}.
A_{1} = (1/2) r_{2} � r_{3}
A_{2} = (1/2) r_{3} � r_{1}
A_{3} = (1/2) r_{1} � r_{2}
Then to get the product of the scalar area A_{1} and the vector r_{1}, we can take the crossproduct of r_{1} � A_{1}, which results in a vector in the xy plane of the right magnitude, but it's pointing perpendicular to r_{1}. To set it right, we cross it with the unit vector in the z direction, which I'll call Z.
Then Z � r_{1} � A_{1} is equal to A_{1} r_{1}.
So the vector sum A_{1} r_{1} + A_{2} r_{2} +
A_{3} r_{3} is given by
Z � r_{1} � A_{1} + Z � r_{2} � A_{2} + Z � r_{3}
� A_{3}, which is
(1/2) Z � (r_{1} � (r_{2} � r_{3}) + r_{2}
� (r_{3} � r_{1}) + r_{3} � (r_{1} � r_{2})
)
r_{1} � (r_{2} � r_{3}) + r_{2} � (r_{3} � r_{1}) + r_{3} � (r_{1} � r_{2}) = 0 by the Jacobi Identity, so this simplifies to
(1/2) Z � 0, which is the zero vector.
> Dear Mr. Graeme McRae
>
> Thanks a lot for your reply.
> As I realized, you have proved this problem only in the case that r_{1},r_{2},r_{3}
> are considered as three vectors on the xy plane but I need to know it for
> the general case which these three vectors are considered in the space. I
> mean r_{1},r_{2},r_{3} have x,y,z components.
> Is there any way to prove it for this case?
>
> I am looking forward hearing from you.
>
> Yours Sincerely,
> Saeed Ebrahimi
If the triangles are not restricted to a plane, then I am puzzled by your statement, "Consider that I have three vectors r_{1},r_{2},r_{3} which have a common initial point and their endpoints form a triangle. So this triangle consists of three triangles which are formed by the r_{1},r_{2},r_{3} and the edges of that triangle" In particular, the phrase "So this triangle consists of three triangles" seems to imply that all four triangles are in the same plane; otherwise how could some triangles "consist" of other triangles.
Even in a plane, the areas of the triangles need to be considered "signed" areas, because it is possible that the three triangles formed by the origin and a pair of vectors could overlap. The only way their sum can be construed as the area of the triangle formed by the endpoints of the three vectors is to treat the area as a signed quantity.
If the three triangles formed by the origin and the three vectors are not restricted to a plane, then the area still adds up, but now instead of a signed quantity, you must consider the area as a vector quantity. The length of the vector is the magnitude of the area, and the direction of the vector is normal to the plane of the triangle, pointing in a direction determined by the righthandrule and the order in which the two vectors are named.
I believe the vector crossproduct solution using the Jacobi identity is not restricted to the xy plane. However, to make the equation work, the area of each triangle needs to be treated as a vector as well, whose length is equal to the magnitude of the area, and whose direction is normal to the plane of the triangle. (In this case, there's no need to cross every vector with "Z", which I did only to rotate that vector back into the xy plane.)
Graeme
Triangle Area Using Vectors, part 1
Triangle Area using Determinant
Cross Product, which gives a number of identities.
Heron's formula for the area of a triangle, if all you know is the lengths of its sides.
The webmaster and author of this Math Help site is Graeme McRae.