Effacer les filtres
Effacer les filtres

How to draw a perpendicular line to another?

8 vues (au cours des 30 derniers jours)
F Z
F Z le 23 Août 2013
Commenté : hadis ensafi le 14 Juin 2022
Hello,
Suppose i have a segment S1 defined by 2 points (x01,z01) and (x02, z02). The center of the segment is defines as (xm0, zm0). I want to plot the perpendicular line PL1 to S1 at (xm0, zm0) and find the intersection point of PL1 and the axis Y=0. The code is detailed below
a1=(z01-z02)/(x01-x02); %z1=a1*x1+b1 b1=z01-a1*x01; a2=-1/a1; % z2 perp to z1--> a2=-1/a1 b2=zm0-a2*xm0; %z2=a2*x2+b2
x=xm0:0.1:10; z2=a2*x+b2; xi=-b2/a2; %solve 0=a2*xi+b2 zi=a2*xi+b2;
h3=line([xm0,xi],[zm0,zi]); %trace du rayon principal
However, when i check the angle between my 2 perpendicular lines, it appears that it's not pi/2
O=(a2-a1)/(1-a1*a2); alpha=atan(O)*180/pi;
Any ideas please?
Fatzo

Réponse acceptée

Iain
Iain le 23 Août 2013
xm0 = (x01+x02)/2; ym0 = (y01+y02)/2
m = (y2 - y1) / (x2-x1);
PLm = tan( atan(m)+pi/2);
PLc = ym0 - PLm*xm0;
%PL line eqn = y_PL = PLm * x_PL + PLc
x_at_yequalto0 = -PLc/PLm;
  1 commentaire
hadis ensafi
hadis ensafi le 14 Juin 2022
Hello.. thanks for your code.. I have this problem in 3D space! how can I calculate a perpendicular of a line in 3D space and in Specified plane? thanks a lot..

Connectez-vous pour commenter.

Plus de réponses (0)

Catégories

En savoir plus sur Simulink PLC Coder dans Help Center et File Exchange

Community Treasure Hunt

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

Start Hunting!

Translated by