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grouping vectors based on z value

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Oday Shahadh 2020-6-8

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关闭： MATLAB Answer Bot 2021-8-20

Hi guys,

If I have how array of vectors a(xi,to,zi) and b(xj,y2k,zj), I want to extract a new group based on the condition zi=zj. And the cross product the resulted vectors. Thanks

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Walter Roberson 2020-6-8

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For each value in intersect(xi, xj) you have a 2D array a(:,:,I) and b(:,:,J) where zi(I) == zj(J) . How do you want to take the cross-product of the 2D arrays? Also, since there are possibly multiple values that match zi(I) == zj(J), how do you want the resulting 2D arrays to be stored ?

Kevin Joshi 2020-6-8

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编辑：Walter Roberson 2020-6-8

Try, https://in.mathworks.com/help/matlab/ref/varfun.html

Oday Shahadh 2020-6-8

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1.JPG
2.JPG

Dear Walter,

you can see in the attached plots that there is a center point which I need to cross product it with the other points at the same plane (same z value), store it and quiver it.

thanks

Walter Roberson 2020-6-8

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You do not have vectors, you have points. Your diagram implies that you might be wanting one vector to be implied by the vertical projection to the x axes: is that what you want to do for all of the vectors?

I am unable to find any possible geometric interpretation for taking the cross-product of so many vectors together with each other. Especially since you have no absolute ordering, and cross-product is dependent on the angle but without ordering the angle is not meaningful.

Oday Shahadh 2020-6-8

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dear Walter,

they are all a position vectors, I did the cross product for just one point with its center as seen, the result is a vector as produced in the plot, pls check it.

as seen there is cylindrical layers with center point per each, what I need is to make a cross product between the center point of each layer with the related points of the same layer.

I know you are very smart Walter, Thanks

Oday Shahadh 2020-6-8

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在 MATLAB Online 中打开

clc
clear
close all
format long
x=[];
y=[];
z=[];
for L=-2:0.5:2
    for r=0:.5:2
        for theta=0:45:360;
            x=[x;r*cos(theta*pi/180)];
            y=[y;r*sin(theta*pi/180)];
            z=[z,L];
            L1=[x-x,x-x,z'];
        end
    end
end
lx=L1(:,1);ly=L1(:,2);lz=L1(:,3);
z=z';
p=[x,y,z];
b=cross(L1(1,:),p(1,:));
   
figure(1)
scatter3(x,y,z);hold on
plot3(lx,ly,lz,'r');hold on
scatter3(b(1),b(2),b(3));hold on

Walter Roberson 2020-6-8

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Why does your L1 not involve y?

Your L1 is just two rows of 0 and the third row is L values each repeated length(0:.5:2) * length(0:45:360) times. Why bother to go through that trouble with it? Why not just build it directly with zeros() and repelem() ?

Your cross() is guaranteed to be (0, 0, 0) each time.

Oday Shahadh 2020-6-8

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编辑：Oday Shahadh 2020-6-8

((Why bother to go through that trouble with it?)) Because I am not Walter Roberson.

to create position vector in cylindrical coordinates we need (L) the central position vectorline,(r) the radiuse for each point and(theta) the angle between any two ponits