Vector representation of triangle interior If O is a triangle ABC, ab = C, AC = B, BC = a, prove a * [0A] + b * [ob] + C * [OC] = [0] ([OA] denotes vector OA) I asked why

Vector representation of triangle interior If O is a triangle ABC, ab = C, AC = B, BC = a, prove a * [0A] + b * [ob] + C * [OC] = [0] ([OA] denotes vector OA) I asked why

Senior one compulsory 4, vector problem,
A person starts from point a, walks 500 meters eastward to point B, then walks 30 degrees north by east to 300 meters to point C, and then walks 45 degrees north by east to 100 meters to point D. try to choose an appropriate scale and use a vector to represent the person's displacement

Three walks are represented by three vectors
a=（500,0）
b=(300*cos30°,300*sin30°)
c=（100*cos45°,100*sin45°）
a+b+c=...
The result takes the root sign, you calculate by yourself

First, draw a graph and observe the shape of the triangle with a, B and C as vertices
A(-1,-4),B(5.2),C(3,4)
Is this formula cos θ = a * B / | a | B | used? The result calculated by this formula is different from the graph drawn. Is there something wrong with my operation process?

Pay attention to the direction of the vector
a. Which two vectors do B refer to
If θ is ∠ C, it should be Ca and CB vectors, otherwise it is wrong

Let a + B = (4, - 2) and a-2b = (1, - 8), find the sine value of the angle between vector 2a and a-b

From a + B = (4, - 2), a-2b = (1, - 8), we can get a = (3, - 4), B = (1,2), so 2A = (6, - 8), A-B = (2, - 6), so | 2A | = 10, | A-B | = 2 times root sign 10, (2a) * (a-b) = 6 * 2 + (- 8) * (- 6) = 60. So cos = (2a) * (a-b) / (| 2A | - A-B |) = 60 / (10 * 2 root sign 10)