On vectors It is known that AC vector is the sum of AB vector and Ad vector, AC vector = a, BD vector = B, AB vector and Ad vector are represented by a respectively

On vectors It is known that AC vector is the sum of AB vector and Ad vector, AC vector = a, BD vector = B, AB vector and Ad vector are represented by a respectively

BD vector = ad-ab vector
therefore
AD+AB=a
AD-AB=b
So ad = (a + b) / 2
AB=(a-b)/2

Given that point O is a point in triangle ABC, satisfying 2oa (vector) + 3ob (vector) + 5oC (vector) = 0, the area of triangle ABC is s, the area of triangle BOC is S1, and S1 = XS, then the value of X is------

Take C as the origin, CB direction as the positive direction of X axis to establish the coordinate system, put a in the first quadrant, then C (0,0), let a (m, n), B (B, 0), O (x, y), (m, N, B are all positive numbers), so OA vector = (M-X, n-y), OB vector = (b-X, - y), OC vector = (- x, - y), substitute 2oa vector + 3ob vector + 5oC vector = 0 vector, get 2 (M-X, n-y

In triangle OAB, M is the midpoint of ob, n is the midpoint of AB, and on and an intersect with P. If vector AP = m vector OA + n vector ob (M and N belong to R), then N-M = ()
A、1
B、2
C、3
D 、4
Guys on the first floor, I'm sorry we haven't learned Menelaus theorem, and this problem doesn't choose a!
I'm sorry. I'm really sorry. I read it wrong. I'm sorry

Connecting Mn, △ OPA ≌ △ NPM,
AP=2PM
Vector AP = (vector om vector OA) × 2 / 3
=OB/3-OA×2/3
∴m= - 2/3,n=1/3
n-m=1

The problem of the combination of triangle heart and vector
Let I be the inner part of the triangle ABC, ab = AC = 5, BC = 6, vector AI = m (Times) vector AB + n (Times) vector BC, the sum value
Let I be the inner part of the triangle ABC, ab = AC = 5, BC = 6, vector AI = m (Times) vector AB + n (Times) vector BC, and find the values of M and n.

Connect AI and extend BC to e'm (multiple) vector AB + n (multiple) vector BC ', that is' vector AD + vector Di'. Because Di and BC, ad and ab are parallel, so m * vector AB = vector ad n * vector BC = vector di. It is easy to know the radius of inscribed circle of triangle is 1.5, so EI = 1.5. From similarity, we know ad = 25 /