Translate vector to a center point

5 次查看（过去 30 天）
Alessandro Ruda
Alessandro Ruda 2022-3-3
评论： Jan 2022-3-4
Dear MatLab comunity,
I have an array of few thousands xyz coordinates of three points in a molecule: A,B and C.
I created two array of vectors:
AB = [(A(:,1)-B(:,1)), (A(:,2)-B(:,2)), (A(:,3)-B(:,3))];
BC = [(C(:,1)-B(:,1)), (C(:,2)-B(:,2)), (C(:,3)-B(:,3))];
The two vectors share one of the point (B). Now I want to translate the two vector - keeping of course the same orientation and magnitude - so that the origin is the center of mass , which in my case is:
COM = [3.9721615314483643 -8.11227798461914 -0.34564414620399475]
How can I do that?
Thank you in advance!
/alex

请先登录，再回答此问题。

回答（1 个）

Jan
Jan 2022-3-3
At first simplify
AB = [(A(:,1)-B(:,1)), (A(:,2)-B(:,2)), (A(:,3)-B(:,3))];
BC = [(C(:,1)-B(:,1)), (C(:,2)-B(:,2)), (C(:,3)-B(:,3))];
to:
AB = A - B;
BC = C - B;
Now I'm not sure if I understand, what you are asking for. Vectors can be seen as directions or location. Moving them while keepinmg the direction is not meaning ful. This works with lines, but not with vectors, which can be move in space freely.
But maybe all you want to do is:
AB - COM
BC - COM
  2 个评论
Alessandro Ruda
Alessandro Ruda 2022-3-3
Okay, not really. I haven't explained clearly.
So I have a molecule and I calculated the center of mass and the principal axis of inertia and these are:
COM = [3.9721615314483643 -8.11227798461914 -0.34564414620399475]
X_ax = [0.43753358721733093 0.719929575920105 -0.5387632250785828]
Y_ax = [0.5724487900733948 0.2390475869178772 0.7843204736709595]
Z_ax = [0.6934455633163452 -0.6515809297561646 -0.3075314462184906]
Then I have these array of points A(x,y,z), B(x,y,z) and C(x,y,z). Each of these array has few thousands of coordinates.
Now I want first to generate the vectors AB and BC and then calculate the angle between these and each of the principal axis separately.
I hope I have explained it clearly, what would be the best way to do it?
/alex

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Mathematics 的更多信息

标签

产品


版本

R2020b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by