Resultant vector from three different location and normal vector of a plane
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Hey, I have three vector in the xyz space.
How to compute the Resultant vector from these three vectors, they are
0.000326 -0.00018 -0.00017
0.000365 -0.00016 -0.00017
-0.00015 -0.00018 -0.00039
and their start coordinates are
2.53493 2.47587 2.47282
2.54089 2.48301 2.47585
2.4791 2.47583 2.46097
these start coordinates can form a plane, how to calculate the normal vector of this plane?
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Torsten
2022 年 9 月 30 日
v1 = [0.000326 -0.00018 -0.00017 ];
v2 = [0.000365 -0.00016 -0.00017 ];
v3 = [-0.00015 -0.00018 -0.00039 ];
% compute resultant
R = v1 + v2 + v3
P1 = [2.53493 2.47587 2.47282 ];
P2 = [2.54089 2.48301 2.47585 ];
P3 = [2.4791 2.47583 2.46097 ];
% compute normal
N = cross(P2-P1,P3-P1)/norm(cross(P2-P1,P3-P1))
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William Rose
2022 年 9 月 30 日
In case it is onf interest to you (and I think it is of interest, based on your other posts):
The resultant vector is not aligned with the normal.
v1 = [0.000326 -0.00018 -0.00017 ];
v2 = [0.000365 -0.00016 -0.00017 ];
v3 = [-0.00015 -0.00018 -0.00039 ];
R = v1 + v2 + v3; % resultant
P1 = [2.53493 2.47587 2.47282 ];
P2 = [2.54089 2.48301 2.47585 ];
P3 = [2.47910 2.47583 2.46097 ];
N = cross(P2-P1,P3-P1)/norm(cross(P2-P1,P3-P1)); % unit normal
The angle between them (in degrees) is
a=acos(dot(R,N)/(norm( R)*norm(N)))*180/pi
An angle a=180-130.6 is equally valid, since the direction of N depends on the side order chosen when computing the cross product, and that choice seems arbitrary.
If you are computing Work = force * displacement = pressure * area * displacement, as you said in a separate thread, you should use the dot product (inner product):
Pr=1; %pressure
A=0.5*norm(cross(P2-P1,P3-P1)); %area
W=Pr*A*dot(N,R) %work (up to a sign)
where Pr=pressure, A=area, N=unit normal, R=displacement. Pr and A are scalars, N and R are vectors. You will determine the sign for W based on your understanding of the physics.
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