MATLAB: How can we solve only for t1 variable in the equation P=R1 which has multiple variables such as t2 t3. Which is possible mathematically.

how can we solve only for t1 variable in the equation p=r1 which has multiple variables such as t2 t3 as well. which is possible mathematically.

R1 = 
[                                                                               cos(t2)*cos(t3) - sin(t2)*sin(t3),                                                                             - cos(t2)*sin(t3) - cos(t3)*sin(t2),                                    0,                                                                           10*cos(t2)]
[ (7344817515977723*cos(t2)*sin(t3))/9223372036854775808 + (7344817515977723*cos(t3)*sin(t2))/9223372036854775808, (7344817515977723*cos(t2)*cos(t3))/9223372036854775808 - (7344817515977723*sin(t2)*sin(t3))/9223372036854775808,    9007196398844849/9007199254740992, (36724087579888615*sin(t2))/4611686018427387904 + 45035981994224245/4503599627370496]
[     - (9007196398844849*cos(t2)*sin(t3))/9007199254740992 - (9007196398844849*cos(t3)*sin(t2))/9007199254740992,       (9007196398844849*sin(t2)*sin(t3))/9007199254740992 - (9007196398844849*cos(t2)*cos(t3))/9007199254740992, 7344817515977723/9223372036854775808, 36724087579888615/4611686018427387904 - (45035981994224245*sin(t2))/4503599627370496]
[                                                                                                               0,                                                                                                               0,                                    0,                                                                                    1]
P =
[  (433*cos(t1))/500, -cos(t1)/2, sin(t1), 0]
[ -(433*sin(t1))/500,  sin(t1)/2, cos(t1), 0]
[               -1/2,    433/500,       0, 0]
[                  0,          0,       0, 1]

Best Answer

With respect to
syms t1 t2 t3
R1 = [
    cos(t2)*cos(t3) - sin(t2)*sin(t3),  - cos(t2)*sin(t3) - cos(t3)*sin(t2), 0, 10*cos(t2);
    (7344817515977723*cos(t2)*sin(t3))/9223372036854775808 + (7344817515977723*cos(t3)*sin(t2))/9223372036854775808, (7344817515977723*cos(t2)*cos(t3))/9223372036854775808 - (7344817515977723*sin(t2)*sin(t3))/9223372036854775808, 9007196398844849/9007199254740992, (36724087579888615*sin(t2))/4611686018427387904 + 45035981994224245/4503599627370496;
     - (9007196398844849*cos(t2)*sin(t3))/9007199254740992 - (9007196398844849*cos(t3)*sin(t2))/9007199254740992,       (9007196398844849*sin(t2)*sin(t3))/9007199254740992 - (9007196398844849*cos(t2)*cos(t3))/9007199254740992, 7344817515977723/9223372036854775808, 36724087579888615/4611686018427387904 - (45035981994224245*sin(t2))/4503599627370496;
     0, 0, 0, 1];
P = [
     (433*cos(t1))/500, -cos(t1)/2, sin(t1),  0;
    -(433*sin(t1))/500,  sin(t1)/2, cos(t1), 0;
    -1/2, 433/500, 0, 0;
    0,  0, 0, 1];
there is no possible solution for R1 == P.
Compare R1(3,3) = 7344817515977723/9223372036854775808 to P(3,3) = 0. These are constants but do not equal each other, so it is not possible to find t1, t2, t3 that will make the matrices equal.
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