Draw Triangle using quiver3 command in Matlab

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iam
iam 2018 年 2 月 28 日
コメント済み: Star Strider 2018 年 3 月 21 日
Following are the position vectors that form vertices of a right angled triangle, a=[2 3 4], b=[6 2 -5], c=[8 1 -4]. I want to plot triangle using quiver3 command in Matlab. Any help will be appreciated. Thanks.

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Star Strider
Star Strider 2018 年 2 月 28 日
編集済み: Star Strider 2018 年 2 月 28 日
Try this:
a=[2 3 4];
b=[6 2 -5];
c=[8 1 -4];
abc = [a; b; c; a];
dabc = diff([abc; a]);
figure(1)
plot3(abc(:,1), abc(:,2), abc(:,3), 'pg', 'MarkerSize',15, 'MArkerFaceColor','g')
hold on
quiver3(abc(:,1), abc(:,2), abc(:,3), dabc(:,1), dabc(:,2), dabc(:,3), 0, 'LineWidth',2)
hold off
grid on
xlabel('X')
ylabel('Y')
zlabel('Z')
view(150, 40)
Experiment to get the result you want. I do not know how well this will generalise to other shapes.
EDIT Improved efficiency of the ‘dabc’ calculation.
  6 件のコメント
Gavin Seddon
Gavin Seddon 2018 年 3 月 21 日
Hello Star strider how do I change the position of the arrowhead so it is at the hypotenuse terminus? Thank you. GS.
Star Strider
Star Strider 2018 年 3 月 21 日
@Gavin Seddon — I do not understand what you are asking.
I encourage you to experiment with the code to get the result you want. In the second plot (link) the points are labeled, so you can follow them more easily.
The code with the labeled vertices is:
a=[2 3 4];
b=[6 2 -5];
c=[8 1 -4];
abc = [a; b; c; a];
dabc = diff([abc; a]);
figure(1)
plot3(abc(:,1), abc(:,2), abc(:,3), 'pg', 'MarkerSize',15, 'MarkerFaceColor','g')
hold on
quiver3(abc(:,1), abc(:,2), abc(:,3), dabc(:,1), dabc(:,2), dabc(:,3), 0, 'LineWidth',2)
hold off
grid on
axis([0 10 0 5 -6 5])
text(abc(1:end-1,1), abc(1:end-1,2), abc(1:end-1,3), {'\bfa\rm','\bfb\rm','\bfc\rm'}, 'HorizontalAlignment','center', 'VerticalAlignment','bottom', 'Color','r', 'FontSize',14)
xlabel('X')
ylabel('Y')
zlabel('Z')

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