Triangle ABC, O is the midpoint of BC, the straight line passing through o intersects straight line AB, AC in M, N. if vector AB = m, vector * am, vector AC = n * vector an, then the value of M + n

Triangle ABC, O is the midpoint of BC, the straight line passing through o intersects straight line AB, AC in M, N. if vector AB = m, vector * am, vector AC = n * vector an, then the value of M + n

Extend Ao to a 'so that Ao = a'o, link a'c to Mn at M'
Triangle OBM and triangle OCM 'congruent, BM = cm'
Triangle Nam is similar to triangle NCM ',
NC/AC = CM'/AM
(AN-AC)/AN = (AB-AM)/AM
n-1 = 1-m
m+n = 2

In △ ABC, ab = 2, BC = 3, ∠ ABC = 60 °, ah ⊥ BC, point m is the midpoint of ah. If vector am = a vector AB + B vector AC, a + B =?

1/3+1/6=1/2=0.5
I won't write
BC=AC-AB
1/3BC=BH=1/3AC-1/3AB
AH=AB+1/3AC-1/3AB=2/3AB+1/3AC
AM=1/2AH=1/3AB+1/6AC
Then a = 1 / 3, B = 1 / 6, a + B = 1 / 3 + 1 / 6 = 1 / 2 = 0.5

In triangle ABC, vector AB = 3, vector AC = 2, and the included angle between vector AB and AC is 60 degrees, then vector ab - vector AC =?

|Vector ab - vector AC | 2 = | vector ab | 2 + | vector AC | 2 - 2 * vector AB * vector AC = | vector ab | 2 + | vector AC | 2 - 2 * | vector ab | * | vector AC | cos = 3 ^ 2 + 2 ^ 2-2 * 3 * 2 * cos60 degrees = 7
So | vector ab - vector AC | = root 7

As shown in the figure, in △ ABC, point O is the midpoint of BC, and the straight line passing through point O intersects straight lines AB and AC respectively at different two points m, N, if AB=m AM, AC=n An, find the value of M + n

Extend Ao to a 'to make Ao = a'o, extend a'c to cross Mn to M', as shown in the figure:
Then △ OBM ≌ △ OCM '≌ BM = cm',
∵△NAM∽△NCM',
∴NC
AN=CM′
Am, i.e. AC − an
AN=AM−AB
AM,

AB=m
AM,
AC=n
AN,
∴|
AB|=m
|AM|,|
AC|=n
|AN|,
Substituting into the above formula, n-1 = 1-m, then M + n = 2

In triangle ABC, O is the midpoint of BC. If AB = 1, AC = 3, < vector AB, vector AC > = 60 °, then | vector Ao | =? In triangle ABC, O is the midpoint of BC. If AB = 1, AC = 3, = 60 °, then | vector Ao | =?

|Vector Ao| ²= 1 / 4 (vector AB + vector AC) ²
=1 / 4 (vector AB) ²+ 2 vector ac * vector AC + vector AC ²)
=1 / 4 (| vector AB)| ²+ 2 | vector AC | * | vector AC | cos + | vector AC| ²
=1/4[(1+2x1x3x(1/2)+9]
=13/4
|Vector Ao | = √ 13 / 2

In △ ABC, the included angle between AB and vector AC is 60 °, M is the midpoint of AB, and the minimum value is obtained In △ ABC, the included angle between AB and vector AC is 60 °, and M is the midpoint of ab (1) If the modulus of vector AB is equal to the modulus of AC vector, find the cosine of the angle between vector AB + 2 and vector AC and vector ab (2) If the module of AB is 2 and the module of BC is 2 radical 3, determine the position of a point D on AC so that the DB vector times the DM vector reaches the minimum, and find the minimum value

M is a point in the plane where △ ABC is located. Connect am and BM, extend AC to D to make ad = 3aC, and extend am to e to make AE ‡ s △ ABM / s △ ABC = (1 / 10) / (1 / 6) = 3 / 5. Is it twice AB = vector AB