Let vector OA = (3,1), vector ob = (- 1,2), vector OC ⊥ vector ob, vector BC parallel vector OA Then satisfy the vector od coordinates of vector od + vector OA = vector OC, (o is the origin)

Let vector OA = (3,1), vector ob = (- 1,2), vector OC ⊥ vector ob, vector BC parallel vector OA Then satisfy the vector od coordinates of vector od + vector OA = vector OC, (o is the origin)

∵ vector OC ⊥ vector ob. ∵ C (2t, t) ∵ BC ∥ OA, (2t + 1) / (T-2) = 3 / 1. T = 7
∴C（14,7）,OD＝OC-OA＝（11,6）.

Given that mnpq is the midpoint of BC, AC, OA and ob in the spatial quadrilateral oabc, if AB = OC, PM vertical QN is proved

Don't be confused by the quadrangle of space. In fact, this is a tetrahedron
Connecting AC and OB to form tetrahedron
Make auxiliary line PN nm MQ QP
Where MQ = PN = 1 / 2 ab (triangle median)
Similarly, PQ = Mn = 1 / 2 OC
And ab = OC
So MQ = PN = PQ = Mn
So the quadrilateral mqpn is a diamond
According to the perpendicular theorem of diamond diagonal
PM vertical QN
Get proof

It is known that m, N, P and Q are the midpoint of BC, AC, OA and ob in o-abc. If AB = OC, it is proved that PM is perpendicular to QN
How urgent is it

Connect m, N, P, Q to form a 4-sided shape. According to the median theorem, it is proved that the 4 edges of the 4-sided shape are equal to rhombus, and the diagonal of rhombus is perpendicular to each other