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

2 Answers
Gió
Jun 13, 2017

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.

Cesareo R.
Jun 13, 2017

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.

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