It can be shown that the length OX also depends on the angle θ between OC and OA, which has not been provided. However, if OC and OA are vectors, as it would appear from the diagram, then the dot product of the vectors is accosθ so cosθ=a.c/ac, where the vectors are shown in bold type and their magnitudes in normal type. It will then be possible to show OX either as a scalar length or as a vector. The question doesn’t specify which is required. Before continuing with a full solution I need clarification please.
The following approach uses geometry and trigonometry.
It’s helpful to draw in CA and AN.
Let θ=CÔA, then ∠ANB=∠OCB=180°-θ (interior and corresponding angles of a parallelogram). AN=OC=c, CN=NB=OA=a, CB=CN+NB=2a. cosθ=-cos(180-θ).
AB²=AN²+NB²-2AN.NBcosANB=c²+a²+2cacosθ.
OB²=OC²+CB²-2OC.CBcosOĈB=c²+4a²+4cacosθ (Cosine Rule),
We know OA=a, AB, OB (from above), so we can find ∠OBA, which we’ll call λ:
OA²=OB²+AB²-2OB.ABcosλ,
a²=4a²+c²+4accosθ+a²+c²+2accosθ-2√((4a²+c²+4accosθ)(a²+c²+2accosθ))cosλ,
cosλ=(2a²+c²+3accosθ)/√((a²+c²+2accosθ)(4a²+c²+4accosθ)).
We know NM=½CA, NB=a, MB=½AB, so we can find ∠NMB, which we’ll call ɸ.
NM=½CA=½√(a²+c²-2accosθ) (NM is a semi-diagonal in the parallelogram with 3 vertices A, N, B corresponding to congruent vertices O, C, N in parallelogram OANC),
MB=½AB=½√(a²+c²+2accosθ),
NB²=NM²+MB²-2NM.MBcosɸ,
a²=¼(a²+c²-2accosθ)+¼(a²+c²+2accosθ)-½√((a²+c²-2accosθ)(a²+c²+2accosθ))cosɸ,
a²=c²-√((a²+c²)²-4a²c²cos²θ)cosɸ,
cosɸ=(c²-a²)/√((a²+c²)²-4a²c²cos²θ).
XB/sinɸ=MB/sin(180-(λ+ɸ))=MB/sin(λ+ɸ) (Sine Rule)
XB=MBsinɸ/sin(λ+ɸ).
OX=OB-XB. Since we know OB and XB, we can calculate OX.