In parallelogram ABCD, a point is marked on side AB so that AK: KB = 2: 1, O is the intersection point of the diagonals

In parallelogram ABCD, a point is marked on side AB so that AK: KB = 2: 1, O is the intersection point of the diagonals. Express the vectors OC and OK in terms of vectors a = AB and b = AD

Vector | AC | is equal to the sum of vectors | AB | | AD |, and since the AC is the diagonal and at the point of intersection is divided in half, then | OC | = (| AB | + | AD |) / 2 = (a + b) / 2.

Since, by condition, AK / KB = 2/1, then KB = AK / 2.

AB = AK + KB = AK + AK / 2 = 3 * AK / 2.

AK = 2 * AB / 3 = 2 * a / 3.

Vector | OK | = | OA | + | AK |.

| OA | = – | OC | = – (a + b) / 2.

Then | OK | = – (a + b) / 2 + 2 * a / 3 = (-3 * a – 3 * b + 4 * a) / 6 = (a – 3 * b) / 6.

Answer: | OC = (a + b) / 2, | OK | = (a – 3 * b) / 6.