solve command
1 次查看(过去 30 天)
显示 更早的评论
[x,y] = meshgrid(-3:1:3);
z = 1.5*x - 3;
mesh(x,y,z);
p = -1.5*pi;
q = -pi/2;
r = 0;
for i = 1:1:10
pause(0.1);
p = p + (0.0173*5);
q = q + (0.0173*5);
r = r + (0.0173*5);
a1 = 3*cos(p); b1 = 3*sin(p); c1 = -3;
a2 = 3*cos(q); b1 = 3*sin(q); c2 = -3;
a3 = 2*cos(r); b3 = 2*sin(r); c3 = 0;
syms a b c d
eq1 = 'a1*a + b1*b +c1*c -d =0';
eq2 = 'a2*a + b2*b + c2*c -d=0';
eq3 = 'a3*a + b3*b +c3*c -d =0';
[a,b,c] = solve(eq1,eq2,eq3,'a,b,c');
a = a/d;b = b/d; c = c/d;
disp(a);disp(b);disp(c);
z = -(a*x + b*y)/c;
mesh(x,y,z);
end
sir with the above code i am not able to get the interger value of a,b,c;the solution for this is
-(-b1*c2+b1*c3+b2*c1-b3*c1+b3*c2-b2*c3)/(-a1*b3*c2+a1*b2*c3+a3*b1*c2-b2*a3*c1+b3*a2*c1-a2*b1*c3)
(-a1*c2+a1*c3-a3*c1-a2*c3+a2*c1+a3*c2)/(-a1*b3*c2+a1*b2*c3+a3*b1*c2-b2*a3*c1+b3*a2*c1-a2*b1*c3)
(-b2*a3+a1*b2-a1*b3+b3*a2+a3*b1-a2*b1)/(-a1*b3*c2+a1*b2*c3+a3*b1*c2-b2*a3*c1+b3*a2*c1-a2*b1*c3)
so can anyone tell me how to access a1,b1,c1 in solve function.can use any other method for this thank you
0 个评论
采纳的回答
Alexander
2012-6-5
Seems to be the same problem as here: http://www.mathworks.de/matlabcentral/answers/40281-substitution-after-differentiation-easy
Try this:
a = double(subs(a)); b = double(subs(b)); c = double(subs(c));
Like here:
[x,y] = meshgrid(-3:1:3);
z = 1.5*x - 3;
mesh(x,y,z);
p = -1.5*pi;
q = -pi/2;
r = 0;
for i = 1:1:10
pause(0.1);
p = p + (0.0173*5);
q = q + (0.0173*5);
r = r + (0.0173*5);
a1 = 3*cos(p); b1 = 3*sin(p); c1 = -3;
a2 = 3*cos(q); b2 = 3*sin(q); c2 = -3;
a3 = 2*cos(r); b3 = 2*sin(r); c3 = 0;
syms a b c d
eq1 = 'a1*a + b1*b +c1*c -d =0';
eq2 = 'a2*a + b2*b + c2*c -d=0';
eq3 = 'a3*a + b3*b +c3*c -d =0';
[a,b,c] = solve(eq1,eq2,eq3,'a,b,c');
a = a/d;b = b/d; c = c/d;
a = double(subs(a)); b = double(subs(b)); c = double(subs(c));
disp(a);disp(b);disp(c);
z = -(a*x + b*y)/c;
mesh(x,y,z);
end
Please note that I renamed one of your b1 to b2.
0 个评论
更多回答(1 个)
Andrei Bobrov
2012-6-5
[x,y] = meshgrid(-3:1:3);
pqr = reshape(bsxfun(@plus,[-1.5;-.5;0]*pi,(1:10)*0.0173*5),3,1,[]);
abc = cat(2,cos(pqr),sin(pqr),bsxfun(@times,[-1;-1;0]*3,ones(1,1,10)));
o3 = ones(3,1);
for jj = 1:size(abc,3)
b = abc(:,:,jj)\o3;
mesh(x,y,-(b(1)*x + b(2)*y)/b(3));
pause(0.5);
end
0 个评论
另请参阅
类别
在 Help Center 和 File Exchange 中查找有关 Assumptions 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!