(cos(t)+sin(t))/√3 cross (-sin(t)-cos(t))/√2 +(cos(t)-sin(t))/√3 cross (-sin(t)+cos(t))/√2

Answer 1

If those are supposed to be vectors, then since they appear to lie in the x-y plane, their cross-products will be parallel to the z-axis. Just plugging in the values, I get (ignoring the scalar factor of 1/√6)
(-cos^2(t)+sin^2(t))k + (cos^2(t)-sin^2(t))k = 0

If they are not i-j vectors, then what do you mean by "cross" involving scalar quantities? Just plain multiplication? If so, then why not just say so? In that case it'd be
(cos(t)+sin(t)) * (-sin(t)-cos(t)) = -(cos(t)+sin(t))^2 = -(1 + sin2t)
(cos(t)-sin(t)) * (-sin(t)+cos(t)) = (cos(t)-sin(t))^2 = (1 - sin2t)

Answer 2

I don;t understand how you got (-cos^2(t)+sin^2(t))k + (cos^2(t)-sin^2(t))k = 0 can explain that part? You cross what with what?

better version: (cos(t)+sin(t))j/√3 cross (-sin(t)-cos(t))i/√2 +(cos(t)-sin(t))i/√3 cross (-sin(t)+cos(t))j/√2

Answer 3

in my cross product, I just evaluated the determinants formed.
Using your second version, the determinant is (again, ignoring the 1/√6 factor)
First term:
|i j k|
|-sint-cost 0 0|
|0 cost+sint 0| = (cos^2t - sin^2t)k
Second term
|i j k|
|cost-sint 0 0|
|0 -sint+cost 0| = (cost-sint)^2 k
So that gives (cos2t + 1-sin2t)k