Cross Product of an Array

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Hollis Williams
Hollis Williams 2019 年 4 月 22 日
コメント済み: James Tursa 2019 年 4 月 22 日
I have an array of size 3 x 100 and I basically want to create a new array where I take each column vector in the array and compute the cross product with the same vector each time to create another array of size 3 x 100, so in this case I take every vector and form the cross product with [0 0 1]'. What would be the easiest way of doing this?

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James Tursa
James Tursa 2019 年 4 月 22 日
編集済み: James Tursa 2019 年 4 月 22 日
M = your 3xN matrix
v = your 3x1 vector
result = cross(M,repmat(v,1,size(M,2)));
  10 件のコメント
Hollis Williams
Hollis Williams 2019 年 4 月 22 日
Yes, I was using the size() command but made a typo, it is working now. I have created three 1 x 100 arrays, is it possible to put these together into one 3 x 100 array? So if A,B,C are all 1 x 100 arrays, I would need
A
B
C
James Tursa
James Tursa 2019 年 4 月 22 日
result = [A;B;C];

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その他の回答 (1 件)

Matt J
Matt J 2019 年 4 月 22 日
編集済み: Matt J 2019 年 4 月 22 日
This way avoids repmatting, which may be desirable when N is large.
M = your 3xN matrix
v = your 3x1 vector
result=xprodmat(v)*M./vecnorm(M);
where
function A=xprodmat(a)
%Matrix representation of a cross product
%
% A=xprodmat(a)
%
%in:
%
% a: 3D vector
%
%out:
%
% A: a matrix such that A*b=cross(a,b)
if length(a)<3, error 'Input must be a vector of length 3'; end
ax=a(1);
ay=a(2);
az=a(3);
A=zeros(3);
A(2,1)=az; A(1,2)=-az;
A(3,1)=-ay; A(1,3)=ay;
A(3,2)=ax; A(2,3)=-ax;
end
  1 件のコメント
Hollis Williams
Hollis Williams 2019 年 4 月 22 日
I don't think repmatting is the thing which is most computationally costly in my code but I will bear this in mind to see if I need to save time later on, thanks a lot.

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