Sym 1x1 in Matrix

syms th1(t) th2(t) th3(t) th1_d(t) th2_d(t) th3_d(t) a1 a2 a3 m1 m2 m3 g

P2 = [a2*cos(th1)*cos(th2)/2; a2*sin(th1)*cos(th2)/2; a2*sin(th2)/2+a1];

P3 = [a2*cos(th1(t))*cos(th2(t)) + (a3*cos(th1(t))*cos(th2(t))*cos(th3(t)))/2 - (a3*cos(th1(t))*sin(th2(t))*sin(th3(t)))/2;...
    a2*cos(th2(t))*sin(th1(t)) + (a3*cos(th2(t))*cos(th3(t))*sin(th1(t)))/2 - (a3*sin(th1(t))*sin(th2(t))*sin(th3(t)))/2;...
    a1 + a2*sin(th2(t)) + (a3*cos(th2(t))*sin(th3(t)))/2 + (a3*cos(th3(t))*sin(th2(t)))/2];

Hi. I'm trying to get P3 in sym matrix, however, it appears 1x1 always.

However, P2 appears 3x1 sym matrix.

I have no idea what's the diffrence between them and how can I convert it to 3x1 matrix?

[Jaehyun Park]

Suprisingly, It's now converted that P2 is sym 1x1 and P3 is sym 3x1.....

Answer 1

[KALYAN ACHARJYA]

That .... dot dot represents the continuation in same line

Continue Long Statements on Multiple Lines

Answer 2

[Paul]

syms th1(t) th2(t) th3(t) th1_d(t) th2_d(t) th3_d(t) a1 a2 a3 m1 m2 m3 g
P2 = [a2*cos(th1)*cos(th2)/2; a2*sin(th1)*cos(th2)/2; a2*sin(th2)/2+a1];
P3 = [a2*cos(th1(t))*cos(th2(t)) + (a3*cos(th1(t))*cos(th2(t))*cos(th3(t)))/2 - (a3*cos(th1(t))*sin(th2(t))*sin(th3(t)))/2;...
    a2*cos(th2(t))*sin(th1(t)) + (a3*cos(th2(t))*cos(th3(t))*sin(th1(t)))/2 - (a3*sin(th1(t))*sin(th2(t))*sin(th3(t)))/2;...
    a1 + a2*sin(th2(t)) + (a3*cos(th2(t))*sin(th3(t)))/2 + (a3*cos(th3(t))*sin(th2(t)))/2];

P2

P2(t) = 

P3

P3 = 

whos P2 P3

  Name      Size            Bytes  Class     Attributes

  P2        1x1                 8  symfun              
  P3        3x1                 8  sym                 

P2 is a symfun but P3 is a sym. P2 does return a 3x1 for a specified value of t:

P2(1)

ans = 

I'm not sure why P3 is not also a symfun. The only difference I see is that the creation of P3 uses th1(t), etc. as explicit functions of t and P2 does not. I didn't think that should matter, but maybe it does. Let's try:

syms th1(t) th2(t) th3(t) th1_d(t) th2_d(t) th3_d(t) a1 a2 a3 m1 m2 m3 g
P2 = [a2*cos(th1(t))*cos(th2(t))/2; a2*sin(th1(t))*cos(th2(t))/2; a2*sin(th2(t))/2+a1];
whos P2

  Name      Size            Bytes  Class    Attributes

  P2        3x1                 8  sym                

So it does matter. If you want P2 and P3 to both be 1x1 symfun objects that are vector valued functions of a scalar variable, then don't use the explicit th1(t), etc. in their defintions. Or if you want both to be 3x1 sym objects, then do use the explicit th1(t) in their definitions. If P2 and P3 should be symfun, it would be clearer to use

P2(t) = ....

P3(t) = ....

in which case it won't matter if the elements on the rhs are shown as explciit functions of t.