Question

We have `{(x+y+z=0),(2xcosalpha+2ysinalpha+z=0),(2xcos2alpha-2ycos2alpha-z=0):}`;How you solve this for `alpha=(61pi)/6`?

Answer 1

You should get:
`x=0`
`y=0`
`z=0`

Explanation:

This is an Homogeneous System (all equations equal to zero) so that you have, as solution, all zeros for the three unknowns meaning that the three planes have the origin in common. It could be possible for the three planes to lie one on top of the other but I do not think this is the case.

Answer 2

See below.

Explanation:

Calling

#M = ((1, 1, 1),(2 Cos(alpha), 2 Sin(alpha), 1),(2 Cos(alpha), -2 Sin(alpha), 1))#

we have

`det(M) = 4 (1 - 2 Cos(alpha)) Sin(alpha)` so

for `4 (1 - 2 Cos(alpha)) Sin(alpha)=0` or

`alpha = (2k pi uu pmpi/3 + 2kpi uu pi + 2kpi),k in ZZ`

we have non null solutions.

For `alpha = 61/6 pi` we have a unique solution which is `(0,0,0)` because

`4 (1 - 2 Cos(61/6 pi)) Sin(61/6 pi) = 2 (1 - sqrt[3])` and thus `M` has inverse.