Cross section plot for 2D plot

I have a code to plot the quiver data.

I want to plot a 1D line plot across the 2D plot to see how the value of quiver changes from left to right alone a line.

Here is the code:

function[d]=DOMAINGENERATION_assocwithDNAPL(alpha, beta, kappa,Yo,CTD,vo)

fid = fopen('domain¶meter.dat');  % open file that will contain grid coordinate system and associated parameters

%   for i = 1:5:45;
     i=1;
% fh=figure;
    m=4+(i-1)*11844;
    n=11843+(i-1)*11844;
    M=dlmread('3d2pct.plt',',',m,0, [m 0 n 9]); %3d2pct.plt  data file

      xarray=M(:,1);
      yarray=M(:,2);
      DV_XX=M(:,8);
      DV_YY=M(:,9);
      FRONT=M(:,3);

      SF=reshape(FRONT,80,148);
      X=reshape(xarray,80,148);
      Y=reshape(yarray,80,148);

% figure;

     v = [1 0.95:1.00];
     contour(X(1,:),Y(:,1),SF,1);
     colormap([0 0 0]);
 hold on 
     quiver(xarray(1:4:end),yarray(1:4:end),DV_XX(1:4:end),DV_YY(1:4:end),1, 'k','filled');  
  hold on 
              xlabel('(m)');
              ylabel('(m)');
                  set(gcf,'color',[1 1 1]);
                  set(gca,'color','none');
              axis equal;
  %*************** Plot area ***********************************************
              axis([0 0.37 0 0.20])
  %*************************************************************************     
   end

Answer 1

[Babak]

Do you need to find the intersection of a 1D line and a curve that has been plotted in a 2D plane?

Did I understand your problem correctly?

The intersection will be a whole bunch of points if I understood you correctly.

[Tan Has]

yes for question. you got it. Could you help me

[Babak]

OK so here's my answer:

You need to have a boundry [xmin xmax] and [ymin ymax] that you are interested in finding the intersection of a line and your curve. Now, assume that the curve is given by its points as X=[x1 x2 x3...] and Y=[y1 y2 y3 ...] and the line you want to intersect it with is given by an equation y=a*x+b, the intersection includes points like (xj,yj) that satisfies yj=a*xj+b and if you find xj in X, it may be located somewhere between x(i) and x(i+1). The corresponding Y for x(i) and x(i+1) are y(i) and y(i+1). If y(i) and y(i+1) are close enough to yj and closer than y(i-1) and y(i+2) to yj then you can conclude that (xj,yj) is an intersection point.